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ChemistryCh 2 Structure of Atom

Structure of Atom: Class 11 Chemistry · Chapter 2

Chemistry Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 120 min
  • 1 simulation
  • 71 practice questions
  • 39 MCQs

What you will learn in Structure of Atom

  • Describe the experiments that revealed electrons, protons and neutrons, and state their charges and masses
  • Explain Thomson's and Rutherford's models, the alpha-scattering evidence and the weaknesses of each model
  • Use atomic number, mass number and A-Z-X symbols for atoms and ions, and tell isotopes from isobars
  • Relate wavelength, frequency, wavenumber and photon energy, and use Planck's and Einstein's equations in photon and photoelectric problems
  • Explain line spectra with Bohr's model and calculate energies, radii and spectral lines for hydrogen and hydrogen-like ions
  • Apply the de Broglie relation and the uncertainty principle, and explain why they rule out Bohr orbits
  • Describe orbitals with the four quantum numbers, sketch s, p and d orbitals and count their nodes
  • Write electronic configurations and orbital diagrams using the aufbau principle, Pauli's principle and Hund's rule, including the Cr and Cu exceptions

Original companion text aligned to NCERT Chemistry Class 11, Chapter 2, pp. 29–73. Not affiliated with or endorsed by NCERT.

Why does sodium burst into flame in water while neon reacts with nothing at all? Why is copper a shiny metal and sulphur a brittle yellow solid? Push any question in chemistry far enough and the answer lies inside the atom.

The idea of a smallest, uncuttable particle is old. In India the Vaisheshika school, linked with the sage Kanad, spoke of the paramanu (परमाणु); in Greece, Leucippus and Democritus called it atomos, 'uncuttable'. These were inspired guesses that nobody could test.

John Dalton made atoms testable in 1808: each element has its own kind of atom with its own mass, and compounds form when atoms join in small whole-number ratios. That explained the laws of chemical combination (Chapter 1), but not why a plastic comb pulled through dry hair picks up scraps of paper. If atoms were featureless balls, where would electric charge come from? Experiments at the end of the 1800s gave the answer: atoms have parts.

2.1 Discovery of Sub-atomic Particles

The parts of the atom were found by passing electricity through gases. One rule runs through everything that follows: like charges repel, and opposite charges attract.

2.1.1 Discovery of Electron

The first hint came from solutions. In the 1830s Michael Faraday showed that a given quantity of charge passed through a dissolved or molten salt always deposits the same amount of substance at an electrode (you will study his laws in Class 12). Electricity, it seemed, might come in particles too.

From the 1850s Faraday and others passed electricity through sealed glass tubes with most of the gas pumped out and a metal electrode at each end. At very low pressure and high voltage a current flows, carried by something that leaves the negative electrode (the cathode) and travels towards the positive one (the anode). These streams were named cathode rays. If the anode has a hole and the far end of the tube is coated with zinc sulphide, the rays that pass through the hole make a bright spot there (Fig. 2.1).

What are these rays? Experiments with many such tubes settled it:

  • They are negative particles. An electric field pulls the beam towards the positive plate, and a magnet swings it just as it would swing negative charges moving away from the cathode. The particles were named electrons.
  • Every atom has them. Change the electrode metal or the gas in the tube and the rays behave the same, so electrons are part of every element's atoms.
  • They fly in straight lines. Away from fields, a metal object in their path casts a sharp shadow.
  • We see only their effects. The rays are invisible, but fluorescent coatings glow where they land, which is how the picture tube of an old (CRT) television worked.

2/4Charged plates bend the beam towards the positive plate: the particles are negative.

Fig. 2.1Cathode rays travel straight from cathode to screen; electric and magnetic fields bend them in opposite directions.NCERT Fig. 2.1 · p. 30

A glass tube has a cathode at the left, an anode with a central hole, a pair of horizontal plates, magnetic coils, and a glowing screen at the right. In step 1 the fields are off and a straight beam lights a spot at the centre of the screen. In step 2 only the plates are charged, top positive, and the beam curves up. In step 3 only the coils are on and the beam curves down. In step 4 both fields act, adjusted so that the beam runs straight and the spot returns to the centre.

2.1.2 Charge to Mass Ratio of Electron

In 1897 J. J. Thomson measured how strongly cathode rays are bent, using an electric field and a magnetic field at right angles to each other and to the beam. The deflection of a moving charged particle depends on three things: a bigger charge feels a bigger force, a lighter particle is easier to deflect, and a stronger field (a higher voltage on the plates or a larger current in the coils) pushes harder.

Each field alone moves the spot, in opposite directions. Thomson adjusted the two until the spot returned to its undeflected position, and from the field strengths needed, together with the deflection by one field alone, he found the ratio of the electron's charge to its mass. The modern value is

eme=1.758820×1011=  C kg−1
(2.1)
where
e
magnitude of the electron's charge (C)
me
mass of the electron (kg)

Since the electron is negative, its charge is written −e; e itself is always positive.Per kilogram, electrons carry about 1800 times more charge than hydrogen ions do: an early sign that the electron is far lighter than any atom.

2.1.3 Charge on the Electron

Robert Millikan measured the charge itself between 1909 and 1913 with his oil-drop experiment (Fig. 2.2). Tiny oil drops falling between two horizontal plates reach a steady speed that reveals their size and mass. X-rays ionise the air, and some drops pick up charge from the ions. A voltage across the plates can then slow, stop or even lift a charged drop. Measuring the field needed, drop after drop, Millikan found that every charge q was a whole-number multiple of one smallest amount, e:

q=nen=1,2,3,…

3/3With the field tuned, the electric force qE balances the weight mg and the drop hangs still.

Fig. 2.2Weighing a drop with drag, then holding it still with an electric field, reveals its charge.NCERT Fig. 2.3 · p. 32

Two horizontal metal plates face each other with one oil drop between them. In step 1 the plates are uncharged and equal arrows labelled weight and drag show the drop falling at a steady speed. In step 2 an X-ray beam enters and minus signs appear on the drop. In step 3 the top plate is positive and an upward arrow labelled electric force balances the weight, so the drop hangs still.

That smallest charge belongs to a single electron. Millikan's value was close to 1.6×10−19 C; since 2019 the elementary charge has been fixed by definition at exactly e=1.602176634×10−19 C. Combining e with Thomson's ratio gives the mass of the electron:

me=e(e/me)=9.1094×10−31 kg
(2.2)

2.1.4 Discovery of Protons and Neutrons

Atoms are neutral, so something positive must balance the electrons. In 1886 Eugen Goldstein used a tube with holes in its cathode and saw rays streaming back through the holes, opposite to cathode rays. These canal rays behaved very differently:

  • fields bend them the way positive charges bend;
  • their mass and charge-to-mass ratio change with the gas in the tube: they are positive ions of that gas, formed when electrons are knocked off its atoms;
  • some ions carry two or more units of charge, always a whole multiple of e.

The lightest positive ion comes from hydrogen. Around 1919 Rutherford showed that this particle can be knocked out of other nuclei too, and it was named the proton (प्रोटॉन). Its charge is equal and opposite to the electron's, but it is about 1836 times heavier.

The masses still did not add up: helium is about four times as heavy as hydrogen yet has only two protons. In 1932 James Chadwick bombarded beryllium with alpha particles and detected uncharged particles slightly heavier than protons: neutrons (न्यूट्रॉन). Table 2.1 compares the three particles.

Table 2.1The three fundamental particles of the atom
ParticleCharge (C)Relative chargeMass (kg)Mass (u)Mass ÷ me
electron, e−1.602×10−19−19.109×10−310.0005491
proton, p+1.602×10−19+11.6726×10−271.007281836
neutron, n001.6749×10−271.008661839
Quick check
1 of 3
Cathode rays from tubes containing hydrogen, neon or air, with iron or aluminium electrodes, all have the same charge-to-mass ratio. What does this show?
Show answer

Answer (D) Electrons are present in every atom

The same particle appears whatever the gas or electrode, so it is part of every atom. Canal rays, by contrast, change with the gas, because they are ions of that gas.

2.2 Atomic Models

By the early 1900s scientists knew that atoms contain electrons and something positive. A good model of how these are arranged had to explain why atoms are stable, why elements differ in their physical and chemical behaviour, how atoms join into molecules, and why each kind of atom absorbs and emits its own particular radiation. We follow two early attempts.

2.2.1 Thomson Model of Atom

Thomson pictured the atom as a sphere about 10−10 m in radius, filled evenly with positive charge, with electrons embedded where the arrangement is electrically most stable: like raisins in a laddoo. In English it is called the plum-pudding model. It gives a neutral atom with its mass spread evenly through the sphere, and that second feature is what the next experiment tested. (Thomson won the 1906 Nobel Prize in Physics for his work on how gases conduct electricity.)

2.2.2 Rutherford's Nuclear Model of Atom

If Thomson was right, a fast, heavy alpha particle should pass through a gold atom like a bullet through a cloud, swerving by at most a fraction of a degree. In 1909 Hans Geiger and Ernest Marsden, in Ernest Rutherford's laboratory, fired alpha particles from a radioactive source at gold foil less than a micrometre thick and counted the flashes they made on a zinc sulphide screen at different angles (Fig. 2.3). They found:

  • almost all the alpha particles went straight through, as if nothing were there;
  • a small fraction were turned aside, mostly through small angles;
  • a very few, roughly one in many thousands, came almost straight back.

2/3Most passed straight through, some swerved, and a rare few bounced back.

Fig. 2.3Most alpha particles pass straight through; the rare ones that bounce back reveal a tiny, dense, positive nucleus.NCERT Fig. 2.5 · p. 34

A beam of alpha particles from a source at the left meets a thin gold foil inside a curved detecting screen. In step 1 an inset shows Thomson's atom and every path passes almost straight through. In step 2 most paths pass straight, a few bend slightly and one turns back towards the source. In step 3 a close-up shows atoms with tiny nuclei marked plus: distant paths are straight, a near miss bends sharply and a head-on path turns back.

Rutherford reasoned that if most particles feel nothing, most of the atom is empty space; and that turning back a heavy, fast alpha particle needs a huge repulsion, possible only if the atom's positive charge, with nearly all its mass, is packed into a tiny region, the nucleus. Its radius is about 10−15 m, against 10−10 m for the atom: a factor of 100 000. Shrink a nucleus to a mustard seed on the centre spot of a cricket ground, and the atom's edge would lie near the boundary rope.

SimulationRepulsion near a nucleus

A fixed positive charge of plus 5 sits left of centre with field lines spreading outwards, and a movable test charge of plus 2 represents an alpha particle. An arrow on the test charge shows the repulsive force, and a readout gives its size. Dragging the test charge closer makes the arrow and the readout grow steeply, by four times when the distance halves.

Fig. 2.4A fixed +5 charge stands for a nucleus and a +2 test charge for an alpha particle. Drag the alpha particle and read the force.
Try this
  • Read the force with the alpha particle far away, then drag it to half the distance. By what factor does the force grow?
  • Bring it right up to the nucleus. Why can only a nearly head-on approach turn an alpha particle back?

Rutherford's nuclear model (1911): the positive charge and nearly all the mass are concentrated in a tiny central nucleus; electrons move round it at high speed in circular orbits, as planets circle the Sun; and electrostatic attraction holds the electrons and nucleus together.

