Atoms: Class 12 Physics · Chapter 12

Physics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

85 min
1 simulation
14 practice questions
26 MCQs

What you will learn in Atoms

Explain how the Geiger–Marsden α-scattering results ruled out Thomson's model and led to Rutherford's small, dense, positive nucleus
Relate the impact parameter to the scattering angle and use energy conservation to find the distance of closest approach
Show that an electron in a circular orbit has K = −E and U = 2E, and explain why a negative total energy means it is bound
Distinguish continuous, emission-line and absorption spectra and explain why a line spectrum identifies an element
Explain why classical physics predicts the collapse of Rutherford's atom, and state Bohr's three postulates
Derive and use rₙ = n²a₀ and Eₙ = −13.6/n² eV for hydrogen, and read its energy-level diagram (ground state, excitation, ionisation)
Calculate the frequency and wavelength of the photon in any transition and identify the lines of the hydrogen spectrum
Explain Bohr's quantisation condition as a standing de Broglie wave and state the limitations of Bohr's model

Original companion text aligned to NCERT Physics Class 12, Chapter 12, pp. 290–305. Not affiliated with or endorsed by NCERT.

12.1 Introduction

Drive along a road lit by low-pressure sodium lamps at night and every car looks the same dull orange-brown. The lamp's light is not a blend of colours: nearly all of it comes out at one sharp wavelength that only sodium atoms give, whether in a lamp in Pune or in a star far away. Wavelengths like this are messages from inside the atom. This chapter follows how physicists read them: first by firing α-particles at thin metal foil to find where an atom keeps its charge and its mass, then through Niels Bohr's bold rules for hydrogen, which predict its light line by line.

The puzzle began with J. J. Thomson's discovery of the electron in 1897. Every element he tested gave out the same tiny negative particle, about $1 / 1836$ of the mass of a hydrogen atom. Since a whole atom carries no net charge, it must also hold an equal amount of positive charge. And since electrons are so light, nearly all of the atom's mass has to belong to that positive part. What shape does the positive part take, and where inside the atom do the electrons sit?

Thomson's own guess, put forward in 1898, was the simplest possible arrangement. Spread the positive charge evenly through a ball the size of the whole atom, and set the electrons into it at points where the forces on them balance, like raisins pressed into a round laddoo. English writers called it the plum pudding model. In this picture the mass is spread out as well, so the atom has no hard core anywhere (Fig. 12.1).

3/3Through Thomson's atom every α-particle goes nearly straight. Through the nuclear atom the rare one that passes close to the nucleus is thrown aside, or even back.

Fig. 12.1Two pictures of the atom. Only a tiny, dense, positive centre can turn an α-particle round. (Not to scale: a real nucleus is thousands of times smaller than drawn.)NCERT p. 290

Two atoms of the same size side by side. Step 1: the left atom, labelled Thomson model, is a green ball filled with plus signs, with eight blue electrons dotted through it; the right atom is faint. Step 2: the right atom, labelled Nuclear model, appears in full: a dashed outer boundary, faint circular orbits with eight electrons on them, and a single tiny dot of positive charge at the centre. Step 3: orange alpha-particle tracks cross both atoms from the left. Through the Thomson atom all five tracks are almost straight. Through the nuclear atom most tracks are straight, one aimed near the centre is bent sharply downwards, and one aimed almost straight at the nucleus turns round and heads back to the left.

Light offered a second clue. Pass the glow of a hydrogen discharge tube through a prism and you see just four visible colours, at wavelengths that form a neat, regular pattern. In 1885 a Swiss schoolteacher, Johann Balmer, found a formula that reproduced them, with no idea why it worked. Every element has its own pattern of lines (Section 12.3). Contrast the filament of a bulb, whose light contains every wavelength. The difference is spacing: in a thin gas each atom radiates undisturbed, so you see the light of single atoms, while in a solid the crowded atoms keep jostling and sharing energy, which smears the light into a continuous band. Whatever an atom looks like inside, it must explain those single-atom patterns.

The decisive evidence came from a different kind of experiment. Instead of studying the light atoms give out, Ernest Rutherford's group in Manchester fired fast α-particles from a radioactive source through thin metal foils and recorded where they went (Section 12.2). The pattern of deflections, measured by Hans Geiger and Ernest Marsden between 1909 and 1913, was impossible for Thomson's atom. It required almost all of an atom's mass and all of its positive charge to sit in a tiny central nucleus (नाभिक), with the electrons far outside, circling like planets round the Sun. This is the nuclear, or planetary, model.

The planetary model was a giant step, but it left two awkward questions. Why should hydrogen, the simplest atom there is, give out a whole family of sharp wavelengths and nothing in between? And why does an orbiting electron not lose energy and fall into the nucleus? Classical physics answers neither (Section 12.4). Bohr's model, in the second half of the chapter, answers both for hydrogen.

Warm-up

1 of 2

An electron has about

1 / 1836

of the mass of a hydrogen atom. What follows at once?

Atoms must contain about 1836 electrons eachElectrons carry most of an atom's chargeThe positive charge must fill the whole atomNearly all of an atom's mass belongs to its positive part

Show answer

Answer (D) Nearly all of an atom's mass belongs to its positive part

Compare the electron's mass with the whole atom's mass.

12.2 Alpha-particle Scattering and Rutherford's Nuclear Model of Atom

An α-particle (ऐल्फ़ा कण) is the nucleus of a helium atom: two protons and two neutrons, with charge $+ 2 e$ and about four times the mass of a hydrogen atom. Radioactive sources fling them out at about a twentieth of the speed of light, with kinetic energies of a few MeV. To an α-particle, a sheet of gold leaf is a forest of atoms, and the way the particles come out of the forest tells you what it is made of.

The goal was simple to state: count how many α-particles leave a thin foil at each angle $θ$ from their original direction (Fig. 12.2). The counter was a small screen coated with zinc sulphide, which gives a faint flash of light, a scintillation, wherever an α-particle strikes it; an observer watched it through a microscope and could swing it to any angle round the foil. The particles came from a radioactive source buried in a lead block, whose narrow channel let out only a thin pencil of them. The target was gold leaf less than a millionth of a metre thick, only several hundred atoms deep.Observers sat in a darkened room for about half an hour first so that their eyes adapted, then counted for only a minute or so at a time, because accuracy fell quickly with fatigue. Everything was kept in a vacuum so that air molecules did not scatter the beam.

3/3Even at $150^{\circ}$ , on the source's side of the foil, a flash appears now and then: roughly one α-particle in eight thousand is turned through more than $90^{\circ}$ .

Fig. 12.2Counting α-particles at different angles. The detector swings round the foil; the count falls steeply as the angle grows, but it never quite reaches zero.NCERT Fig. 12.2 · p. 292

Seen from above: on the left a radioactive source sits inside a lead block with a narrow slot, and an orange beam runs right to a thin gold foil at the centre of a dashed circular track marked every 30 degrees. Step 1: a fan of nearly straight orange rays continues past the foil to a detector, a zinc sulphide screen with a microscope behind it, placed straight ahead at 0 degrees; the label reads most here. Step 2: the first detector fades and the detector sits at 60 degrees, catching a single deflected ray, labelled a few. Step 3: the detector sits at 150 degrees, on the same side as the source, catching one ray that came back, labelled very rare. A note says everything is inside a vacuum.

Suppose Thomson's picture were right. Then every gold atom would be a soft, spread-out ball of charge whose electric field is never very strong anywhere. The α-particle is also about 7300 times heavier than an electron, so the electrons cannot push it aside either. Each atom could nudge an α-particle by a tiny fraction of a degree at most, and even after crossing hundreds of atoms the total deflection should stay within a degree or two.

The counts said otherwise. Most α-particles did go almost straight through, but about one in seven hundred was turned through more than a degree, and roughly one in eight thousand came back out on the side of the foil it had entered, deflected by more than $90^{\circ}$ (Fig. 12.3). Rutherford later said it was as startling as a heavy artillery shell bouncing back off a sheet of tissue paper.

3/3Geiger and Marsden's counts followed the nuclear curve over more than five powers of ten, including the rare large angles.

Fig. 12.3Counts at each angle (relative to the count at

180^{\circ}

). A nucleus scatters a few α-particles to every angle; Thomson's atom would scatter none beyond a few degrees.NCERT Fig. 12.3 · p. 293

A graph of counts, on a logarithmic scale from 1 to 100 000, against scattering angle from 0 to 180 degrees. Step 1: a solid blue curve labelled nuclear atom falls steeply from the top left, through about 1000 at 20 degrees and 4 at 90 degrees, then levels off to 1 at 180 degrees. Step 2: a dashed vermillion curve labelled Thomson atom plunges from the top almost vertically and disappears off the bottom of the graph before 5 degrees. Step 3: the region beyond 90 degrees is shaded and labelled rare, not zero.

Estimate how hard each kind of atom can push. Spread a gold atom's $+ 79 e$ through a ball $10^{- 10} m$ in radius and the strongest field anywhere, at its surface, is about $10^{13} N / C$ . Crossing the whole ball would change an α-particle's energy by only about a kiloelectronvolt, against the several MeV it carries, so it would barely notice. Squeeze the same charge into a ball of radius $7 f m$ and, because the field goes as $1 / r^{2}$ , the field at its surface is about $(10^{- 10} / 7 \times 10^{- 15})^{2} \approx 2 \times 10^{8}$ times stronger. Only such a concentrated, heavy centre can turn an α-particle round. Rutherford calculated how many α-particles a point-like nucleus should send to each angle. The predicted count fell smoothly but steeply with angleFor a point nucleus the count per unit area of screen is proportional to $1 / \sin^{4} (θ / 2)$ . This formula is beyond the syllabus., and Geiger and Marsden's 1913 counts followed it from $5^{\circ}$ out to $150^{\circ}$ .

Seen by an α-particle, then, a gold foil is a thin fog of tiny, hard specks. Later measurements put nuclear radii between about $1 f m$ and $7 f m$ ( $1 f m = 10^{- 15} m$ , a femtometre or fermi), against about $10^{- 10} m$ for a whole atom, so the specks are ten to a hundred thousand times smaller than the spaces they occupy (Example 12.1). A particle crossing the fog usually misses every speck and comes out almost undisturbed. The rare one that happens to pass very close to a speck is flung aside, or back.

Definition

Nucleus

The tiny, dense, positively charged centre of an atom, holding all its positive charge and over 99.9 % of its mass. Its radius is a few femtometres, about $10^{- 4}$ to $10^{- 5}$ of the atom's.

ExampleA gold nucleus has charge $+ 79 e$ and radius about $7 f m$ .

Predicting a path calls for three idealisations, each backed by the experiment:

The nucleus stays put. A gold nucleus (mass $197 u$ ) outweighs an α-particle ( $4 u$ ) roughly fifty-fold, so its recoil can be neglected.
The electrons do not matter. They are thousands of times lighter than the α-particle.
One close encounter at most. The foil is so thin that an α-particle rarely passes near more than one nucleus.

With these, the only force on the α-particle is the Coulomb repulsion of a nucleus of charge $Z e$ ( $Z = 79$ for gold):

Force on an α-particle near a nucleus

F = \frac{1}{4 π ε_{0}} \frac{(2 e) (Z e)}{r^{2}}

(12.1)

where

$r$: the distance between the centres of the α-particle and the nucleus
$Z e$: the charge of the nucleus ( $Z$ is the atomic number)
$2 e$: the charge of the α-particle

The force acts along the line joining the two, and both its size and its direction change continuously as the α-particle approaches and then moves away. Newton's second law with this force gives the path: one branch of a hyperbola, bending away from the nucleus.

Try this

Find a target you cannot see

You’ll need

a stiff board or tray about 50 cm wide
eight books of equal thickness
a round tin about 8 cm across
20 marbles
a ruler

Predict first

The tin blocks about a fifth of the board's width. Out of 20 marbles rolled in evenly spaced lanes, about how many do you expect to be knocked aside?

None of themAbout 4About 10All 20

Rest a stiff board, about 50 cm wide, on four equal stacks of books so that a marble can roll underneath.
While you look away, a partner hides a round tin, about 8 cm across, somewhere under the board.
Mark ten launch lanes, 5 cm apart, along one edge. Roll two marbles straight across along each lane, at the same gentle speed.
For each marble, note whether it came out straight or was knocked aside, and roughly in which direction.
Count the deflected marbles. The fraction deflected, times the board's width, estimates the tin's width plus one marble's width.

Table 12.1Was each marble deflected? (yes or no)

	Marble 1	Marble 2
Lane 1 (0 cm)
Lane 2 (5 cm)
Lane 3 (10 cm)
Lane 4 (15 cm)
Lane 5 (20 cm)
Lane 6 (25 cm)
Lane 7 (30 cm)
Lane 8 (35 cm)
Lane 9 (40 cm)
Lane 10 (45 cm)

Your entries are saved automatically.

Usually one or two lanes cross the tin, so 2 or 4 marbles are knocked aside and the rest roll straight through. Each lane stands for a 5 cm strip of the board, so the tin's width is roughly (lanes hit × 5 cm) minus a marble's width; shift the lanes by 2.5 cm and repeat to sharpen the estimate. You have sized the tin without ever seeing it. Rutherford did the same with α-particles: the rarity of large deflections told him how small the nucleus is. One difference matters: a marble bounces only on contact, while an α-particle is turned by a force acting across empty space, so its path bends smoothly.

12.2.1 Alpha-particle trajectory

Begin with the simplest case: an α-particle aimed dead centre at a nucleus. It slows as it climbs the electric 'hill', stops for an instant and is driven straight back along its own line, so $θ = 180^{\circ}$ . Now shift the aim sideways. The perpendicular distance between the particle's original line and the centre of the nucleus is called the impact parameter $b$ (संघट्ट प्राचल), and the angle between the incoming and outgoing directions is the scattering angle $θ$ .

Definition

Impact parameter

The perpendicular distance $b$ between a particle's initial line of motion and the centre of the target nucleus.

ExampleA head-on collision has $b = 0$ .

With $b$ just above zero, the particle swings round the nucleus and leaves at a large angle. With a larger $b$ it keeps its distance, feels a weaker push and is bent less; far enough out, it hardly notices the nucleus at all (Fig. 12.4).

3/4 $b = d / 4$ : the particle swings round the nucleus, $θ \approx 127^{\circ}$ .

Fig. 12.4Paths computed from Coulomb's law for four impact parameters. Distances are in units of

d

, the closest an α-particle of this energy can get in a head-on collision (Example 12.2).NCERT Fig. 12.4 · p. 294

A nucleus sits on a dashed horizontal axis, and alpha-particles arrive from the left above the axis. Step 1: a particle with impact parameter b equal to 2d bends gently away and leaves at about 28 degrees; a bracket on the left marks b. Step 2: a particle with b equal to d leaves at about 53 degrees; the first path fades. Step 3: a particle with b equal to a quarter of d swings round the nucleus and leaves at about 127 degrees, heading up and back to the left. Step 4: a particle aimed straight at the nucleus stops a distance d away and returns along its own line, at 180 degrees.

A beam, however, is not aimed. Its particles all have nearly the same energy but arrive at random positions across the foil, and almost all of them have impact parameters far larger than any in the figure. Only those that happen to pass through a tiny disc round some nucleus are turned through large angles.For a fixed nucleus, $\tan (θ / 2) = d / 2 b$ , so only particles with $b < d / 2$ are turned through more than $90^{\circ}$ . You can check this in the lab below. The fraction that bounce back therefore measures the size of that disc, and so how small the charged centre is. It gives an upper limit to the size of a nucleus: if even the closest-approaching α-particles still obey Coulomb's law, the nucleus must be smaller than their distance of closest approach.