2.2.3 Atomic Number and Mass Number

The charge of a nucleus comes from its protons. Their number is the atomic number, Z (परमाणु क्रमांक): every carbon nucleus has 6, every iron nucleus 26. A neutral atom has as many electrons as protons, so

in a neutral atom
Z=no. of protons=no. of electrons
(2.3)

Protons and neutrons are together called nucleons, and they carry almost all the mass. Their total is the mass number, A (द्रव्यमान संख्या):

A=Z+N
(2.4)
where
A
mass number (total nucleons)
N
number of neutrons

2.2.4 Isobars and Isotopes

A nucleus, or its atom, is written XZAX2Z2AX: the element symbol with the mass number as a left superscript and the atomic number as a left subscript. Uranium-235 is X92235X2922235U, with 92 protons and 235−92=143 neutrons. Since the symbol already fixes Z, the subscript is often dropped: X235X22235U or 'U-235'.

Definition
Isotopes

Atoms of the same element (same Z) with different numbers of neutrons, and hence different mass numbers.

ExampleX816X28216O, X817X28217O and X818X28218O

Definition
Isobars

Atoms of different elements that have the same mass number.

ExampleX1840X218240Ar, X1940X219240K and X2040X220240Ca

Hydrogen shows isotopes (समस्थानिक) most clearly. About 99.985% of its atoms are protium, X11X2121H, with no neutron; about 0.015% are deuterium, X12X2122H or D, with one; and a trace is radioactive tritium, X13X2123H or T, with two. Natural chlorine is roughly three parts X35X2235Cl to one part X37X2237Cl, and carbon includes X12X2212C, X13X2213C and the radioactive X14X2214C used in dating. Isobars (समभारिक) belong to different elements, so they behave quite differently.

Why do isotopes react alike? Chemistry is about electrons, and their number is fixed by the number of protons. Neutrons add mass but leave the electron count unchanged, so isotopes have practically identical chemical properties. Small differences show up in rates and physical properties, most of all for hydrogen, since D is twice as heavy as H: heavy water, DX2O, freezes at 3.8 °C.

Example 2.1NCERT Problem 2.1

Counting the particles in a nuclide

Find the numbers of protons, electrons and neutrons in a neutral atom of X53127X2532127I.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.2NCERT Problem 2.2

Naming a species from its particles

A species contains 12 protons, 12 neutrons and 10 electrons. Write its complete symbol.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
Which pair are isobars?
Show answer

Answer (C) X2658X226258Fe and X2858X228258Ni

Isobars share A but differ in Z: iron-58 and nickel-58. The iron and hydrogen pairs are isotopes; the oxygen pair is one nuclide with different electron counts.

2.2.5 Drawbacks of Rutherford Model

Rutherford's picture borrows the solar system. A planet held by gravity, F=Gm1m2/r2, follows a steady orbit that Newton's mechanics predicts precisely, and the electrical attraction between electron and nucleus, F=kq1q2/r2, has the same form. So electrons ought to orbit just as smoothly.

The catch: anything moving in a circle is accelerating, because its direction keeps changing. Maxwell's theory says an accelerating charge radiates electromagnetic waves (planets have no net charge, so they escape this). An orbiting electron should lose energy, spiral inwards and fall into the nucleus in well under a microsecond (Fig. 2.5); the usual estimate is about 10−8 s. Real atoms last indefinitely. Nor can electrons stand still, since the nucleus would then pull them straight in, collapsing the atom into a tiny Thomson-like one.

2/2An accelerating charge radiates, so the electron should spiral in.

Fig. 2.5Classical physics predicts that an orbiting electron radiates away its energy and spirals into the nucleus.

A nucleus marked plus sits at the centre. In step 1 an electron marked minus moves on a circular orbit, as in Rutherford's model. In step 2 a spiral winds inwards to the nucleus, with wavy arrows along it showing radiation given off as the electron loses energy.

The model also says nothing about how the electrons are arranged or what energies they have, which is exactly what chemistry needs.

Quick check
1 of 3
Why could Thomson's model not explain the alpha particles that bounced back from gold foil?
Show answer

Answer (B) Its positive charge was spread too thinly to exert a large force

Spread over the whole atom, the positive charge can only push weakly. Electrons are far too light to turn back an alpha particle.

2.3 Developments Leading to the Bohr's Model of Atom

In 1913 Niels Bohr rescued Rutherford's atom by combining two lines of evidence: electromagnetic radiation behaves both as a wave and as a stream of particles, and atoms emit and absorb light only at sharply defined wavelengths. We take them in turn.

2.3.1 Wave Nature of Electromagnetic Radiation

Newton thought light was a stream of tiny particles (corpuscles), but experiments on interference in the early 1800s convinced most physicists that it is a wave. In the 1860s and 1870s James Clerk Maxwell showed what kind: an accelerating charge creates changing electric and magnetic fields, each regenerating the other, so the disturbance travels outwards as an electromagnetic wave. Heinrich Hertz made and detected such waves in 1887–88, and their speed matched that of light. The thermal radiation given out by hot objects turned out to be electromagnetic waves too.

Three properties matter here (Fig. 2.6):

  • the electric and magnetic fields oscillate at right angles to each other and to the direction of travel;
  • no medium is needed: sound needs air, but sunlight crosses empty space;
  • wavelengths span an enormous range, the electromagnetic spectrum (Fig. 2.7), and our eyes respond only to a narrow band from about 400 nm (violet) to 750 nm (red).

3/3One wavelength λ runs crest to crest; ν crests pass each second, so c=νλ.

Fig. 2.6The electric and magnetic fields oscillate in step, at right angles to each other and to the direction of travel.NCERT Fig. 2.6 · p. 37

A horizontal axis shows the direction of travel. In step 1 a wave in a vertical plane shows the electric field. In step 2 a wave drawn flat in the horizontal plane shows the magnetic field, in step with the electric field but at right angles to it. In step 3 a bracket marks one wavelength from crest to crest, with a note that the speed equals frequency times wavelength.

A wave has a wavelength λ, the distance from one crest to the next (in m, nm, pm or the older ångström, 1 Å = 10−10 m), and a frequency ν, the number of crests passing a point each second (in hertz, 1 Hz = 1 s⁻¹). In a vacuum every electromagnetic wave travels at c=2.998×108 m s−1, usually rounded to 3.0×108 m s−1. Each second ν crests pass, each λ long, so

c=νλ
(2.5)

Spectroscopists also use the wavenumber νˉ: how many wavelengths fit into one metre (SI unit m⁻¹) or, more often in practice, into one centimetre (cm⁻¹).

νˉ=1λ=νc
Fig. 2.7Frequency rises and wavelength falls from left to right; visible light is a thin slice near 1015 Hz.NCERT Fig. 2.7 · p. 38

A long band is divided into radio, microwave, infrared, visible, ultraviolet, X-ray and gamma regions. A frequency arrow above runs from ten to the six hertz to ten to the twenty hertz, and a wavelength arrow below runs the other way, from 100 metres to 1 picometre. The thin visible slice is enlarged underneath, from red at 750 nanometres to violet at 400.

Example 2.3NCERT Problem 2.3

Wavelength of an FM broadcast

An FM radio channel transmits at 102.6 MHz. Find the wavelength and the region of the spectrum.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.4NCERT Problem 2.4

Frequencies of two LEDs

A blue LED emits at 450 nm and a red LED at 650 nm. Find the frequency of each.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.5NCERT Problem 2.5

The red line of hydrogen

Hydrogen's brightest visible line has a wavelength of 6563 Å. Find (a) its wavenumber in cm⁻¹ and m⁻¹ and (b) its frequency.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
A Wi-Fi router transmits at 2.4 GHz. What is the wavelength, in centimetres?
cm
Show answer

Answer 12.5 cm

λ=c/ν=3.0×108/2.4×109=0.125 m=12.5 cm: microwaves.

2.3.2 Particle Nature of Electromagnetic Radiation: Planck's Quantum Theory

The wave picture explains interference and diffraction (the bending of waves round obstacles). Yet by 1900 several observations refused to fit it: the light given out by hot objects (black-body radiation), the release of electrons from metals by light (the photoelectric effect), the fall in the heat capacity of solids at low temperature, and the line spectra of atoms. All pointed to one strange conclusion: energy changes hands only in fixed amounts.

An electric heater's coil glows dull red as it warms, then orange, then yellow-white, and stars hotter than the Sun look blue-white: a hot body's light shifts to shorter wavelengths as its temperature rises. To study this cleanly, physicists imagined a black body, an ideal object that absorbs every wavelength and reflects none; in equilibrium it emits as much as it absorbs, and its radiation depends only on its temperature. Soot comes close, and a hollow box with a tiny hole is almost perfect, because light entering the hole is absorbed before it can escape. At a fixed temperature the intensity rises with wavelength to a peak and then falls; a hotter body emits more at every wavelength and peaks at a shorter one (Fig. 2.8).

3/3At 6000 K, close to the Sun's surface, the peak is near 480 nm.

Fig. 2.8A hotter black body emits more at every wavelength, and its peak moves to shorter wavelengths.NCERT Fig. 2.8 · p. 40

Axes show intensity against wavelength, with the visible band shaded. Step 1 shows a low, broad curve for 4000 kelvin peaking at the red edge of the visible band. Step 2 adds a taller curve for 5000 kelvin peaking in the middle of the visible band. Step 3 adds a much taller curve for 6000 kelvin peaking in the blue-green. A dot marks each peak, and the peaks move to shorter wavelengths as the temperature rises.

Wave theory could not reproduce these curves. In 1900 Max Planck found a formula that fitted them perfectly by assuming that the vibrating atoms in the walls can absorb or emit energy only in whole packets, which he called quanta (singular quantum). The energy of one quantum is proportional to the frequency:

E=hν
(2.6)
where
h
Planck's constant, 6.626×10−34 J s

An oscillator of frequency ν can hold energy 0,hν,2hν,… but nothing in between, rather like paying only with ₹1 coins: ₹7 or ₹8, never ₹7.50. Because h is so tiny, the steps go unnoticed in daily life; for atoms they decide everything.

Photoelectric effect

In 1887 Heinrich Hertz noticed that ultraviolet light helped sparks jump between electrodes. The reason, found later, is that light falling on a clean metal can knock electrons out of it: the photoelectric effect. Caesium and potassium respond even to visible light; zinc and copper need ultraviolet. Experiments in evacuated tubes (Fig. 2.9) showed four things:

  • No delay. Electrons leave the moment the light arrives, however dim it is.
  • A threshold. Each metal has a minimum frequency ν0, its threshold frequency (देहली आवृत्ति); below it no electrons escape, however bright or long the light.
  • Brightness sets the number. Above ν0, brighter light ejects more electrons per second.
  • Frequency sets the energy. The electrons' maximum kinetic energy rises with frequency and does not depend on brightness.

Wave theory predicts the opposite at almost every point: bright red light should eventually free electrons, dim light should need time, and brighter light should give faster electrons. Instead, potassium gives off no electrons in red light of any brightness, while even faint violet light ejects them at once.

2/3Each violet photon has enough energy: electrons escape at once.

Fig. 2.9Frequency decides whether electrons escape and how fast; brightness decides only how many.NCERT Fig. 2.9 · p. 41

A potassium plate is lit from the upper left. In step 1 many red photons hit it and no electrons come out. In step 2 a few violet photons hit it and a few electrons fly off, each with an arrow for its kinetic energy. In step 3 many violet photons free many electrons, with arrows of the same length as before.

In 1905 Albert Einstein took Planck's idea literally: light travels as packets of energy hν, now called photons (फोटॉन). One photon gives all its energy to one electron at once. At least hν0 is needed to pull the electron out; this minimum is the metal's work function, W0 (कार्यफलन). The rest becomes kinetic energy:

hν=hν0+12mev2
(2.7)
where
hν0=W0
work function of the metal
v
speed of the fastest ejected electrons

Every observation follows. Below threshold no single photon can free an electron, and an electron cannot pool energy from several photons. Above it, the surplus h(ν−ν0) becomes kinetic energy, which therefore rises with frequency. Brighter light means more photons, hence more electrons, each with the same maximum energy. One photon acts at once, so there is no delay.Electrons from deeper inside the metal lose some energy on the way out, so Eq. (2.7) gives the maximum kinetic energy.

Work functions are usually given in electron volts: 1 eV, the energy an electron gains crossing 1 V, is 1.602×10−19 J. A handy shortcut: photon energy in eV ≈ 1240 ÷ wavelength in nm (Table 2.2).

Table 2.2Approximate work functions (values vary with the state of the surface)
MetalW0 (eV)Threshold wavelength (nm)
caesium2.1590
potassium2.3539
calcium2.9428
zinc4.3288
iron4.5276
gold5.1243
InteractivePhotoelectric lab

A lamp shines on a metal plate in a vacuum tube. You choose one of six metals and set the wavelength (200 to 700 nanometres) and the brightness with sliders. Readouts give the photon energy, the threshold wavelength, the maximum kinetic energy and speed of the electrons, and the relative number per second. Shine animates photons arriving and electrons leaving, or none below threshold. A graph of maximum kinetic energy against frequency shows the metal's straight line of slope h with a dot for the present setting.

Fig. 2.10Choose a metal, set the wavelength and brightness, and press Shine to see whether electrons escape and how fast.

So is light a wave or a stream of particles? It shows dual behaviour: it interferes and diffracts like a wave as it travels, but is absorbed or emitted in whole photons, like particles. Which face we see depends on the experiment, and in Section 2.5.1 you will see that electrons, too, behave like waves.

Example 2.6NCERT Problem 2.6

Energy in a mole of photons

Find the energy of one photon, and of one mole of photons, of frequency 6.0×1014 Hz (green light, 500 nm).

Try it yourself first. Then reveal the solution one step at a time.

Example 2.7NCERT Problem 2.7

Photons from a laser pointer

A green laser pointer emits 5.0 mW of light of wavelength 532 nm. How many photons does it emit per second?

Try it yourself first. Then reveal the solution one step at a time.

Example 2.8NCERT Problem 2.8

Work function from an experiment

Light of wavelength 250 nm ejects electrons from a metal with a maximum kinetic energy of 0.66 eV. Find the work function and the threshold wavelength.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.9NCERT Problem 2.9

Kinetic energy and speed of photoelectrons

A metal's threshold frequency is 5.5×1014 Hz. Light of 8.0×1014 Hz falls on it. Find the electrons' maximum kinetic energy and speed.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
Light above the threshold frequency falls on a metal. If its intensity is doubled, what happens?
Show answer

Answer (B) Twice as many electrons are ejected per second

Twice the intensity means twice the photons, so twice the electrons, each with the same maximum energy, which depends only on frequency.

2.3.3 Evidence for the Quantized Electronic Energy Levels: Atomic Spectra

A glass prism spreads white light into a band of colours. Light slows down in glass, shorter wavelengths more, so violet bends most and red least. Every colour from red to violet is present, merging into the next: a continuous spectrum, like a rainbow.

Atoms can also emit light. Heat them in a flame or pass a discharge through a gas and they are excited to higher-energy states; dropping back, they give out the energy as radiation, forming an emission spectrum. Shine continuous light through a cool sample instead, and the atoms absorb certain wavelengths, leaving dark gaps: an absorption spectrum, like a photographic negative of the emission spectrum. The study of spectra is spectroscopy (Fig. 2.11).

The crucial clue: free atoms emit only certain sharp wavelengths, seen as bright lines with darkness between, a line spectrum or atomic spectrum (रेखा स्पेक्ट्रम). An element's absorption lines fall exactly where its emission lines are. Each element's pattern is as personal as a fingerprint: the yellow of sodium street lamps and the red of neon signs are two familiar examples. Around 1860 Robert Bunsen and Gustav Kirchhoff discovered caesium and rubidium from new lines in spectra, and thallium, indium, gallium and scandium were also first noticed this way.

3/3Cool hydrogen removes the same four wavelengths from white light.

Fig. 2.11Hydrogen's bright emission lines and dark absorption lines fall at the same wavelengths.NCERT Fig. 2.10 · p. 45

Three strips run from violet at the left to red at the right. Step 1 shows a smooth band of colours from white light. Step 2 adds a dark strip with four bright hydrogen lines labelled 410, 434, 486 and 656 nanometres. Step 3 adds the continuous band crossed by four dark lines at the same positions, from white light passed through cool hydrogen.

Line spectrum of hydrogen

Hydrogen, with one electron, has the simplest spectrum. A discharge splits HX2 into excited atoms, which emit light at a set of separate frequencies. In 1885 Johann Balmer, a Swiss schoolteacher, found that hydrogen's visible lines, written as wavenumbers, fit

n = 3, 4, 5, …
νˉ=109 677×(122−1n2) cm−1
(2.8)

These form the Balmer series. Johannes Rydberg then showed that every series of hydrogen lines fits one expression:

n₂ > n₁
νˉ=109 677×(1n12−1n22) cm−1
(2.9)

where n1=1,2,3,… and n2=n1+1,n1+2,…; 109 677 cm−1 is the Rydberg constant for hydrogen. Each n1 gives one series (Table 2.3). Every element's line spectrum is unique and regular, but why should whole numbers appear at all? Bohr gave the answer.

Table 2.3Series of lines in the hydrogen spectrum
Seriesn1n2RegionLongest wavelength
Lyman12, 3, …ultraviolet122 nm
Balmer23, 4, …visible, near UV656 nm
Paschen34, 5, …infrared1876 nm
Brackett45, 6, …infrared4052 nm
Pfund56, 7, …infrared7460 nm
Quick check
1 of 3
How does the absorption spectrum of sodium vapour compare with its emission spectrum?
Show answer

Answer (A) Its dark lines lie at the same wavelengths as the bright emission lines

An atom absorbs exactly the photons it can emit, since both match the same energy gaps.

2.4 Bohr's Model for Hydrogen Atom

In 1913 Niels Bohr accepted that classical physics fails inside atoms and laid down new rules, chosen to fit hydrogen's spectrum while keeping Rutherford's nucleus. It is not modern quantum mechanics, but its numbers for hydrogen are still right. His postulates:

  1. The electron moves only in certain circular orbits of fixed radius and energy, the stationary states or allowed energy levels (स्थिर अवस्था), arranged concentrically round the nucleus.
  2. In a stationary state the electron neither gains nor loses energy, so the atom does not radiate or collapse.
  3. The electron can jump between states, absorbing exactly the energy difference ΔE to move up and emitting it as one photon on dropping down, with frequency given by Bohr's frequency rule, Eq. (2.10).
  4. Only orbits in which the angular momentum is a whole-number multiple of h/2π are allowed, Eq. (2.11).
ν=ΔEh=E2−E1h
(2.10)
where
E2,E1
energies of the higher and lower states
n = 1, 2, 3, …
mevr=n h2π
(2.11)

The fourth postulate is the new physics: orbits in between are forbidden, and Maxwell's rule that orbiting charges radiate is set aside. Its consequences (derived below) are three. The orbits are numbered by the principal quantum number n=1,2,3,… (Section 2.6.1). The radii grow as n2:

rn=n2a0
(2.12)
where
a0
the Bohr radius, 52.9 pm

so the electron normally sits 52.9 pm from the nucleus, and the second and third orbits are 4 and 9 times as far out. The energies are negative and crowd together as n grows:

En=−RH(1n2)
(2.13)
where
RH
2.18×10−18 J, the Rydberg constant in energy units

The ground state has E1=−2.18×10−18 J, and E2=−5.45×10−19 J. Drawn as lines, these make an energy-level diagram (Fig. 2.12). As n→∞ the energy rises to zero: the electron is free and the atom ionised.

The theory works for any hydrogen-like species, an ion with one electron such as HeX+, LiX2+ or BeX3+. For nuclear charge Ze:

En=−2.18×10−18×Z2n2 J
(2.14)
rn=52.9 n2Z pm
(2.15)

A larger Z binds the electron more tightly and closer in. The electron's speed rises with Z and falls with n; in hydrogen's first orbit it is about 2.2×106 m s−1.

DerivationWhere Bohr's numbers come from

The Coulomb attraction provides the centripetal force (k=1/4πε0):

mev2r=kZe2r2

Eliminating v with Bohr's condition, v=nh/2πmer:

rn=n2h24π2mekZe2=52.9 n2Z pm

Kinetic energy is kZe2/2r and potential energy −kZe2/r, so

En=−kZe22rn=−2.18×10−18×Z2n2 J

The total energy is minus the kinetic energy, and the speed is vn=2.18×106 Z/n m s−1. Class 12 Physics (Atoms) returns to this.

2.4.1 Explanation of Line Spectrum of Hydrogen

When the electron jumps from level ni to level nf, the atom's energy changes by

ΔE=Ef−Ei
(2.16)
ΔE=RH(1ni2−1nf2)=2.18×10−18×(1ni2−1nf2) J
(2.17)

Moving up (nf>ni), ΔE is positive: energy is absorbed (an absorption line). Falling (ni>nf), ΔE is negative: a photon is emitted (an emission line). The photon's frequency is set by the gap; writing n1 for the lower level and n2 for the upper:

ν=∣ΔE∣h
(2.18)
ν=3.29×1015×(1n12−1n22) Hz
(2.19)
νˉ=νc
(2.20)
νˉ=1.097×107×(1n12−1n22) m−1
(2.21)

Equation (2.21) is Rydberg's formula, since 109 677 cm−1=1.097×107 m−1, now derived rather than fitted: Bohr's triumph. Each line is one jump: jumps down to n=1 give the Lyman series, down to n=2 the Balmer series, and so on (Fig. 2.12). One atom makes one jump at a time, but a tube holds countless atoms making every possible jump, so all lines appear together; a line is bright when many atoms make that jump.

3/4Jumps down to n=2: the visible lines.

Fig. 2.12Each series is the set of jumps that end on the same lower level.NCERT Fig. 2.11 · p. 46

Horizontal lines mark levels n equals 1 to 6 and infinity, with energies from minus 2.18 times ten to the minus eighteen joules for n equals 1 up to zero, crowding together near the top (spacing compressed, not to scale). Step 1 shows the levels alone. Step 2 adds downward arrows ending on n equals 1, labelled Lyman, ultraviolet. Step 3 adds arrows ending on n equals 2, labelled Balmer, visible. Step 4 adds arrows ending on n equals 3, labelled Paschen, infrared.

InteractiveHydrogen spectrum lab

An energy-level diagram for a one-electron species shows levels 1 to 7 and infinity. You choose hydrogen, the helium ion or the doubly charged lithium ion, an upper and a lower level, and emission or absorption. An arrow joins the levels and readouts give the energy change in joules and electron volts, the frequency, wavelength and wavenumber, the series and the region. A spectrum strip marks where the line falls, and a counter gives how many different lines can be emitted between the two levels.

Fig. 2.13Pick two levels to see the photon's energy, frequency, wavelength and series; switch to HeX+ or LiX2+ to see the effect of Z2.
Example 2.10NCERT Problem 2.10

A violet line of hydrogen

Find ΔE, the frequency and the wavelength for the transition n=6→n=2 in hydrogen, and name the series.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.11NCERT Problem 2.11

A hydrogen-like lithium ion

Find the energy and radius of the n=2 orbit of LiX2+.

Try it yourself first. Then reveal the solution one step at a time.