InteractiveScattering lab

A nucleus sits at the centre of a panel, with a dashed circle showing where an alpha-particle would touch it. You choose the foil (gold, silver, copper or aluminium), the alpha-particle's energy from 1 to 10 MeV and its impact parameter from 0 to 200 femtometres. The computed path is drawn, with a bracket showing b, and readouts give the head-on closest distance d, the scattering angle and the closest approach on this path. A box can draw a family of paths for b from a quarter of d to four times d. The Fire button moves the alpha-particle along its path, and a message says whether it is thrown back and whether it would touch the nucleus.

Fig. 12.5Choose a foil, an energy and an impact parameter, then press Fire to send an α-particle past the nucleus.

Example 12.1NCERT Example 12.1 · p. 294

How empty is an atom?

In a hydrogen atom the proton has a radius of about $0.84 f m$ and the electron is typically $5.3 \times 10^{- 11} m$ away from it. (a) Imagine enlarging the whole atom until the proton is the size of a cricket ball (radius $3.6 c m$ ). How far away would the electron be? (b) The Earth's radius is $6.37 \times 10^{6} m$ and the Moon circles it at $3.84 \times 10^{8} m$ . Which is emptier, the hydrogen atom or the Earth–Moon system, and by roughly what factor?

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

In terms of volume the contrast is even starker: the proton fills only about $4 \times 10^{- 15}$ of the sphere inside the electron's orbit. Even the solar system, the emptiest arrangement most of us can picture, is crowded by comparison, as the simulation below shows.

SimulationHow empty is the solar system?

The Sun and the four inner planets, Mercury, Venus, Earth and Mars, on their orbits, drawn with true ratios of sizes and distances, so the Sun and planets shrink to tiny dots. Pressing Play sets the planets moving round the Sun. A switch changes to a schematic view in which sizes are exaggerated so that the planets can be seen.

Fig. 12.6The inner planets drawn to true scale. Press Play to set them moving, and use the scale switch to compare with the squeezed schematic view.

Try this

At true scale, compare the Sun's dot with the width of the Earth's orbit: the ratio of radii is about 1 to 216.
Switch to the schematic view. What had to be exaggerated to make the planets visible?
A hydrogen atom scaled up so that its nucleus matched this Sun would put its electron about 290 times farther out than the Earth: nearly ten times as far as Neptune.

Example 12.2NCERT Example 12.2 · p. 295

Closest approach to a lead nucleus

Polonium-210 gives out α-particles with kinetic energy $5.30 M e V$ . One of them heads straight for a lead nucleus ( $Z = 82$ ) in a thin lead foil. (a) How close does it get before it stops and turns back? (b) The lead nucleus and the α-particle have radii of about $7.1 f m$ and $1.7 f m$ . What kinetic energy would an α-particle need to just touch the nucleus? Treat the lead nucleus as fixed: it is about 52 times heavier.

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

So in Rutherford's experiments the α-particles never touched a gold or lead nucleus. They turned round well outside it, under the electric force alone. That is why the counts obeyed Coulomb's law so faithfully, and also why scattering could only set an upper limit on the size of the nucleus, not measure it.

12.2.2 Electron orbits

Rutherford's model still has to say what the electrons are doing. They cannot sit still, or the nucleus would pull them straight in. So, like planets, they must move in orbits, with the Coulomb attraction supplying the centripetal force that keeps them curving. Take the simplest atom, hydrogen: one electron of charge $- e$ and mass $m$ , moving at speed $v$ in a circle of radius $r$ round a proton.

DerivationEnergy of an electron in a circular orbit

For a circular orbit, the Coulomb attraction must equal mass times centripetal acceleration:

\frac{1}{4 π ε_{0}} \frac{e^{2}}{r^{2}} = \frac{m v^{2}}{r}

(12.2)

Rearranged, this links the radius of an orbit to the speed in it. A faster electron circles closer in:

r = \frac{e^{2}}{4 π ε_{0} m v^{2}}

(12.3)

Multiplying (12.2) by $r / 2$ gives the kinetic energy directly. The potential energy of the electron–proton pair is zero when they are far apart and negative here, because they attract:

K = \frac{1}{2} m v^{2} = \frac{e^{2}}{8 π ε_{0} r}, U = - \frac{e^{2}}{4 π ε_{0} r}

Adding the two gives the total energy:

E = K + U = - \frac{e^{2}}{8 π ε_{0} r}

(12.4)

Two results are worth remembering (Fig. 12.7). First, $K = - E$ and $U = 2 E$ : the potential energy is twice the kinetic energy in size, and opposite in sign. Second, the total energy is negative. You would have to supply at least $∣ E ∣$ to pull the electron away to infinity and leave it at rest there, so until then it is bound to the proton. With a positive total energy the electron could reach infinity with energy to spare: it would swing past the proton once and leave, like a comet that never returns.

3/3The total $E = K + U = - K$ is negative: the electron is bound.

Fig. 12.7In a circular orbit the energies are locked together:

U = - 2 K

, so

E = K + U = - K

.NCERT p. 296

On the left a dashed circular orbit surrounds a proton, with an electron on the orbit, its velocity arrow along the orbit and the radius r drawn to it. On the right a horizontal zero line. Step 1: a green bar of height K stands above the line. Step 2: a vermillion bar twice as long hangs below the line, labelled U equals minus 2K. Step 3: a blue bar as long as the green one hangs below the line, labelled E equals minus K, with a dashed guide showing it reaches exactly half as far down as U.

Example 12.3NCERT Example 12.3 · p. 296

Size and speed from a binding energy

It takes $1.51 e V$ to remove the electron from a hydrogen atom in a certain excited state. Treating the orbit as a circle, find (a) its radius, (b) the electron's speed and (c) its kinetic and potential energies.

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

Quick check

1 of 3

An electron in a circular orbit round a proton has kinetic energy

3.40 e V

. What is its total energy?

Show answer

Answer -3.4 eV

In a circular Coulomb orbit

E = - K

, so

E = - 3.40 e V

. (Its potential energy is

U = 2 E = - 6.80 e V

12.3 Atomic Spectra

A prism or a grating spreads light out by wavelength into a spectrum, and the three kinds of spectrum in Fig. 12.8 tell three different stories. The glowing filament of a bulb gives an unbroken band of colour, a continuous spectrum. Hydrogen at low pressure in a discharge tube, glowing because a current passes through it, gives just a few bright, narrow lines with darkness between them: an emission line spectrum (उत्सर्जन रेखीय स्पेक्ट्रम).

Definition

Emission line spectrum

The set of sharp bright lines, at wavelengths characteristic of the element, in the light from an excited gas or vapour at low pressure.

ExampleSodium vapour glows almost entirely at about $589 n m$ , the yellow of sodium street lamps.

Every element has its own arrangement of lines, as distinctive as the barcode on a packet. Heat a pinch of a sample in a flame or a plasma, read its lines, and you know which elements it contains, even in traces too small to weigh; laboratories test water and soil for lead or arsenic this way. Astronomers use the same lines to find out what stars billions of kilometres away are made of.

The third story runs in reverse. Send white light through cool hydrogen and examine what comes out. Almost every colour survives, but thin dark gaps appear, and each gap sits exactly where hydrogen has a bright emission line. A gas, it seems, removes from passing light precisely the wavelengths it would give out when glowing. This pattern of gaps is the gas's absorption spectrum (अवशोषण स्पेक्ट्रम); Bohr's model will explain why the two sets match (Section 12.5).

Definition

Absorption spectrum

A continuous spectrum crossed by dark lines, produced when white light passes through a cool gas that absorbs its own characteristic wavelengths.

ExampleThe dark lines in sunlight, made by gases in the Sun's cooler outer layers.

4/4The missing wavelengths are exactly the emitted ones.

Fig. 12.8Continuous, emission and absorption spectra. Hydrogen absorbs exactly the wavelengths it emits.NCERT Fig. 12.5 · p. 297

Three horizontal strips share a wavelength scale from 400 to 700 nanometres. Step 1: the top strip, labelled hot filament: every colour, is a smooth band shading from violet through blue, green and yellow to red. Step 2: the middle strip, labelled glowing hydrogen: bright lines, is plain except for four narrow coloured lines, violet near 410, blue near 434, sky blue near 486 and red near 656 nanometres. Step 3: the bottom strip, labelled through cool hydrogen: gaps, is the smooth band again with four thin gaps cut out. Step 4: dashed vertical guides run through all three strips, showing the gaps sit exactly under the bright lines.