2.4.2 Limitations of Bohr's Model

Bohr's model explains hydrogen's stability and predicts the spectra of hydrogen and one-electron ions such as HeX+ and LiX2+ remarkably well. Beyond that it fails:

  • it cannot predict the spectrum of any atom with two or more electrons, not even helium;
  • seen at high resolution, many hydrogen lines are close pairs (doublets), a fine structure the model misses;
  • it cannot account for the Zeeman and Stark effects, in which lines split when the atom sits in a magnetic or an electric field;
  • it says nothing about how atoms bond into molecules.
Quick check
1 of 4
What is the radius of the third Bohr orbit of hydrogen, in pm?
pm
Show answer

Answer 476 pm

r3=32×52.9=476 pm: radius grows as n2, not n.

2.5 Towards Quantum Mechanical Model of the Atom

Two discoveries of the 1920s showed why Bohr's model had to fail, and pointed to its replacement. Matter, like light, has a dual nature; and there is a basic limit to how precisely we can know both where a particle is and how it is moving.

2.5.1 Dual Behaviour of Matter

In 1924 Louis de Broglie turned the photon idea round: if light waves can act like particles with momentum, perhaps particles such as electrons can act like waves. A photon's momentum is p=h/λ; de Broglie proposed the same link for any particle of mass m and speed v:

λ=hmv=hp
(2.22)
where
p
momentum of the particle
λ
its de Broglie wavelength

In 1927 Clinton Davisson and Lester Germer in the USA, and George Paget Thomson in Scotland, found that electron beams are diffracted by crystals just as X-rays are. (G. P. Thomson was J. J. Thomson's son: the father showed the electron is a particle, the son that it is also a wave.) The electron microscope uses this wave behaviour; electron wavelengths are far shorter than those of light, so it reveals detail thousands of times finer, down to single atoms.

Every moving object has a de Broglie wavelength, but for everyday objects it is absurdly small: a cricket ball's is about 10−34 m (Example 2.12), far too small to detect. Only for very light particles such as electrons is it comparable to the size of atoms.

It also explains Bohr's condition. A wave running round an orbit survives only if it joins up with itself, so a whole number of wavelengths must fit the circumference, 2πr=nλ (Fig. 2.14). With λ=h/mev this gives mevr=nh/2π, exactly Eq. (2.11).

1/3Three wavelengths fit exactly round the orbit: 2πr=3λ.

Fig. 2.14Only orbits that hold a whole number of electron wavelengths are allowed, which is Bohr's condition.

A nucleus sits inside a dashed circular orbit with a wave running round it. In step 1 the wave has three wavelengths and joins itself smoothly (n equals 3). In step 2 a larger orbit holds four wavelengths that also join up (n equals 4). In step 3 three and a half wavelengths fail to meet: the end is out of step with the start, marked with a cross.

Example 2.12NCERT Problem 2.12

The wavelength of a cricket ball

Find the de Broglie wavelength of a 160 g cricket ball bowled at 40 m s⁻¹.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.13NCERT Problem 2.13

The wavelength of a slow electron

An electron has a kinetic energy of 4.0×10−19 J (about 2.5 eV). Find its de Broglie wavelength.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.14NCERT Problem 2.14

The mass equivalent of a photon

A photon has no rest mass, but writing its momentum p=h/λ as mc defines a mass equivalent m=h/λc. Find m for an X-ray photon of wavelength 1.0 nm.

Try it yourself first. Then reveal the solution one step at a time.

2.5.2 Heisenberg's Uncertainty Principle

If an electron is a wave, where exactly is it? In 1927 Werner Heisenberg showed there is no sharp answer. His uncertainty principle (अनिश्चितता सिद्धांत) says that the exact position and the exact momentum (or velocity) of a particle such as an electron cannot both be determined at the same time:

Δx Δpx≥h4π
(2.23)
where
Δx
uncertainty in position
Δpx
uncertainty in momentum along the same direction

Since px=mvx, also Δx Δvx≥h/4πm: the more sharply the position is known, the more blurred the velocity, and vice versa.

Imagine finding a ping-pong ball in a dark room by throwing tennis balls at it: each hit tells you where it was, but knocks it somewhere new. To locate an electron sharply we need light of very short wavelength, and such photons carry a large momentum, p=h/λ, so each one kicks the electron hard. We learn where it was and lose track of how it is moving (Fig. 2.15).

Fig. 2.15Measuring position precisely disturbs momentum; measuring gently leaves position blurred.

Two panels. On the left a short-wavelength photon strikes an electron drawn as a sharp dot, and a long arrow shows it knocked away: position sharp, velocity disturbed. On the right a long-wavelength photon meets an electron drawn as a fuzzy blob with only a short arrow: velocity barely changed, position blurred.

So electrons have no definite paths, or trajectories: a path needs position and velocity known together. For everyday objects the limit is irrelevant: locate a 0.02 g rice grain to within a micrometre and its speed is uncertain by only about 3×10−24 m s−1. Locate an electron to within 0.1 nm, the size of an atom, and its speed is uncertain by about 6×105 m s−1, a large fraction of its orbital speed. For electrons in atoms, exact positions must give way to probabilities.

Example 2.15NCERT Problem 2.15

An electron in a quantum dot

An electron is trapped in a tiny semiconductor crystal (a quantum dot) about 2.0 nm across, so Δx≈2.0 nm. Find the minimum uncertainty in its velocity.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.16NCERT Problem 2.16

A table-tennis ball

A 2.7 g table-tennis ball moves at 20 m s⁻¹, and its speed is measured to within 1%. Find the minimum uncertainty in its position.

Try it yourself first. Then reveal the solution one step at a time.

Reasons for the failure of the Bohr model

Now we see why Bohr's model fails. It treats the electron as a particle on a sharply defined orbit, ignoring its wave nature, and such an orbit needs exact position and velocity at every moment, which the uncertainty principle forbids. Ignoring duality and contradicting the uncertainty principle, the model could not be extended to other atoms; a theory built on duality was needed.

Quick check
1 of 3
A proton and an electron move at the same speed. Which has the longer de Broglie wavelength?
Show answer

Answer (B) The electron, because it has less momentum

λ=h/mv: at equal speed the lighter electron has less momentum, so a longer wavelength (1836 times the proton's).

2.6 Quantum Mechanical Model of Atom

Newton's classical mechanics works for stones, cricket balls and planets, whose wave nature is negligible, but fails for electrons, atoms and molecules because it ignores duality and uncertainty. Quantum mechanics includes them; for large objects it gives the same answers as classical mechanics.

It was developed in 1925–26 in two equivalent forms, by Heisenberg and by Erwin Schrödinger. In Schrödinger's wave form, for a system of constant energy (such as an atom left to itself), the central equation is written

Schrödinger equation
Hˆψ=Eψ

where ψ is the wave function and Hˆ the Hamiltonian operator, built from the total energy: kinetic energies of electrons and nuclei, electron–nucleus attractions, and electron–electron and nucleus–nucleus repulsions. Solving it gives the allowed energies E and wave functions ψ; the mathematics comes in higher classes.

For hydrogen it can be solved exactly, giving the energy levels and, for each, a wave function labelled by three quantum numbers, n, l and ml, which emerge from the solution rather than being assumed as Bohr's n was. Such a one-electron wave function is an atomic orbital (कक्षक). ψ itself has no physical meaning, but ∣ψ∣2 at a point is proportional to the probability of finding the electron there. The results reproduce every feature of hydrogen's spectrum, including details Bohr missed.

With more than one electron the equation cannot be solved exactly, because the electrons repel one another. Approximate computer methods show orbitals like hydrogen's, somewhat contracted by the larger nuclear charge, with energies that depend on both n and l (Section 2.6.3).

Put together, these results make up the quantum mechanical model of the atom:

  • each electron is described by an orbital, a wave function ψ that holds everything knowable about it; an orbital has a definite energy and room for at most two electrons, and electrons take the lowest-energy orbitals first;
  • we cannot follow an electron's path, because its position and velocity are never both sharp; what ψ gives instead is ∣ψ∣2, a never-negative probability density that maps where the electron is likely to be;
  • a bound electron can have only certain energies, and this quantisation is not imposed from outside: it follows from the electron behaving as a wave, as the allowed solutions of Schrödinger's equation.

2.6.1 Orbitals and Quantum Numbers

Definition
Atomic orbital

The wave function ψ of an electron in an atom, labelled by the quantum numbers n, l and ml; ∣ψ∣2 gives the probability density of finding the electron at each point.

ExampleThe 2p orbital with ml=0.

Orbitals differ in size (how close the electron stays to the nucleus), shape and orientation (the directions in which it is more likely to be found). Each is labelled by three quantum numbers (क्वांटम संख्याएँ); a fourth describes the electron itself.

Principal quantum number, n (1, 2, 3, …) fixes the orbital's size and largely its energy; in one-electron species, energy and size depend on n alone. Orbitals with the same n form a shell: n = 1, 2, 3, 4 are the K, L, M, N shells. A larger n puts the electron further out on average, and pulling it away from the nucleus takes energy, so energy rises with n. The nth shell has n2 orbitals.

Azimuthal quantum number, l (orbital angular momentum or subsidiary quantum number) fixes the shape. For a given n it takes the n values 0,1,…,(n−1), each defining a subshell. Values l = 0, 1, 2, 3, 4, 5 are called s, p, d, f, g, h, and a subshell is named by n and the letter: n=3, l=1 is 3p.

Magnetic quantum number, ml fixes the orientation relative to chosen axes. For a given l it takes the 2l+1 values from −l to +l, so s, p, d and f subshells have 1, 3, 5 and 7 orbitals. n limits l, and l limits ml (Table 2.4).

Table 2.4How shells divide into subshells and orbitals
Shell (n)Allowed lSubshellsOrbitals (n2)Max. electrons (2n2)
K (1)01s12
L (2)0, 12s, 2p1 + 3 = 48
M (3)0, 1, 23s, 3p, 3d1 + 3 + 5 = 918
N (4)0, 1, 2, 34s, 4p, 4d, 4f1 + 3 + 5 + 7 = 1632

Spin quantum number, ms. Closely spaced pairs and triplets of lines in many-electron spectra need more levels than three numbers give. George Uhlenbeck and Samuel Goudsmit suggested in 1925 that the electron carries its own angular momentum, spin, loosely pictured as spinning on its axis like the Earth. It can point only two ways: ms=+12 (↑) or −12 (↓). An orbital holds at most two electrons, with opposite spins, which gives the electron counts in Table 2.4.

In short: n gives the shell, size and most of the energy; l the subshell, shape and (in many-electron atoms) part of the energy; ml the orientation; ms the spin.

Compare

Orbit or orbital?

Bohr orbit

  • a definite circular path round the nucleus
  • needs the exact position and velocity at every instant, which the uncertainty principle forbids
  • has no real meaning and can never be observed
  • labelled by one number, n

Atomic orbital

  • a one-electron wave function ψ
  • describes where the electron is likely to be, not a path
  • ∣ψ∣2 is a probability density: probability per unit volume
  • labelled by three numbers, n, l and ml

Since ∣ψ∣2 varies from place to place, the probability in a region is found by adding ∣ψ∣2 ΔV over tiny volumes ΔV in which it is nearly constant.

Example 2.17NCERT Problem 2.17

Orbitals and electrons in the N shell

How many orbitals are there in the shell with n=4, and how many electrons can it hold?

Try it yourself first. Then reveal the solution one step at a time.

Example 2.18NCERT Problem 2.18

Naming subshells

Name the subshells with (a) n=3, l=1; (b) n=5, l=2; (c) n=4, l=3; (d) n=6, l=0.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
Which set of quantum numbers is not allowed?
Show answer

Answer (D) n=2, l=2, ml=0

l goes only up to n−1, so n=2 allows l=0 or 1, never 2.

2.6.2 Shapes of Atomic Orbitals

Plotted against distance r from the nucleus, ψ differs from orbital to orbital (Fig. 2.16). For 1s it is largest at the nucleus and falls smoothly; for 2s it falls, crosses zero, turns negative and creeps back towards zero.

Max Born showed that ∣ψ∣2 is the meaningful quantity, the probability density. For 1s it is greatest at the nucleus and drops sharply; for 2s it falls to zero at one distance, rises to a smaller maximum and fades. A surface of zero probability density is a node (nodal surface, नोड); an ns orbital has n−1 spherical nodes.Entrance papers also use the radial probability, 4πr2ψ2, the chance of finding the electron in a thin shell at distance r. For 1s it is zero at the nucleus and peaks at r=52.9 pm, Bohr's radius.

3/3Squared, both are positive; the 2s probability density is zero at the node.

Fig. 2.16The 2s wave function changes sign once, so its probability density falls to zero at one distance: a node.NCERT Fig. 2.12 · p. 57

Two graphs share a distance axis. In step 1 the upper graph shows psi for 1s, high at the nucleus and falling smoothly. In step 2 a 2s curve is added that crosses the axis at a point marked node and dips below zero before rising back. In step 3 the lower graph shows psi squared: the 1s curve falls steadily, while the 2s curve touches zero at the node and then forms a small second hump.

A charge-cloud picture shows dots crowded where the density is high. Shape is shown more cleanly by a boundary surface of constant ∣ψ∣2 enclosing about 90% of the probability. Not 100%, because ∣ψ∣2 never quite reaches zero, so such a surface would be infinite.

s orbitals have spherical boundary surfaces centred on the nucleus: the probability depends only on how far you are from the nucleus, not on the direction. They grow with n: 1s < 2s < 3s < 4s.

p orbitals (l=1, n≥2) have two lobes on opposite sides of the nucleus, with a nodal plane between them (Fig. 2.17). The three p orbitals of a subshell are identical in size, shape and energy but point along perpendicular axes: px, py, pz (there is no simple match between these labels and ml = −1, 0, +1). They grow in size and energy with n (4p > 3p > 2p) and have n−2 spherical nodes.

d orbitals (l=2, n≥3) are five: dxy, dyz, dxz, dx2−y2 and dz2. The first four have four lobes like a clover leaf; dz2 has two lobes along z and a ring round the middle. All five 3d orbitals of a free atom have equal energy; 4d and 5d look similar but are larger and higher in energy.

2/3Three p orbitals point along x, y and z, each with a nodal plane through the nucleus.

Fig. 2.17s orbitals are spheres; p orbitals are pairs of lobes along x, y and z; most d orbitals have four lobes.NCERT Fig. 2.14 · p. 58

Step 1 shows the 1s and larger 2s boundary surfaces as circles round the nucleus. Step 2 shows px, py and pz, each a pair of lobes along one axis, with the nodal plane of pz dashed. Step 3 shows d xy with lobes between the axes, d x squared minus y squared with lobes along them, and d z squared with two lobes along z and a ring.

Radial nodes are spheres (as in 2s and 3p); angular nodes are planes or cones through the nucleus, such as the xy plane of pz or the xz and yz planes of dxy. Altogether:

numbers of nodes
angular=lradial=n−l−1total=n−1
Quick check
1 of 3
How many radial nodes does a 4d orbital have?
node
Show answer

Answer 1 node

n−l−1=4−2−1=1, plus l=2 angular nodes: 3 in all.

2.6.3 Energies of Orbitals

In a hydrogen atom the energy of an orbital depends only on n (Eq. 2.13), so the order is

1s<2s=2p<3s=3p=3d<4s=4p=4d=4f<…

2s and 2p differ in shape but not in energy: orbitals of equal energy are degenerate (अपभ्रष्ट). An electron in 1s, held most strongly, is in the ground state; in 2s, 2p or higher it is in an excited state (Fig. 2.18, left).

In a many-electron atom energy depends on l as well: within a shell, s < p < d < f, and shells can overlap: 4s lies below 3d, 6s below 4f and 5d, and 4f below 6p.

The cause is electron–electron repulsion, absent in hydrogen. Inner electrons partly shield an outer electron from the nucleus, so it feels a reduced charge, the effective nuclear charge Zeffe (प्रभावी नाभिकीय आवेश). Still, its attraction grows with nuclear charge, so orbital energies become more negative as Z increases.

Shielding depends on shape. In one shell an s electron spends more time near the nucleus than a p electron, and p more than d. So s electrons shield better than p, and p better than d; and within a shell Zeff falls from s to p to d, so s is held most tightly and lies lowest. This splits each shell into subshells (Fig. 2.18, right).

Fig. 2.18In hydrogen energy depends only on n; in many-electron atoms subshells split, and 4s drops below 3d.NCERT Fig. 2.16 · p. 60

Two energy ladders stand side by side, energy increasing upwards. The left ladder, for hydrogen, has 1s at the bottom, then 2s and 2p on one shared line, 3s, 3p and 3d on one shared line, and 4s, 4p, 4d and 4f on another, showing that orbitals with the same n are degenerate. The right ladder, for a many-electron atom, has separate lines in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, with the 4s line drawn below the 3d line, showing that subshells split and that 4s lies below 3d.

There is no simple formula, but a useful rule of thumb is the (n + l) rule: the lower the value of n+l, the lower the energy; when two orbitals have the same n+l, the one with the lower n has the lower energy (Table 2.5).

Table 2.5Orbitals grouped by n + l (within each group, the lower n comes first)
n+lOrbitals, in order of increasing energy
11s
22s
32p, 3s
43p, 4s
53d, 4p, 5s
64d, 5p, 6s
74f, 5d, 6p, 7s

A given orbital also drops in energy as nuclear charge rises: helium's 1s lies far below hydrogen's, and lithium's lower still.

2.6.4 Filling of Orbitals in Atom

Which orbitals do an atom's electrons actually occupy? Three rules decide, together with the energy order above.

PrincipleAufbau principle

In the ground state of an atom, electrons occupy orbitals in order of increasing energy: each new electron enters the lowest-energy orbital still available.

Aufbau is German for 'building up'. Orbital energies shift from atom to atom, so no order is right for all atoms, but the (n + l) order works remarkably well for the outermost electrons:

1s, 2s, 2p, 3s, 3p,4s, 3d, 4p, 5s, 4d,5p, 6s, 4f, 5d, 6p,7s, 5f, 6d

This is Table 2.5 read row by row. Potassium's last electron goes to 4s, not 3d, since 4s has the lower n+l (4 against 5). Where two subshells are very close in energy, exceptions occur (Section 2.6.6).

PrinciplePauli exclusion principle

Within one atom, each electron has its own unique combination of the four quantum numbers n, l, ml and ms. Equivalently, an orbital holds at most two electrons, and their spins must be opposite.

Wolfgang Pauli stated it in 1925. Two electrons in one orbital share n, l and ml, so they differ in ms. Hence s, p, d and f subshells hold 2, 6, 10 and 14 electrons, and a shell at most 2n2.

PrincipleHund's rule of maximum multiplicity

Electrons entering orbitals of equal energy (one subshell) do not pair up until every orbital of the subshell holds one electron, and the single electrons have parallel spins.

Pairing therefore begins with the 4th electron in a p subshell, the 6th in d and the 8th in f, like passengers on an empty bus who each take a double seat alone before anyone sits beside a stranger.

Quick check
1 of 4
Arrange these subshells of a many-electron atom in order of increasing energy.
  1. 3d
  2. 3p
  3. 5s
  4. 4s
  5. 4p
Show answer
Answer
  1. 3p
  2. 4s
  3. 3d
  4. 4p
  5. 5s
3p and 4s both have n+l=4, and 3p has the lower n, so it comes first. Then 3d, 4p and 5s all have n+l=5, so they follow in order of increasing n.

2.6.5 Electronic Configuration of Atoms

The way an atom's electrons are distributed among its orbitals is its electronic configuration (इलेक्ट्रॉनिक विन्यास). It can be written in two ways:

  • spdf notation: each occupied subshell is written as n and its letter, with the number of electrons as a superscript; nitrogen is 1s2 2s2 2p3;
  • orbital (box) diagram: each orbital is a box and each electron an arrow, ↑ for ms=+12 and ↓ for ms=−12; because it shows the spins, it carries information on all four quantum numbers.

Hydrogen is 1s1 and helium 1s2, its two electrons of opposite spin. With 1s full, lithium's third electron goes to 2s (1s22s1) and beryllium fills it (1s22s2). From boron to neon 2p fills by Hund's rule: carbon (2p2) has two unpaired electrons, nitrogen (2p3) three, and pairing starts at oxygen (2p4). Neon, 1s22s22p6, completes the second shell.

Sodium to argon repeat this with 3s and 3p. Filled inner shells are written as the matching noble gas: sodium is [Ne] 3s1, argon [Ne] 3s23p6. Inner-shell electrons are core electrons; those in the outermost shell are valence electrons, which decide the chemistry.

Potassium is [Ar] 4s1 and calcium [Ar] 4s2; then 3d fills from scandium to zinc, before 4p. Two break the pattern: chromium is [Ar] 3d54s1, not 3d44s2, and copper [Ar] 3d104s1, not 3d94s2, because half-filled and filled subshells are especially stable (Section 2.6.6).

Gallium to krypton fill 4p; rubidium to xenon repeat the pattern with 5s, 4d and 5p. Caesium and barium fill 6s; lanthanum to mercury fill 4f and 5d; then come 6p, 7s, 5f and 6d. Almost all elements beyond uranium are artificial and short-lived; all 118 known elements are now named. Tabulated configurations come mostly from spectroscopy.

InteractiveElectron configuration builder

Rows of orbital boxes for 1s to 4p are drawn in filling order. You pick an element from hydrogen to krypton with a slider or previous and next buttons, or press Add electron to place electrons one by one. Arrows fill the boxes singly before pairing. Readouts give the element, the full and noble-gas configurations and the numbers of unpaired and outer-shell electrons, and a message names the rule that placed the last electron, with a note for the chromium and copper exceptions.

Fig. 2.19Step through the elements from hydrogen to krypton, or add electrons one at a time, and watch the aufbau principle, Pauli's principle and Hund's rule at work.

Configurations explain why atoms combine, why some elements are metals, and why helium is inert while the halogens are fiercely reactive, questions Dalton's atoms could not touch. You will use them in every chapter that follows, starting with the periodic table in Chapter 3.

2.6.6 Stability of Completely Filled and Half Filled Subshells

The ground state is the configuration of lowest total energy. When two subshells are very close in energy, as 4s and 3d are in chromium and copper, an electron moves from 4s to 3d if that leaves 3d exactly half full (d5) or full (d10). Two effects make these subshells extra stable.

Symmetry. The electrons are spread evenly over all the orbitals. Electrons of one subshell occupy different regions of space, so they shield each other little and are held more strongly.

Exchange energy. Electrons of the same spin in degenerate orbitals can exchange places, and each possible exchange lowers the energy (the exchange energy). k parallel-spin electrons allow k(k−1)/2 exchanges: 6 for d4, 10 for d5 (Fig. 2.20). Half-filled and filled subshells maximise this, and it also underlies Hund's rule.

2/3Five parallel spins: 4 + 3 + 2 + 1 = 10 exchanges.

Fig. 2.20Five parallel spins allow 10 exchanges against 6 for four, which is why chromium prefers 3d54s1.NCERT Fig. 2.18 · p. 65

Five boxes represent a d subshell. In step 1 four boxes hold up arrows joined in pairs by six curved links. In step 2 a fifth arrow raises the links to ten (four plus three plus two plus one). In step 3 two chromium configurations are compared: 3d four 4s two with six exchanges, and the ticked 3d five 4s one with ten.

So half-filled and filled subshells gain from smaller shielding, smaller repulsion and larger exchange energy. Similar exceptions include molybdenum (4d55s1) and silver (4d105s1), though not every case is this simple.