Try this

A spectroscope from an old CD

You’ll need

an old CD or DVD you can spare
a small cardboard box such as a toothpaste carton
scissors and tape
lamps: a filament bulb, a tube light or CFL, a white LED, and a sodium street lamp at night

Predict first

Which lamp do you expect to show separate bright lines rather than a smooth band?

An old filament bulbA candle flameA tube light or CFLA white wall in sunlight

Ask an adult to cut a piece of an old CD or DVD about 3 cm square. Handle the cut edges carefully.
Cut a narrow slit, about 1 mm wide, in one end of a small box (a toothpaste carton works), and a viewing hole in the top near the other end.
Tape the CD piece inside the box under the viewing hole, shiny side up, tilted so that light from the slit falls on it.
Point the slit at each lamp in turn, look through the hole, and tilt the box until a band of colours appears.
Record whether you see a smooth band or separate lines, and which colours.

Table 12.2What does each lamp's light contain?

	Smooth band or lines?	Colours you see
Filament bulb
Tube light or CFL
White LED
Yellow sodium street lamp
Candle flame

Your entries are saved automatically.

A filament bulb and a candle give a smooth band, because hot solids (the filament, the glowing soot in a flame) radiate every wavelength. A sodium lamp puts nearly all its light into one bright yellow line. A tube light shows separate violet, blue and green lines from its mercury vapour, on top of light from its coating. A white LED gives a broad blue peak and a broad yellow band, smooth rather than lined, because its light comes from solids.

12.4 Bohr Model of the Hydrogen Atom

In Electromagnetic Waves you saw that any accelerated charge radiates; that is how a radio antenna broadcasts. An electron in a circular orbit is accelerating all the time, since the direction of its velocity keeps changing, so by the same rule it should broadcast continuously at its frequency of revolution, losing energy as it does. Planets escape this fate because they carry no net charge; electrons cannot.

Now chain the equations together (Fig. 12.9). Radiating makes $E$ more negative; by (12.4) a more negative $E$ means a smaller $r$ ; by (12.3) a smaller $r$ means a larger $v$ ; and the revolution frequency $v / 2 π r$ therefore climbs. So the electron winds inwards ever faster, its broadcast sliding steadily up in frequency, until it reaches the nucleus. Real atoms contradict both halves of this story: they are stable, and they emit sharp lines rather than a smear.

2/3Losing energy, the electron spirals in; its revolution, and the light it emits, get faster and faster.

Fig. 12.9What classical physics predicts for Rutherford's atom. Real atoms are stable and give sharp lines, so something in the classical picture must fail.NCERT Fig. 12.6 · p. 298

On the left a proton with an electron on a blue circular orbit; on the right a horizontal axis carrying the emitted wave. Step 1: the electron circles on a fixed orbit and the wave on the right is a steady sine, labelled one steady frequency. Step 2: the orbit is replaced by a blue spiral winding inwards to the proton, and the wave on the right becomes a chirp whose crests crowd closer and closer, labelled frequency keeps rising. Step 3: the electron has reached the proton, the spiral fades, and the label atom collapses appears with a time of about 10 to the minus 11 seconds.

Example 12.4NCERT Example 12.4 · p. 298

The classical frequency of an orbit

Take the electron of Example 12.3, with orbit radius $4.77 \times 10^{- 10} m$ and speed $7.29 \times 10^{5} m / s$ . According to classical electromagnetism, at what frequency would it begin to radiate, and in which part of the spectrum is that? What would happen to this frequency as the electron spiralled in?

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

Bohr's response, in 1913, was pragmatic. Keep what works, Newton's laws and Coulomb's law for the orbit, and add three rules, inspired by Planck's and Einstein's quantum ideas, that simply overrule classical electromagnetism inside the atom. He could not justify the rules; he could only test them, and hydrogen's spectrum was the test.

PostulateBohr's first postulate: stationary states

There are certain special orbits in which an electron does not radiate, whatever classical electromagnetism says. While its electron is in one of them, the atom keeps a fixed total energy; such a state is called a stationary state (स्थायी अवस्था).

Bohr offered no mechanism for this. He simply declared that the radiation law does not apply to these special orbits, and let the predictions be the judge. His second rule says which orbits are special.

PostulateBohr's second postulate: quantised angular momentum

The allowed orbits are those in which the electron's angular momentum is a whole-number multiple of $h / 2 π$ , where $h$ is Planck's constant.

Bohr's quantisation condition

L = m v r = \frac{n h}{2 π}, n = 1, 2, 3, \dots

(12.5)

where

$h$: Planck's constant, $6.63 \times 10^{- 34} J s$
$n$: a positive whole number, the principal quantum number

Why angular momentum, of all quantities? Check the units of $h$ : $J s = k g m^{2} / s$ , exactly the units of angular momentum $m v r$ . For a circular orbit, angular momentum is the natural quantity to quantise, and $h / 2 π = 1.055 \times 10^{- 34} J s$ is the natural step size. The whole number $n$ is called the principal quantum number (मुख्य क्वांटम संख्या).

PostulateBohr's third postulate: one photon per jump

Light is given out only when the electron jumps from a stationary state to one of lower energy. One photon carries off the whole energy difference, so its frequency $ν$ is set by the energies $E_{i}$ before and $E_{f}$ after the jump.

Frequency of the emitted photon

h ν = E_{i} - E_{f}, E_{i} > E_{f}

(12.6)

Run backwards, the same rule describes absorption: an atom can take in a photon only if $h ν$ exactly matches the gap to a higher state. Notice what the frequency of the light is not. It is not the frequency of revolution in either orbit, which was the classical expectation.

The postulates fix the allowed orbits completely. Combining the orbit condition (12.2) with the quantisation rule (12.5) gives the radius, speed and energy of every stationary state of hydrogen.

DerivationRadius, speed and energy of the nth orbit

From (12.5), the speed in an allowed orbit is $v = n h / 2 π m r$ . Put this into $m v^{2} = e^{2} / 4 π ε_{0} r$ , which is (12.2) multiplied by $r$ :

m {(\frac{n h}{2 π m r})}^{2} = \frac{e^{2}}{4 π ε_{0} r}

Solving for $r$ gives the allowed radii:

r_{n} = \frac{n^{2}}{m} {(\frac{h}{2 π})}^{2} \frac{4 π ε_{0}}{e^{2}} = \frac{ε_{0} h^{2}}{π m e^{2}} n^{2}

(12.7)

Dividing the quantisation rule by $m r_{n}$ gives the speed in the $n$ th orbit:

v_{n} = \frac{n h}{2 π m r_{n}} = \frac{e^{2}}{2 ε_{0} h} \frac{1}{n}

Finally, put $r_{n}$ into (12.4):

E_{n} = - \frac{e^{2}}{8 π ε_{0} r_{n}} = - \frac{m e^{4}}{8 ε_{0}^{2} h^{2}} \frac{1}{n^{2}}

(12.8)

With $m = 9.11 \times 10^{- 31} k g$ , $e = 1.602 \times 10^{- 19} C$ , $ε_{0} = 8.854 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ and $h = 6.626 \times 10^{- 34} J s$ , the constants give

E_{n} = - \frac{2.18 \times 10^{- 18} J}{n^{2}}

(12.9)

or, dividing by $1.602 \times 10^{- 19} J / e V$ ,

E_{n} = - \frac{13.6 e V}{n^{2}}

(12.10)

The same constants give the radii and speeds:

r_{n} = (5.29 \times 10^{- 11} m) n^{2}, v_{n} = \frac{2.19 \times 10^{6} m / s}{n}

The smallest orbit, $r_{1} = 5.29 \times 10^{- 11} m$ , is the Bohr radius $a_{0}$ (बोर त्रिज्या). It matches the size of a hydrogen atom known from the kinetic theory of gases, a first success for the model. The radii grow as $n^{2}$ (Fig. 12.10) and the speeds fall as $1 / n$ . Even in the innermost orbit the electron moves at less than 1 % of the speed of light, so ignoring relativity is safe.