Quick check
1 of 4
Which is the ground-state configuration of chromium (Z=24)?
Show answer

Answer (B) [Ar] 3d54s1

A half-filled d5 subshell has more exchange energy; the plain aufbau order would wrongly give 3d44s2.
Recap

Structure of Atom: key points and summary

  1. Discharge-tube experiments showed that atoms contain electrons (e/me=1.76×1011 C kg⁻¹; charge 1.602×10−19 C from Millikan's oil drops), protons and neutrons (Section 2.1).
  2. Thomson's evenly spread positive sphere failed the alpha-scattering test; Rutherford concluded that the positive charge and nearly all the mass sit in a nucleus about 10−15 m across (Section 2.2.2).
  3. Z is the number of protons and A=Z+N; isotopes share Z, isobars share A, and isotopes have the same chemistry (Section 2.2.4).
  4. An orbiting electron would radiate and spiral into the nucleus, so Rutherford's model cannot explain why atoms are stable (Section 2.2.5).
  5. Electromagnetic waves obey c=νλ and νˉ=1/λ (Section 2.3.1); black-body radiation showed that energy comes in quanta, E=hν (Section 2.3.2).
  6. Einstein's equation hν=hν0+12mev2 explains the photoelectric effect: frequency sets the electrons' energy, brightness only their number (Photoelectric effect).
  7. Free atoms give line spectra; hydrogen's lines fit Rydberg's formula with 109 677 cm⁻¹ (Section 2.3.3).
  8. Bohr's quantised angular momentum, mevr=nh/2π, gives rn=52.9 n2/Z pm and En=−2.18×10−18 Z2/n2 J; every line is a jump between levels (Section 2.4).
  9. Matter has wave character, λ=h/mv, and Δx Δpx≥h/4π rules out definite orbits (Section 2.5.1, Section 2.5.2).
  10. In the quantum mechanical model an orbital is a wave function labelled by n, l and ml, ∣ψ∣2 is the probability density, and spin adds ms=±12 (Section 2.6, Section 2.6.1).
  11. s orbitals are spheres, p orbitals have two lobes and most d orbitals four; there are n−l−1 radial and l angular nodes (Section 2.6.2).
  12. In many-electron atoms energy depends on n and l (the (n + l) rule), and orbitals fill by the aufbau principle, Pauli's principle and Hund's rule (Section 2.6.3, Section 2.6.4).
  13. Half-filled and filled subshells are extra stable, giving Cr 3d54s1 and Cu 3d104s1 (Section 2.6.5, Section 2.6.6).
Practice · 63 questions

Exercises: questions with answers

  1. NumericalEasyNCERT Exercises 2.1

    Electron mass: 9.109×10−31 kg.

    1. How many electrons together have a mass of 1.0 mg?

      electrons
      Show solution

      n=1.0×10−6 kg9.109×10−31 kg=1.10×1024

    2. Find the total mass and the total charge of 2.5 mol of electrons.

      Show solution

      1.506×1024 electrons: mass 1.37×10−6 kg, charge −2.41×105 C.

  2. NumericalMediumNCERT Exercises 2.2

    Proton mass 1.6726×10−27 kg, neutron mass 1.675×10−27 kg.

    1. How many electrons are there in 9.0 g of water?

      electrons
      Show solution

      0.50 mol HX2O × 10 electrons =5.0 mol =3.01×1024.

    2. Find the number and total mass of neutrons in 3.2 mg of X16X2216O atoms.

      Show solution

      2.0×10−4 mol atoms × 8 =9.64×1020 neutrons, mass 1.61×10−6 kg.

    3. Find the number and total mass of protons in 4.4 g of COX2.

      Show solution

      0.10 mol × 22 protons =1.32×1024, mass 2.22×10−3 kg.

    4. Would the answer to (c) change if the gas were heated or compressed?

      Show solution

      No: it depends only on the mass of gas.

  3. Short answerEasyNCERT Exercises 2.3

    Find the numbers of protons and neutrons in X919X29219F, X1327X213227Al, X2963X229263Cu and X56138X2562138Ba.

    Show solution

    (p, n): F (9, 10); Al (13, 14); Cu (29, 34); Ba (56, 82).

  4. Short answerEasyNCERT Exercises 2.4

    Give the full nuclear symbol for each atom: (a) Z=19, A=39; (b) Z=82, A=208; (c) Z=5, A=11.

    Show solution

    (a) X1939X219239K (b) X82208X2822208Pb (c) X511X25211B

  5. NumericalEasyNCERT Exercises 2.5

    A red laser emits light of wavelength 650 nm. Find its frequency and its wavenumber.

    Show solution

    ν=3.0×1086.50×10−7=4.62×1014 Hz; νˉ=16.50×10−7 m=1.54×106 m⁻¹ =1.54×104 cm⁻¹

  6. NumericalEasyNCERT Exercises 2.6

    Find the energy of a photon (a) of frequency 1.5×1016 Hz; (b) of wavelength 0.20 nm.

    Show solution

    (a) E=(6.626×10−34)(1.5×1016)=9.94×10−18 J; (b) E=(6.626×10−34)(3.0×108)2.0×10−10=9.94×10−16 J

  7. NumericalEasyNCERT Exercises 2.7

    An infrared wave has a period of 5.0×10−15 s. Find its frequency, wavelength and wavenumber.

    Show solution

    ν=1/T=2.0×1014 Hz; λ=c/ν=1.5×10−6 m; νˉ=6.7×105 m⁻¹.

  8. NumericalEasyNCERT Exercises 2.8

    How many photons of wavelength 600 pm are needed to deliver 1.0 J of energy?

    photons
    Show solution

    E=hc/λ=(6.626×10−34)(3.0×108)6.0×10−10=3.31×10−16 J; N=1.0/3.31×10−16=3.02×1015

  9. NumericalMediumNCERT Exercises 2.9

    Light of wavelength 350 nm falls on potassium, whose work function is 2.30 eV (1 eV=1.602×10−19 J).

    1. Find the photon energy in eV.

      eV
      Show solution

      E=hc/λ=5.68×10−19 J =3.55 eV

    2. Find the maximum kinetic energy of the photoelectrons, in eV.

      eV
      Show solution

      3.55−2.30=1.25 eV =1.99×10−19 J

    3. Find their maximum speed.

      Show solution

      v=2Ek/me, so v=6.6×105 m s⁻¹

  10. NumericalMediumNCERT Exercises 2.10

    Radiation of wavelength 286 nm is just enough to ionise a potassium atom. Find the ionisation energy of potassium in kJ mol⁻¹.

    kJ mol⁻¹
    Show solution

    hc/λ=6.95×10−19 J; × NA = 419 kJ mol⁻¹.

  11. NumericalEasyNCERT Exercises 2.11

    A 10 W lamp puts all its power into red light of wavelength 0.62 μm. How many photons does it emit per second?

    s⁻¹
    Show solution

    E=hc/λ=3.21×10−19 J; N=10/3.21×10−19=3.1×1019 s⁻¹

  12. NumericalEasyNCERT Exercises 2.12

    Light of 5400 Å just releases electrons, with zero speed, from a metal. Find the threshold frequency and work function.

    Show solution

    ν0=c/λ0=3.0×108/5.4×10−7=5.56×1014 Hz; W0=hν0=3.68×10−19 J =2.30 eV

  13. NumericalMediumNCERT Exercises 2.13

    The electron of a hydrogen atom drops from n=5 to n=3. Find the wavelength of the photon given out and its region of the spectrum.

    Show solution

    ∣ΔE∣=2.18×10−18(1/9−1/25)=1.55×10−19 J; λ=hc/∣ΔE∣=1280 nm, infrared (Paschen).

  14. NumericalEasyNCERT Exercises 2.14

    How much energy is needed to ionise a hydrogen atom whose electron is in the n=4 level? Compare it with the energy needed from n=1.

    Show solution

    2.18×10−18/16=1.36×10−19 J, a sixteenth of the ground-state value.

  15. Quick answerEasyNCERT Exercises 2.15

    Hydrogen atoms excited to n=5 return to the ground state by all routes. How many different lines can appear?

    lines
    Show answer

    Answer 10 lines

    n(n−1)/2=5×4/2=10.
  16. NumericalEasyNCERT Exercises 2.16

    (a) The ground-state energy of hydrogen is −2.18×10−18 J. Find the energy of the third orbit. (b) Find the radius of the third Bohr orbit.

    Show solution

    (a) E3=−2.18×10−18/9=−2.42×10−19 J; (b) r3=9×52.9=476 pm

  17. Quick answerMediumNCERT Exercises 2.17

    Find the wavenumber, in cm⁻¹, of the longest-wavelength line in the Paschen series of hydrogen.

    cm⁻¹
    Show answer

    Answer 5332 cm⁻¹

    The smallest jump is 4→3: νˉ=109 677(1/9−1/16)=5332 cm⁻¹ (λ=1876 nm).
  18. NumericalMediumNCERT Exercises 2.18

    Hydrogen's ground-state energy is −13.6 eV. Find the energy, in joules, to raise the electron from n=1 to n=4, and the wavelength emitted when it falls straight back.

    Show solution

    ΔE=13.6(1−1/16)=12.75 eV =2.04×10−18 J; λ=1240/12.75=97.3 nm (ultraviolet, Lyman series)

  19. NumericalMediumNCERT Exercises 2.19

    Find the energy needed to remove the electron of hydrogen from n=3, and the longest wavelength (in cm) that can do it.

    Show solution

    E=2.18×10−18/9=2.42×10−19 J; λ=hc/E=8.21×10−7 m =8.21×10−5 cm

  20. Quick answerEasyNCERT Exercises 2.20

    Find the de Broglie wavelength, in pm, of an electron moving at 3.0×106 m s⁻¹.

    pm
    Show answer

    Answer 242 pm

    λ=6.626×10−34/[(9.11×10−31)(3.0×106)]=2.42×10−10 m.
  21. NumericalMediumNCERT Exercises 2.21

    A neutron (mass 1.675×10−27 kg) has a kinetic energy of 1.0×10−20 J. Find its de Broglie wavelength.

    Show solution

    v=2E/m=3.46×103 m s⁻¹; λ=h/mv=114 pm

  22. Quick answerEasyNCERT Exercises 2.22

    Which of these species are isoelectronic with neon (10 electrons)?

    Select all that apply
    Show answer
    Answer
    • (A) FX−
    • (B) NaX+
    • (D) AlX3+
    FX− 9+1, NaX+ 11−1, AlX3+ 13−3: all 10. ClX− and KX+ have 18, like argon.
  23. Short answerMediumNCERT Exercises 2.23

    Configurations. (a) Write the configurations of LiX+, MgX2+, NX3− and ClX−. (b) Give the atomic numbers of the elements whose configurations end in (i) 2p4, (ii) 3s2, (iii) 3p1. (c) Name the atoms (i) [He] 2s22p2, (ii) [Ne] 3s23p5, (iii) [Ar] 3d54s2.

    Show solution

    (a) LiX+ 1s2; MgX2+, NX3− [Ne]; ClX− [Ar]. (b) (i) 8, oxygen (ii) 12, magnesium (iii) 13, aluminium; (c) (i) carbon (ii) chlorine (iii) manganese.

  24. Short answerEasyNCERT Exercises 2.24

    What is the smallest value of n for which f orbitals exist, and how many f orbitals does that shell contain?

    Show solution

    l=3 needs n≥4; 7 f orbitals.

  25. Short answerEasyNCERT Exercises 2.25

    An electron is in a 4f orbital. Give its possible values of n, l and ml.

    Show solution

    n=4, l=3, ml any of −3 to +3.