3/3 $r_{4} = 16 a_{0}$ , with angular momentum $4 h / 2 π$ .

Fig. 12.10Bohr orbits of hydrogen to scale. Each step out in

n

adds one unit of

h / 2 π

to the angular momentum, but the radius grows as

n^{2}

.NCERT p. 299

Concentric blue circles round a proton, drawn to scale, with an electron dot where each circle crosses a dashed ray to the right and the value of n written under it. Step 1: the n equals 1 orbit is a tiny circle close to the proton and the n equals 2 orbit is four times wider. Step 2: the n equals 3 orbit appears, nine times the first. Step 3: the n equals 4 orbit appears, sixteen times the first, filling most of the drawing. The formulas beside it state that the radius equals n squared times a zero and the angular momentum equals n h over 2 pi.

Example

The n = 2 orbit in numbers

For the $n = 2$ state of hydrogen, find the orbit radius, the electron's speed, the time for one revolution and the total energy.

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

Quick check

1 of 4

Match each of Bohr's postulates with what it says.

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

Show answer

Answer

First postulate : Certain orbits do not radiate
Second postulate : $L = n h / 2 π$
Third postulate : $h ν = E_{i} - E_{f}$

The first postulate allows stable, non-radiating orbits; the second picks them out by quantising angular momentum; the third says light is emitted only in jumps between them.

12.4.1 Energy levels

Equation (12.10) lists every energy a hydrogen atom can have while its electron is bound: $- 13.6$ , $- 3.40$ , $- 1.51$ , $- 0.85 e V$ and so on. Drawing them as horizontal lines on a vertical energy scale gives an energy-level diagram (Fig. 12.11). Read it from the bottom up.

Think of the diagram as a ladder. The bottom rung, $n = 1$ , is the ground state (निम्नतम अवस्था), with the most tightly bound electron in the orbit of radius $a_{0}$ . Left alone, every hydrogen atom ends up there, which is why a jar of hydrogen at room temperature is almost entirely in its ground state. Climbing from that rung right off the top of the ladder, to $E = 0$ , costs $13.6 e V$ : the ionisation energy (आयनन ऊर्जा). Experiments on real hydrogen give the same figure, one of the model's most convincing successes.

Definition

Ground state

The lowest-energy state of an atom. For hydrogen it is $n = 1$ , with $E_{1} = - 13.6 e V$ and the electron in the orbit of radius $a_{0}$ .

Definition

Ionisation energy

The least energy needed to remove an electron from an atom in its ground state, leaving it free and at rest.

ExampleFor hydrogen it is $13.6 e V$ .

The rungs above, $n \geq 2$ , are excited states (उत्तेजित अवस्थाएँ). A collision in a discharge tube, or an absorbed photon of exactly the right energy, can lift an atom onto one of them. The first step up is the biggest, $10.2 e V$ from $n = 1$ to $n = 2$ ; from $n = 2$ , a further $3.40 e V$ is enough to set the electron free. On a high rung the electron is held only loosely.

4/4Dropping from $n = 3$ to $n = 2$ , the atom emits a photon of $1.89 e V$ .

Fig. 12.11The energy levels of hydrogen,

E_{n} = - 13.6 e V / n^{2}

. The levels crowd towards

E = 0

; above it the electron is free and can have any energy.NCERT Fig. 12.7 · p. 300

Horizontal lines on a vertical energy scale, with the quantum number n on the left and the energy in electronvolts on the right: n equals 1 at minus 13.6 at the bottom, n equals 2 at minus 3.40, n equals 3 at minus 1.51, and more lines crowding together just below zero, with a hatched band above zero labelled free. A blue electron sits on the n equals 1 line. Step 1: the levels only. Step 2: green arrows go up from n equals 1 to n equals 2, labelled 10.2 eV, and to n equals 3, labelled 12.09 eV, and the electron rises to n equals 3. Step 3: a green arrow goes from n equals 1 right up to zero, labelled 13.6 eV, and the electron rises into the free band. Step 4: a dashed vermillion arrow goes down from n equals 3 to n equals 2, labelled 1.89 eV, and the electron sits on n equals 2.

The diagram shows two more features. The levels crowd together as $n$ grows, because $1 / n^{2}$ changes less and less, and they converge on $E = 0$ , the limit $n \to \infty$ : an electron at rest infinitely far away. Above $E = 0$ the electron is free and can have any energy at all, so the allowed energies there form a continuous band rather than separate lines. And an excited atom does not stay excited for long. Usually within a few nanoseconds it drops to lower levels, giving out a photon at each step.

InteractiveEnergy-level lab

An energy-level diagram for hydrogen, singly ionised helium or doubly ionised lithium, with levels from n equals 1 to 7 and a free band above zero. Choosing a starting level and a final level draws an arrow, downwards for emission or upwards for absorption, and shows the photon's energy in electronvolts, its frequency and its wavelength, with a message naming the region of the spectrum and, for hydrogen, the series. Emitted lines are added to a spectrum strip below, on a logarithmic wavelength scale from 10 to 5000 nanometres with the visible band shaded. A slider sets the energy of a bombarding electron beam; the Show every line button finds the highest level the beam can reach and draws every line the atoms can then emit.

Fig. 12.12Pick two levels to see the photon a jump emits or needs. Then excite the gas with an electron beam and see which lines appear.

Quick check

1 of 3

How much energy is needed to ionise a hydrogen atom that is already in the

n = 3

state?

Show answer

Answer 1.51 eV

The electron must go from

E_{3} = - 13.6 / 9 = - 1.51 e V

0

, so

1.51 e V

is needed, far less than the

13.6 e V

from the ground state.

12.5 The Line Spectra of the Hydrogen Atom

Bohr's third postulate turns the energy-level diagram into a spectrum. When the electron drops from an upper state $n_{i}$ to a lower state $n_{f}$ , the atom emits one photon whose frequency $ν_{i f}$ is given by

h ν_{i f} = E_{n_{i}} - E_{n_{f}}

(12.11)

The levels form a discrete set, so the gaps between them do too, and each gap fixes one frequency. A hydrogen spectrum can therefore only be a set of lines. Putting (12.8) into (12.11) and using $ν = c / λ$ gives the wavelength of every line at once:

Wavelengths of hydrogen's lines

\frac{1}{λ} = R (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}), R = \frac{m e^{4}}{8 ε_{0}^{2} h^{3} c} = 1.097 \times 10^{7} m^{- 1}

where

$R$: the Rydberg constant
$n_{f}, n_{i}$: the final and initial principal quantum numbers, $n_{i} > n_{f}$

A quick way to convert is $λ = h c / Δ E$ with $h c = 1240 e V n m$ : a photon of $Δ E$ electronvolts has a wavelength of $1240 / Δ E$ nanometres.

Lines that end on the same lower level form a family, called a series. Jumps down to $n = 2$ give hydrogen's visible lines, the very ones Balmer had fitted in 1885 (Fig. 12.13). Bohr's theory reproduced Balmer's formula, with the constant $R$ worked out from $m$ , $e$ , $h$ and $ε_{0}$ , and matched the measured wavelengths to better than $0.1$ %.

4/4A jump from $E = 0$ would release $3.40 e V$ : the series limit at $365 n m$ , in the ultraviolet.

Fig. 12.13The Balmer series of hydrogen: every jump that ends on

n = 2

. Longer jumps give shorter wavelengths, and the lines crowd towards the series limit at

365 n m

.NCERT Fig. 12.5 · p. 297

The upper part shows energy levels n equals 2 to 8, crowding upwards, with a dashed line at zero energy. The lower part is a wavelength strip from 350 to 700 nanometres. Each jump is a black arrow down to the n equals 2 line, placed directly above the line it produces, with a dotted guide down to the strip. Step 1: the jump from n equals 3 gives a red line at 656 nanometres. Step 2: the jump from n equals 4 adds a sky-blue line at 486. Step 3: jumps from n equals 5 and 6 add a blue line at 434 and a violet line at 410. Step 4: a dashed arrow from zero energy marks the series limit at 365 nanometres, where the lines bunch together.