  26. Short answerMediumNCERT Exercises 2.26

    An atom contains 24 electrons and 28 neutrons. Find (a) its number of protons and mass number, and (b) its electronic configuration.

    Show solution

    (a) 24 protons, A=52 (chromium). (b) [Ar] 3d54s1.

  27. Quick answerEasyNCERT Exercises 2.27

    How many electrons are there in each species?

    1. NX2X+

      electrons
      Show answer

      Answer 13 electrons

      7+7−1=13
    2. CO

      electrons
      Show answer

      Answer 14 electrons

      6+8=14
    3. OX2X2−

      electrons
      Show answer

      Answer 18 electrons

      8+8+2=18
  28. Short answerMediumNCERT Exercises 2.28

    Quantum numbers.

    1. For n=2, list the possible values of l and ml.

      Show solution

      l=0 (ml=0); l=1 (ml=−1,0,+1).

    2. Give l and the values of ml for a 4p orbital.

      Show solution

      l=1; ml=−1,0,+1.

    3. Which of these orbitals can exist?

      Select all that apply
      Show answer
      Answer
      • (B) 3p
      • (C) 4f
      • (D) 1s
      2d needs l=2≤n−1=1, and 3f needs l=3≤2: both impossible.
  29. Quick answerEasyNCERT Exercises 2.29

    Match each pair of quantum numbers with its subshell.

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • n=2, l=0 : 2s
    • n=4, l=1 : 4p
    • n=5, l=2 : 5d
    • n=6, l=3 : 6f
    Write n, then the letter for l.
  30. Quick answerMediumNCERT Exercises 2.30

    Which of these sets of quantum numbers are not possible?

    Select all that apply
    Show answer
    Answer
    • (A) n=2, l=2, ml=0, ms=+12
    • (B) n=3, l=1, ml=−2, ms=−12
    • (D) n=1, l=0, ml=0, ms=0
    • (E) n=0, l=0, ml=0, ms=+12
    A: l<n fails. B: ∣ml∣>l. D: ms must be ±12. E: n≥1.
  31. Quick answerMediumNCERT Exercises 2.31

    What is the largest number of electrons in one atom that can share these quantum numbers?

    1. n=2, ms=−12

      electrons
      Show answer

      Answer 4 electrons

      Four orbitals (2s and three 2p), one electron of each spin per orbital.
    2. n=4, l=2

      electrons
      Show answer

      Answer 10 electrons

      Five 4d orbitals, two electrons each.
  32. ProofHardNCERT Exercises 2.32

    Combine Bohr's condition (Eq. 2.11) with de Broglie's relation to show that the circumference of the nth Bohr orbit holds exactly n electron wavelengths. Then find the electron's wavelength in the third orbit of hydrogen.

    Show solution

    Eq. (2.11) gives mev=nh/2πr, so λ=h/mev=2πr/n, i.e. 2πr=nλ. For n=3, λ=2π(476)/3=997 pm.

  33. NumericalHardNCERT Exercises 2.33

    Which transition in the hydrogen spectrum has the same wavelength as the n=6→n=3 transition of LiX2+?

    Show solution

    LiX2+: 9(1/9−1/36)=3/4=1−1/4, so in hydrogen n=2→1 (first Lyman line).

  34. Quick answerMediumNCERT Exercises 2.34

    The ionisation energy of a ground-state hydrogen atom is 2.18×10−18 J. Find the energy, in joules, needed for LiX2+(g)→LiX3+(g)+eX−.

    J
    Show answer

    Answer 1.962e-17 J

    Z2×2.18×10−18=9×2.18×10−18=1.96×10−17 J per ion.
  35. NumericalEasyNCERT Exercises 2.35–2.37

    Atoms in a row.

    1. Sodium atoms are about 0.38 nm across. How many fit side by side along 1.0 cm?

      atoms
      Show solution

      1.0×10−2/3.8×10−10=2.6×107

    2. 5.0×107 copper atoms in a row stretch 1.28 cm. Find the radius of a copper atom in pm.

      pm
      Show solution

      Diameter =1.28×10−2/5.0×107=2.56×10−10 m, so the radius is 128 pm.

    3. A gold atom is 2.88 Å across. Find its radius in pm and the number of atoms along 2.0 cm.

      Show solution

      Radius =1.44 Å =144 pm; number =2.0×10−2/2.88×10−10=6.9×107.

  36. NumericalEasyNCERT Exercises 2.38–2.39

    Charge comes in whole electrons.

    1. A dust particle carries −4.8×10−17 C. How many extra electrons does it hold?

      electrons
      Show answer

      Answer 300 electrons

      4.8×10−17/1.602×10−19=300
    2. An oil drop carries −8.01×10−19 C. How many extra electrons is that?

      electrons
      Show answer

      Answer 5 electrons

      8.01×10−19/1.602×10−19=5
  37. Short answerMediumNCERT Exercises 2.40

    A student repeats Rutherford's experiment with a copper foil (Z=29) instead of gold (Z=79), with the same number of atoms in the path of the beam. How would the results differ, and why?

    Show solution

    Most would still pass straight through. Large-angle deflections would be much rarer: the smaller nuclear charge repels less, and a lighter nucleus recoils instead of turning a particle back.

  38. Short answerEasyNCERT Exercises 2.41

    Explain why X614X26214C and X14X2214C are both acceptable symbols, but X146X21426C and X6X262C are not.

    Show solution

    Mass number goes top left, atomic number bottom left. 'C' already means Z=6, so the subscript can go, but A cannot be deduced, so X6X262C names no isotope. X146X21426C claims Z=14 (silicon) and A<Z: impossible.

  39. NumericalMediumNCERT Exercises 2.42

    An atom has mass number 65 and 24.1% more neutrons than protons. Identify it and write its symbol.

    Show solution

    2.241p=65, so p=29, N=36: X2965X229265Cu.

  40. NumericalHardNCERT Exercises 2.43–2.44

    Ions with extra neutrons. (a) An anion of mass number 80 carries one negative charge and has 25.0% more neutrons than electrons. Write its symbol. (b) A cation of mass number 59 carries two positive charges and has 28.0% more neutrons than electrons. Write its symbol.

    Show solution

    (a) e=x, p=x−1, N=1.25x: 2.25x−1=80, x=36: X3580X235280BrX−. (b) p=x+2, N=1.28x: 2.28x+2=59, x=25: X2759X227259CoX2+.

  41. Quick answerEasyNCERT Exercises 2.45

    Arrange in order of increasing frequency.

    1. AM radio waves
    2. medical X-rays
    3. TV-remote infrared
    4. green signal light
    5. Wi-Fi microwaves
    Show answer
    Answer
    1. AM radio waves
    2. Wi-Fi microwaves
    3. TV-remote infrared
    4. green signal light
    5. medical X-rays
    Radio < microwave < infrared < visible < X-rays.
  42. Quick answerMediumNCERT Exercises 2.46

    An argon-ion laser emits 2.5×1018 photons per second at 488 nm. What is its power, in watts?

    W
    Show answer

    Answer 1.02 W

    E=hc/λ=4.07×10−19 J; P=2.5×1018×4.07×10−19=1.02 W.
  43. NumericalEasyNCERT Exercises 2.47

    A sodium street lamp emits strongly at 589 nm.

    1. Find the frequency.

      Show solution

      ν=c/λ=5.09×1014 Hz

    2. How far does this light travel in 1.0 minute?

      Show solution

      (3.0×108)(60)=1.8×1010 m

    3. Find the energy of one quantum.

      Show solution

      E=hν=3.37×10−19 J

    4. How many quanta carry 5.0 J?

      quanta
      Show solution

      5.0/3.37×10−19=1.48×1019

  44. Quick answerEasyNCERT Exercises 2.48

    A telescope's detector receives 1.99×10−17 J of light of wavelength 500 nm from a faint star. How many photons arrived?

    photons
    Show answer

    Answer 50 photons

    E=hc/λ=3.98×10−19 J; N=1.99×10−17/3.98×10−19=50.
  45. NumericalMediumNCERT Exercises 2.49

    A pulsed laser emits 2.0×1016 photons of wavelength 1064 nm in a pulse lasting 5.0 ns. Find the energy of the pulse and its average power during the pulse.

    Show solution

    hc/λ=1.87×10−19 J, so the pulse carries 3.7×10−3 J; P=3.7×10−3/5.0×10−9=7.5×105 W.

  46. NumericalMediumNCERT Exercises 2.50

    Potassium vapour absorbs strongly at two close wavelengths, 766.5 nm and 769.9 nm. Find the frequency of each and the energy difference between the two excited states.

    Show solution

    3.914×1014 and 3.897×1014 Hz; ΔE=h Δν=1.15×10−21 J.

  47. NumericalMediumNCERT Exercises 2.51

    Potassium has a work function of 2.3 eV. Light of wavelength 400 nm falls on it.

    1. Find the threshold wavelength.

      nm
      Show solution

      λ0=hc/W0=539 nm

    2. Find the threshold frequency.

      Show solution

      ν0=W0/h=5.56×1014 Hz

    3. Find the maximum kinetic energy of the photoelectrons.

      Show solution

      E=1240/400=3.10 eV; 3.10−2.3=0.80 eV =1.28×10−19 J

    4. Find their maximum speed.

      Show solution

      v=2Ek/me, so v=5.3×105 m s⁻¹

  48. NumericalHardNCERT Exercises 2.52

    Light of three wavelengths falls on a metal, and the maximum speeds of the photoelectrons are 400 nm: 6.23×105 m s⁻¹; 450 nm: 5.16×105 m s⁻¹; 500 nm: 4.12×105 m s⁻¹. Find (a) Planck's constant from these data and (b) the threshold wavelength.

    Show solution

    12mv2 = 1.768, 1.213, 0.773 (×10−19 J) at ν = 7.50, 6.67, 6.00 (×1014 Hz). (a) h=0.995×10−19/1.50×1014=6.63×10−34 J s; the middle point fits. (b) W0=4.97×10−19−1.77×10−19=3.20×10−19 J; λ0=hc/W0=621 nm.

  49. Quick answerMediumNCERT Exercises 2.53

    Photoelectrons from a metal lit by 230 nm light are just stopped by a retarding potential of 1.20 V. Find the work function, in eV.

    eV
    Show answer

    Answer 4.19 eV

    E=hc/λ=5.39 eV; W0=5.39−1.20=4.19 eV.
  50. NumericalMediumNCERT Exercises 2.54

    An X-ray photon of wavelength 100 pm knocks an inner electron out of an atom, and the electron leaves at 2.0×107 m s⁻¹. With what energy was the electron bound?

    Show solution

    hc/λ=1.99×10−15 J; 12mv2=1.82×10−16 J; binding energy 1.81×10−15 J.

  51. NumericalMediumNCERT Exercises 2.55

    A Balmer line of hydrogen, ν=3.29×1015 (1/22−1/n2) Hz, has a wavelength of 486 nm. Find the upper level n and the region of the spectrum.

    Show solution

    ν=6.17×1014 Hz, so 1/n2=0.25−0.1875=0.0625: n=4, visible (blue-green).

  52. NumericalMediumNCERT Exercises 2.56

    The electron of a hydrogen atom drops from an orbit of radius 846.4 pm to one of radius 52.9 pm. Find the wavelength emitted, the series and the region.

    Show solution

    846.4/52.9=16, so n=4→1: νˉ=1.097×107(15/16) m⁻¹, λ=97.2 nm, Lyman, ultraviolet.

  53. Quick answerEasyNCERT Exercises 2.57

    Electrons in an electron microscope move at 5.0×106 m s⁻¹. Find their de Broglie wavelength, in pm.

    pm
    Show answer

    Answer 145 pm

    λ=6.626×10−34/[(9.11×10−31)(5.0×106)]=1.45×10−10 m.
  54. Quick answerEasyNCERT Exercises 2.58

    Neutron diffraction uses neutrons of wavelength 200 pm. Find their speed, in m s⁻¹ (neutron mass 1.675×10−27 kg).

    m s⁻¹
    Show answer

    Answer 1978 m s⁻¹

    v=h/mλ=6.626×10−34/[(1.675×10−27)(2.0×10−10)]=1.98×103 m s⁻¹.
  55. NumericalMediumNCERT Exercises 2.59

    An electron in the second Bohr orbit of hydrogen moves at about 1.09×106 m s⁻¹. Find its de Broglie wavelength and compare it with the circumference of the orbit.

    Show solution

    λ=667 pm; circumference 2π(211.6)=1330 pm ≈2λ, as 2πr=nλ requires.

  56. NumericalEasyNCERT Exercises 2.60

    An electron accelerated through 100 V moves at 5.93×106 m s⁻¹. Find the de Broglie wavelength of (a) the electron and (b) a 0.15 kg cricket ball moving at the same speed.

    Show solution

    (a) 1.23×10−10 m (b) 7.4×10−40 m, far too small to detect.

  57. NumericalMediumNCERT Exercises 2.61

    An electron's position is measured to within 0.005 nm. (a) Find the minimum uncertainty in its momentum. (b) Its momentum is quoted as 2.0×10−24 kg m s⁻¹. Is that value meaningful?

    Show solution

    (a) Δp≥h/4πΔx=1.05×10−23 kg m s⁻¹. (b) No: the uncertainty is five times the value.

  58. Quick answerMediumNCERT Exercises 2.62

    Arrange electrons with these quantum numbers in order of increasing energy in a many-electron atom. (Electrons with the same n and l but different ml or ms have the same energy.)

    1. n=3, l=2
    2. n=5, l=0
    3. n=4, l=1
    4. n=3, l=1
    5. n=4, l=0
    Show answer
    Answer
    1. n=3, l=1
    2. n=4, l=0
    3. n=3, l=2
    4. n=4, l=1
    5. n=5, l=0
    3p (4) < 4s (4, higher n) < 3d (5) < 4p (5) < 5s (5): lower n+l first, ties to the lower n.
  59. Quick answerMediumNCERT Exercises 2.63

    An iodine atom has electrons in 3d, 4d and 5p orbitals. Which of these electrons feels the lowest effective nuclear charge?

    Show answer

    Answer (C) 5p

    The 5p electrons are in the outermost shell, shielded by all the inner electrons, so they feel the smallest Zeff.
  60. Quick answerMediumNCERT Exercises 2.64

    In each pair, which orbital experiences the larger effective nuclear charge?

    1. 3p or 2p

      Show answer

      Answer (B) 2p

      2p is closer to the nucleus, with fewer electrons inside it.
    2. 3s or 3p

      Show answer

      Answer (A) 3s

      In the same shell, s penetrates closer than p.
    3. 5f or 5d

      Show answer

      Answer (B) 5d

      In the same shell, d penetrates more than f.
  61. Short answerMediumNCERT Exercises 2.65

    The unpaired electrons of phosphorus and sulphur are both in 3p orbitals. Which feels the larger effective nuclear charge, and why?

    Show solution

    Sulphur's: one more proton, while the extra electron is in the same subshell and shields poorly.

  62. Quick answerMediumNCERT Exercises 2.66

    Match each atom with its number of unpaired electrons.

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • N : 3
    • S : 2
    • Mn : 5
    • Cu : 1
    • Ar : 0
    N 2p3: 3. S 3p4: 2. Mn 3d54s2: 5. Cu 3d104s1: 1. Ar: all paired.
  63. Quick answerEasyNCERT Exercises 2.67

    Subshells and spins.

    1. How many subshells does the shell with n=5 have?

      subshells
      Show answer

      Answer 5 subshells

      5s, 5p, 5d, 5f and 5g.
    2. How many electrons in the n=3 shell can have ms=+12?

      electrons
      Show answer

      Answer 9 electrons

      Nine orbitals, one spin-up electron each.
Practice · 8 questions

Practice questions with answers (NCERT “Board and entrance practice”)

  1. Quick answerMedium

    Assertion (A): In a hydrogen atom, a 2s electron has the same energy as a 2p electron. Reason (R): In a hydrogen atom, orbital energy depends only on n.

    Show answer

    Answer (A) Both A and R are true, and R explains A

    With one electron there is no shielding, so energy depends on n alone, which is exactly why 2s and 2p are degenerate.
  2. Quick answerMedium

    Assertion (A): Above the threshold frequency, the photoelectric current increases with the intensity of light. Reason (R): The kinetic energy of the photoelectrons increases with intensity.

    Show answer

    Answer (C) A is true but R is false

    More intense light has more photons, so more electrons per second (A is true), but their maximum kinetic energy depends only on frequency (R is false).
  3. Quick answerMedium

    Assertion (A): One orbital can hold two electrons with the same spin. Reason (R): Two electrons in one orbital have the same n, l and ml.

    Show answer

    Answer (D) A is false but R is true

    R is true, and that is why the two electrons must differ in ms: by Pauli's principle they need opposite spins, so A is false.
  4. Quick answerHard

    A proton and an alpha particle are accelerated from rest through the same potential difference. What is the ratio λproton/λα of their de Broglie wavelengths?

    Show answer

    Answer (A) 22

    λ=h/2mqV; the alpha particle has 4× the mass and 2× the charge, so λp/λα=8=22.
  5. Quick answerEasy

    A metal has a work function of 3.0 eV. Find the maximum kinetic energy, in eV, of electrons ejected by 200 nm light. Use E≈1240/λ.

    eV
    Show answer

    Answer 3.2 eV

    1240/200=6.2 eV; 6.2−3.0=3.2 eV.
  6. Quick answerEasy

    The ionisation energy of a hydrogen atom is 13.6 eV. What is the ionisation energy of a ground-state HeX+ ion?

    Show answer

    Answer (D) 54.4 eV

    Z2=4, so 4×13.6=54.4 eV. Doubling instead of squaring gives 27.2 eV.
  7. Quick answerMedium

    Find the shortest wavelength in the Balmer series of hydrogen, in nm (R=109 677 cm⁻¹).

    nm
    Show answer

    Answer 364.7 nm

    νˉ=109 677/4=27 419 cm⁻¹, so λ=3.647×10−5 cm =364.7 nm.
  8. Quick answerHard

    The uncertainty in the position of a moving electron equals its de Broglie wavelength. What is the minimum uncertainty in its velocity, compared with its velocity v?

    Show answer

    Answer (B) Δv≥v/4π

    (h/mv) m Δv≥h/4π gives Δv≥v/4π.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT Exercises 2.1Practice Q1
NCERT Exercises 2.2Practice Q2
NCERT Exercises 2.3Practice Q3
NCERT Exercises 2.4Practice Q4
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NCERT Exercises 2.22Practice Q22
NCERT Exercises 2.23Practice Q23
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NCERT Exercises 2.27Practice Q27
NCERT Exercises 2.28Practice Q28
NCERT Exercises 2.29Practice Q29
NCERT Exercises 2.30Practice Q30
NCERT Exercises 2.31Practice Q31
NCERT Exercises 2.32Practice Q32
NCERT Exercises 2.33Practice Q33
NCERT Exercises 2.34Practice Q34
NCERT Exercises 2.35–2.37Practice Q35
NCERT Exercises 2.38–2.39Practice Q36
NCERT Exercises 2.40Practice Q37
NCERT Exercises 2.41Practice Q38
NCERT Exercises 2.42Practice Q39
NCERT Exercises 2.43–2.44Practice Q40
NCERT Exercises 2.45Practice Q41
NCERT Exercises 2.46Practice Q42
NCERT Exercises 2.47Practice Q43
NCERT Exercises 2.48Practice Q44
NCERT Exercises 2.49Practice Q45
NCERT Exercises 2.50Practice Q46
NCERT Exercises 2.51Practice Q47
NCERT Exercises 2.52Practice Q48
NCERT Exercises 2.53Practice Q49
NCERT Exercises 2.54Practice Q50
NCERT Exercises 2.55Practice Q51
NCERT Exercises 2.56Practice Q52
NCERT Exercises 2.57Practice Q53
NCERT Exercises 2.58Practice Q54
NCERT Exercises 2.59Practice Q55
NCERT Exercises 2.60Practice Q56
NCERT Exercises 2.61Practice Q57
NCERT Exercises 2.62Practice Q58
NCERT Exercises 2.63Practice Q59
NCERT Exercises 2.64Practice Q60
NCERT Exercises 2.65Practice Q61
NCERT Exercises 2.66Practice Q62
NCERT Exercises 2.67Practice Q63

Quick check: MCQs on Structure of Atom

  1. 2.1.4 Discovery of Protons and Neutrons3 questions
  2. 2.2.4 Isobars and Isotopes3 questions
  3. 2.2.5 Drawbacks of Rutherford Model3 questions
  4. 2.3.1 Wave Nature of Electromagnetic Radiation3 questions
  5. Photoelectric effect3 questions
  6. Line spectrum of hydrogen3 questions
  7. 2.4.2 Limitations of Bohr's Model4 questions
  8. Reasons for the failure of the Bohr model3 questions
  9. 2.6.1 Orbitals and Quantum Numbers3 questions
  10. 2.6.2 Shapes of Atomic Orbitals3 questions
  11. 2.6.4 Filling of Orbitals in Atom4 questions
  12. 2.6.6 Stability of Completely Filled and Half Filled Subshells4 questions
Glossary · 15

Key terms (English–Hindi)

atomic numberपरमाणु क्रमांकparamanu kramank
The number of protons in the nucleus, Z.First seen in the chapter →
atomic orbitalकक्षकkakshak
A one-electron wave function ψ; ∣ψ∣2 gives the probability density.First seen in the chapter →
de Broglie wavelengthदे ब्रॉग्ली तरंगदैर्ध्यde Broglie tarangdairghya
The wavelength λ=h/mv of a moving particle.First seen in the chapter →
degenerate orbitalsअपभ्रष्ट कक्षकapabhrasht kakshak
Orbitals of equal energy, such as the three 2p orbitals.First seen in the chapter →
effective nuclear chargeप्रभावी नाभिकीय आवेशprabhavi nabhikiya avesh
The net nuclear charge an electron feels after shielding.First seen in the chapter →
electronic configurationइलेक्ट्रॉनिक विन्यासelectronic vinyas
How an atom's electrons are distributed among its orbitals.First seen in the chapter →
isobarसमभारिकsambharik
Atoms of different elements that have the same mass number.First seen in the chapter →
isotopeसमस्थानिकsamasthanik
Atoms of one element with different numbers of neutrons.First seen in the chapter →
line spectrumरेखा स्पेक्ट्रमrekha spectrum
A spectrum of sharp lines at wavelengths typical of an element.First seen in the chapter →
nodeनोडnod
A surface of zero probability density: radial (sphere) or angular (plane or cone).First seen in the chapter →
photonफोटॉनphoton
A packet (quantum) of electromagnetic energy, E=hν.First seen in the chapter →
quantum numbersक्वांटम संख्याएँquantum sankhyayen
n, l, ml and ms, which label an orbital and an electron's spin.First seen in the chapter →
stationary stateस्थिर अवस्थाsthir avastha
In Bohr's model, an allowed orbit of fixed energy.First seen in the chapter →
uncertainty principleअनिश्चितता सिद्धांतanishchitata siddhant
Position and momentum cannot both be known exactly: Δx Δpx≥h/4π.First seen in the chapter →
work functionकार्यफलनkaryaphalan
The minimum energy, hν0, that frees an electron from a metal.First seen in the chapter →