Table 12.3Three series of hydrogen lines, named after the physicists who found them

	Ends on	Wavelengths (nm)
Lyman (ultraviolet)	$n_{f} = 1$	121.6 down to 91.2
Balmer (visible)	$n_{f} = 2$	656.3 down to 364.7
Paschen (infrared)	$n_{f} = 3$	1875 down to 820

Example

Which jump made this line?

A hydrogen lamp shows a sky-blue line at $486 n m$ . (a) What is the photon energy? (b) Which jump produces it? (c) What is its frequency?

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

The same diagram explains absorption. A hydrogen atom in its ground state can take in a photon only if the photon's energy matches the gap to some higher level: $10.2 e V$ , $12.09 e V$ , $12.75 e V$ and so on, or any amount above $13.6 e V$ , which ionises it. Send continuous light through cool hydrogen and those wavelengths go missing, leaving the dark lines of Section 12.3. Since cool hydrogen sits in its ground state, its absorption lines are the ultraviolet Lyman lines. The visible Balmer wavelengths are absorbed only by atoms already in $n = 2$ , such as those in the hot outer layers of stars.

From a handful of constants of nature, Bohr's model accounted for every line of the hydrogen spectrum, and it earned him the 1922 Nobel Prize in Physics. Its success did much to launch the quantum theory of the atom.

Quick check

1 of 4

Find the wavelength of the photon emitted when hydrogen's electron drops from

n = 5

n = 2

Show answer

Answer 434 nm

Δ E = 13.6 (1 / 4 - 1 / 25) = 2.86 e V

, so

λ = 1240 / 2.86 = 434 n m

, a blue Balmer line.

12.6 De Broglie's Explanation of Bohr's Second Postulate of Quantisation

Bohr's second rule gave the right answers, but it looked like a rule plucked from the air. Louis de Broglie showed in 1923 where it comes from, using the idea you met in the previous chapter, Dual Nature of Radiation and Matter: a moving electron has a wavelength $λ = h / p$ , a claim that electron-diffraction experiments confirmed in 1927.

Try drawing a wave round a circle. Start anywhere and trace it once round. When you arrive back at the start, the wave must carry on exactly as it began, at the same height and rising the same way; otherwise the second lap lands out of step with the first, the third with the second, and lap after lap the mismatched waves cancel one another. A wave that persists round an orbit must therefore fit it a whole number of times. It is the circular version of the standing waves on a stretched string you met in Class 11 (Fig. 12.14).

1/3Three wavelengths fit exactly: crest meets crest, and a standing wave forms.

Fig. 12.14Only whole numbers of wavelengths can form a standing wave round an orbit. (For clarity one orbit is used throughout; in hydrogen each value of

n

has its own orbit and its own wavelength.)NCERT Fig. 12.8 · p. 301

A proton sits at the centre of a dashed circular orbit of radius r, and a blue wave is wrapped round the circle, rising and falling about it. Step 1: exactly three wavelengths fit round the circle and the wave joins smoothly onto itself. Step 2: three and a half wavelengths are tried; the wave, now dashed vermillion, meets its own start going the wrong way, and a circle marks the clash, labelled ends clash. Step 3: exactly five wavelengths fit and the wave again joins smoothly.

DerivationFrom standing waves to Bohr's rule

For the $n$ th orbit, of radius $r_{n}$ , the fitting condition is

2 π r_{n} = n λ, n = 1, 2, 3, \dots

(12.12)

For an electron moving much slower than light, $p = m v_{n}$ , so $λ = h / m v_{n}$ . Substituting,

2 π r_{n} = \frac{n h}{m v_{n}} ⟹ m v_{n} r_{n} = \frac{n h}{2 π}

This is exactly Bohr's second postulate (12.5). The quantised orbits, and with them the energy levels, appear because the electron behaves as a wave, and only standing waves that fit can persist.

Example

The electron's wave in the n = 2 orbit

Find the de Broglie wavelength of the electron in the $n = 2$ orbit of hydrogen (a) from its speed and (b) from the size of the orbit. Check that the two answers agree.

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

Bohr's model, backed by de Broglie's picture, gets the main features of hydrogen right: its size, its ionisation energy, and the frequencies it emits and absorbs. It works equally well for any hydrogenic atom, a nucleus of charge $+ Z e$ with a single electron, such as $H e^{+}$ or $L i^{2 +}$ . But it has serious limits:

It breaks down as soon as there is a second electron: helium already defeats it. The model includes only the pull of the nucleus. In a real atom each electron also repels every other electron, and these forces are comparable with the nuclear attraction, because the charges and distances are similar. In the solar system, by contrast, the planets' pulls on one another are tiny next to the Sun's, since the Sun is about 750 times heavier than all the planets together, so the planetary picture survives there.
It cannot predict intensities. Some hydrogen lines are bright and others faint, so some jumps must happen far more often than others. Bohr's rules give the frequencies of the lines but say nothing about how likely each jump is.
Its sharp orbits are not real. By the uncertainty principle, an electron's position and momentum cannot both be known exactly, so it cannot follow a definite circular path.

Quantum mechanics, developed from 1925 by Werner Heisenberg, Erwin Schrödinger and others, replaced orbits with clouds of probability and answers all these questions. Why, then, learn a model known to be wrong? Because with nothing but Newton, Coulomb and one quantum rule it gets hydrogen's energies exactly right; because its numbers ( $a_{0} = 0.053 n m$ , $13.6 e V$ , $2.2 \times 10^{6} m / s$ ) still set the scale of atomic physics; and because it is a lesson in method. Bohr set aside an objection he could not answer, made sharp predictions, and let experiment rule.

Definition

Hydrogenic atom

An atom or ion with a nucleus of charge $+ Z e$ and a single electron, so that there are no electron–electron forces.

Example $H$ , $H e^{+}$ and $L i^{2 +}$ .

Quick check

1 of 3

According to de Broglie, how many electron wavelengths fit round the

n = 3

orbit of hydrogen?

1369

Show answer

Answer (B) 3

Look at (12.12).

Flash cards · 10

Formula cards

1 / 10

Recap

Atoms: key points and summary

An atom is neutral: its light electrons are balanced by an equal positive charge that carries almost all the mass. Thomson spread that charge through the whole atom (Section 12.1).
Geiger and Marsden found that a few α-particles bounce back from thin gold foil. Only a tiny, massive, positive nucleus, $10^{4}$ to $10^{5}$ times smaller than the atom, can do that (Section 12.2).
A smaller impact parameter $b$ gives a larger scattering angle; a head-on α-particle stops at $d = 2 Z e^{2} / 4 π ε_{0} K$ and turns back (Section 12.2.1).
For an electron in a circular orbit, $E = - e^{2} / 8 π ε_{0} r$ , with $K = - E$ and $U = 2 E$ ; the negative energy means it is bound (Section 12.2.2).
A thin glowing gas gives an emission line spectrum that identifies the element; cool gas absorbs the same wavelengths from white light (Section 12.3).
Classically, an orbiting electron radiates, spirals into the nucleus in about $10^{- 11} s$ and emits a continuous spectrum, so Rutherford's atom cannot be the whole story (Section 12.4).
Bohr's postulates: stationary orbits do not radiate; $L = n h / 2 π$ ; a jump emits one photon with $h ν = E_{i} - E_{f}$ (postulates).
For hydrogen, $r_{n} = n^{2} a_{0}$ with $a_{0} = 0.0529 n m$ , $v_{n} \propto 1 / n$ and $E_{n} = - 13.6 e V / n^{2}$ (derivation).
The ground state is $n = 1$ ; the ionisation energy is $13.6 e V$ ; excited states crowd towards $E = 0$ , above which the electron is free (Section 12.4.1).
Each spectral line is one jump, $h ν_{i f} = E_{n_{i}} - E_{n_{f}}$ ; jumps ending on $n = 1$ , $2$ , $3$ give the Lyman (ultraviolet), Balmer (visible) and Paschen (infrared) series (Section 12.5).
De Broglie explained the quantisation rule: a whole number of electron wavelengths must fit round the orbit, $2 π r_{n} = n λ$ (Section 12.6).
Bohr's model works only for one-electron (hydrogenic) atoms and cannot predict line intensities; quantum mechanics replaced it (limitations).

Practice · 14 questions

Practice questions with answers (NCERT “Exercises”)

1
Short answerEasyNCERT Exercises 12.1
For each observation or prediction, decide which model of the atom it belongs to: Thomson's, Rutherford's, both, or neither.
1. With the detector at $150^{\circ}$ behind a gold foil, a few flashes appear every minute.
  Thomson's modelRutherford's modelBoth modelsNeither model
  Show answer
  Answer (B) Rutherford's model
  Turning an α-particle back needs the intense field close to a concentrated nucleus. A spread-out Thomson atom would send essentially nothing to $150^{\circ}$ .
2. An electron placed at rest at the centre of the atom stays there, and if nudged it oscillates about the centre.
  Thomson's modelRutherford's modelBoth modelsNeither model
  Show answer
  Answer (A) Thomson's model
  Inside a uniform ball of positive charge the field points back towards the centre and grows with distance from it, so an electron there is in stable equilibrium. In the nuclear atom a resting electron would fall straight into the nucleus.
3. Inside the atom, the density is everywhere about that of solid gold, roughly $2 \times 10^{4} k g / m^{3}$ .
  Thomson's modelRutherford's modelBoth modelsNeither model
  Show answer
  Answer (A) Thomson's model
  Thomson's atom spreads its mass through its whole volume, so its density is just the bulk density of the metal. In Rutherford's atom almost all the mass is in a nucleus whose density is about $10^{17} k g / m^{3}$ , while the rest of the atom is nearly empty.
4. Classical electromagnetism predicts that the atom collapses in about $10^{- 11} s$ .
  Thomson's modelRutherford's modelBoth modelsNeither model
  Show answer
  Answer (B) Rutherford's model
  Only orbiting, and therefore accelerating, electrons radiate their energy away. That is the nuclear model's weakness, and Bohr's postulates were invented to cure it.
5. Hydrogen gives out light only at a few sharp wavelengths.
  Thomson's modelRutherford's modelBoth modelsNeither model
  Show answer
  Answer (D) Neither model
  Neither classical model can explain line spectra. Thomson's atom gives no reason for particular wavelengths, and Rutherford's would emit a continuous sweep as its electron spiralled in. Explaining the lines needed Bohr's quantum rules.
2
Long answerMediumNCERT Exercises 12.2
A student repeats the scattering experiment with $5 M e V$ α-particles, but in place of gold uses a thin polythene film, which contains only carbon nuclei ( $Z = 6$ , mass $12 u$ ) and hydrogen nuclei ( $Z = 1$ , mass $1 u$ ). The α-particle's mass is $4 u$ .
1. Can a hydrogen nucleus (a proton) turn an α-particle back towards the source?
  Yes, in a head-on collisionYes, but only for slow α-particlesNo: a proton is lighter than an α-particle and cannot turn it round
  Show answer
  Answer (C) No: a proton is lighter than an α-particle and cannot turn it round
  A heavy particle striking a lighter one at rest always carries on forwards, the way a cricket ball barely notices a table-tennis ball. Momentum and energy conservation limit its deflection to $\sin θ_{\max} = m_{p} / m_{α} \approx 1 / 4$ , about $15^{\circ}$ ; the proton is knocked forwards instead.
2. Compared with gold, how common will large-angle scattering from the carbon nuclei be?
  More common, because carbon nuclei are smallerAbout as common, because carbon nuclei are also heavier than the α-particleMuch rarer, because a carbon nucleus carries far less charge
  Show answer
  Answer (C) Much rarer, because a carbon nucleus carries far less charge
  Carbon is heavier than the α-particle, so it can turn one back, but its charge $6 e$ is about 13 times smaller than gold's $79 e$ . Its field is correspondingly weaker, so an α-particle must pass much closer to be deflected strongly, and such close passes are far rarer.
3. What would the experiment show about the hydrogen in the film, and what would it fail to show?
  Show solution
  It would show protons shooting forwards out of the film, knocked on by α-particles that passed very close. A proton that can be struck that hard must be a small, massive, charged object, so hydrogen too has a nucleus.
  It could not show back-scattering from hydrogen, and with only light, weakly charged nuclei the dramatic large-angle evidence that gold gave would be largely missing. Gold was a good choice because its nucleus is both heavy and highly charged, and because it can be beaten into very thin leaf.
3
NumericalEasyNCERT Exercises 12.3
Two energy levels of a sodium atom are $2.10 e V$ apart. The atom drops from the upper to the lower level.
1. Find the frequency of the emitted light.
  Hz
  Show answer
  Answer 508000000000000 Hz
  $ν = Δ E / h = (2.10 \times 1.602 \times 10^{- 19} J) / (6.626 \times 10^{- 34} J s) = 5.08 \times 10^{14} H z$ .
2. Find its wavelength.
  nm
  Show answer
  Answer 590 nm
  $λ = 1240 / 2.10 = 590 n m$ , or $c / ν = (3.00 \times 10^{8}) / (5.08 \times 10^{14}) = 5.90 \times 10^{- 7} m$ .
3. What colour is it?
  Show solution
  AnswerYellow.
  $590 n m$ lies in the yellow part of the visible spectrum. It is the familiar glow of sodium street lamps.
4
NumericalEasyNCERT Exercises 12.4
The total energy of hydrogen's electron in the first excited state ( $n = 2$ ) is $- 3.40 e V$ .
1. What is the electron's kinetic energy?
  eV
  Show answer
  Answer 3.4 eV
  $K = - E = + 3.40 e V$ .
2. What is its potential energy?
  eV
  Show answer
  Answer -6.8 eV
  $U = 2 E = - 6.80 e V$ . Check: $K + U = 3.40 - 6.80 = - 3.40 e V$ .
5
NumericalMediumNCERT Exercises 12.5
A hydrogen atom in the $n = 2$ state absorbs a photon and ends up in the $n = 5$ state.
1. Find the photon's energy.
  eV
  Show answer
  Answer 2.856 eV
  $E_{5} - E_{2} = - 0.544 - (- 3.40) = 2.86 e V$ .
2. Find its wavelength.
  nm
  Show answer
  Answer 434 nm
  $λ = 1240 / 2.856 = 434 n m$ (blue).
3. Find its frequency.
  Hz
  Show answer
  Answer 691000000000000 Hz
  $ν = (2.856 \times 1.602 \times 10^{- 19} J) / (6.626 \times 10^{- 34} J s) = 6.91 \times 10^{14} H z$ .
6
NumericalMediumNCERT Exercises 12.6
Use Bohr's model of hydrogen.
1. Find the electron's speed in the $n = 2$ and $n = 4$ orbits.
  Show solution
  $v_{2} = 2.19 \times 10^{6} / 2 = 1.09 \times 10^{6} m / s$ and $v_{4} = 2.19 \times 10^{6} / 4 = 5.47 \times 10^{5} m / s$ .
2. Find the time for one revolution in each orbit.
  Show solution
  $r_{2} = 4 a_{0} = 2.12 \times 10^{- 10} m$ , so $T_{2} = 2 π (2.12 \times 10^{- 10}) / (1.09 \times 10^{6}) = 1.22 \times 10^{- 15} s$ .
  $r_{4} = 16 a_{0} = 8.46 \times 10^{- 10} m$ , so $T_{4} = 2 π (8.46 \times 10^{- 10}) / (5.47 \times 10^{5}) = 9.72 \times 10^{- 15} s$ .
  Check: $T \propto r / v \propto n^{3}$ , and $T_{4} / T_{2} = 2^{3} = 8$ .
3. An atom typically stays in the $n = 2$ state for about $1 n s$ before it emits a photon. Roughly how many times does the electron go round in that time?
  revolutions
  Show answer
  Answer 820000 revolutions
  $(10^{- 9} s) / (1.22 \times 10^{- 15} s) \approx 8.2 \times 10^{5}$ : nearly a million laps, so the stationary state is long-lived on the scale of its own motion.
7
NumericalEasyNCERT Exercises 12.7
The Bohr radius is $a_{0} = 5.29 \times 10^{- 11} m$ .
1. What is the radius of the $n = 5$ orbit?
  m
  Show answer
  Answer 1.32e-9 m
  $r_{5} = 25 \times 5.29 \times 10^{- 11} = 1.32 \times 10^{- 9} m$ .
2. Physicists can excite hydrogen atoms into enormous 'Rydberg' states. Which $n$ gives an orbit of radius $1.32 \times 10^{- 7} m$ , about the size of a small bacterium?
  Show answer
  Answer 50
  $n^{2} = (1.32 \times 10^{- 7}) / (5.29 \times 10^{- 11}) = 2495$ , so $n = 49.95 \approx 50$ .
8
Long answerMediumNCERT Exercises 12.8
In a discharge tube, electrons are accelerated through $13.0 V$ and then collide with hydrogen atoms at room temperature.
1. Up to which level can the beam excite ground-state atoms?
  Show answer
  Answer 4
  $E_{4} - E_{1} = 12.75 e V \leq 13.0 e V$ , but $E_{5} - E_{1} = 13.06 e V$ is too much. So $n = 4$ .
2. List every line the excited atoms can emit as they return to the ground state, with its wavelength.
  Show solution
  From $n = 4$ the electron can come down by any route, so every jump between levels 1 to 4 occurs: $4 \times 3 / 2 = 6$ lines.
  Table 12.4
  Jump $Δ E$ (eV) $λ$ (nm) Series
  4 → 1 12.75 97.2 Lyman
  3 → 1 12.09 102.6 Lyman
  2 → 1 10.20 121.6 Lyman
  4 → 2 2.55 486.2 Balmer
  3 → 2 1.89 656.4 Balmer
  4 → 3 0.661 1875 Paschen
3. (c)
  Which of these lines can you see with your eyes?
  $97.2 n m$ $121.6 n m$ $486.2 n m$ $656.4 n m$ $1875 n m$
  Select all that apply
  Show answer
  Answer
  (C) $486.2 n m$
  (D) $656.4 n m$
  Only the two Balmer lines, sky blue at $486 n m$ and red at $656 n m$ , fall in the visible range of about $400$ to $700 n m$ .
9
NumericalHardNCERT Exercises 12.9
Treat the Moon's orbit round the Earth as if it were a Bohr orbit. The Moon's mass is $7.35 \times 10^{22} k g$ , its orbit radius $3.84 \times 10^{8} m$ and its orbital speed $1.02 \times 10^{3} m / s$ .
1. Find the quantum number $n$ .
  Show answer
  Answer 2.73e+68
  $L = m v r = (7.35 \times 10^{22}) (1.02 \times 10^{3}) (3.84 \times 10^{8}) = 2.88 \times 10^{34} k g m^{2} / s$ , so $n = 2 π L / h = 2 π (2.88 \times 10^{34}) / (6.63 \times 10^{- 34}) = 2.73 \times 10^{68}$ .
2. What does this tell you about quantisation in everyday systems?
  Show solution
  With $n$ about $10^{68}$ , neighbouring allowed orbits differ by one part in roughly $10^{68}$ in angular momentum. No measurement could ever detect such steps, so for large objects the allowed values form a continuum and classical mechanics works perfectly. Quantisation matters only when $L$ is a few times $h / 2 π$ , as for electrons in atoms.
Quick answerEasy
How much energy is needed to ionise a hydrogen atom in the $n = 4$ state?
eV
Show answer
Answer 0.85 eV
$E_{4} = - 13.6 / 16 = - 0.85 e V$ , so $0.85 e V$ frees the electron.
11
Quick answerEasy
Match each quantity for hydrogen's $n$ th orbit with how it depends on $n$ .
Tap an item on the left, then its match on the right.
Show answer
Answer
Orbit radius : $\propto n^{2}$
Electron speed : $\propto 1 / n$
Total energy : $\propto - 1 / n^{2}$
Time for one revolution : $\propto n^{3}$
$T = 2 π r / v \propto n^{2} / (1 / n) = n^{3}$ , and $E \propto - 1 / r \propto - 1 / n^{2}$ .
12
Quick answerEasy
Put the steps of Rutherford's reasoning in order.
1. So the atom is mostly empty, with a tiny, dense, positive nucleus
2. Bouncing back needs a huge repulsive force over a short distance
3. Most α-particles pass straight through a thin foil, but a few bounce back
4. Such a force needs the positive charge and mass packed into a tiny region
Show answer
Answer
Most α-particles pass straight through a thin foil, but a few bounce back
Bouncing back needs a huge repulsive force over a short distance
Such a force needs the positive charge and mass packed into a tiny region
So the atom is mostly empty, with a tiny, dense, positive nucleus
Observation, then the force it requires, then what can supply that force, then the model.
Quick answerMedium
A hydrogen atom in its ground state can absorb a photon of energy $11.0 e V$ .
TrueFalse
Show answer
Answer False
The gaps from $n = 1$ are $10.2$ , $12.09$ , $12.75 e V$ and so on, and ionisation needs $13.6 e V$ . A photon of $11.0 e V$ matches none of these, so it passes through. (A colliding electron of $11.0 e V$ could excite the atom, because it can keep the extra energy.)
Quick answerEasy
The energy needed to remove the electron from the ground state of hydrogen, $13.6 e V$ , is called its energy.
Show answer
Answer ionisation
Removing the electron leaves a positive ion, so this is the ionisation energy.

Jump	$Δ E$ (eV)	$λ$ (nm)	Series
4 → 1	12.75	97.2	Lyman
3 → 1	12.09	102.6	Lyman
2 → 1	10.20	121.6	Lyman
4 → 2	2.55	486.2	Balmer
3 → 2	1.89	656.4	Balmer
4 → 3	0.661	1875	Paschen

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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.

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NCERT Exercises 12.1	Practice Q1
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NCERT Exercises 12.3	Practice Q3
NCERT Exercises 12.4	Practice Q4
NCERT Exercises 12.5	Practice Q5
NCERT Exercises 12.6	Practice Q6
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NCERT Exercises 12.8	Practice Q8
NCERT Exercises 12.9	Practice Q9

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Warm-up2 questions
12.2.1 Alpha-particle trajectory4 questions
12.2.2 Electron orbits3 questions
12.3 Atomic Spectra3 questions
12.4 Bohr Model of the Hydrogen Atom4 questions
12.4.1 Energy levels3 questions
12.5 The Line Spectra of the Hydrogen Atom4 questions
12.6 De Broglie's Explanation of Bohr's Second Postulate of Quantisation3 questions

Glossary · 12

Key terms (English–Hindi)

absorption spectrumअवशोषण स्पेक्ट्रमavashoshan spectrum: A continuous spectrum with dark lines where a cool gas has absorbed its own characteristic wavelengths.First seen in the chapter →

Bohr radiusबोर त्रिज्याBohr trijya: The radius of hydrogen's smallest orbit, $a_{0} = 5.29 \times 10^{- 11} m$ .First seen in the chapter →

emission line spectrumउत्सर्जन रेखीय स्पेक्ट्रमutsarjan rekhiya spectrum: Bright lines at wavelengths characteristic of an element, from its excited gas at low pressure.First seen in the chapter →
excited stateउत्तेजित अवस्थाuttejit avastha: Any state of an atom above the ground state ( $n \geq 2$ in hydrogen).First seen in the chapter →

ground stateनिम्नतम अवस्थाnimnatam avastha: An atom's lowest-energy state; for hydrogen $n = 1$ , $E_{1} = - 13.6 e V$ .First seen in the chapter →

hydrogenic atomहाइड्रोजन-सम परमाणुhydrogen-sam paramanu: A nucleus of charge $+ Z e$ with a single electron, such as $H e^{+}$ or $L i^{2 +}$ .First seen in the chapter →

impact parameterसंघट्ट प्राचलsanghatt prachal: The perpendicular distance between a particle's initial line of motion and the centre of the target nucleus.First seen in the chapter →
ionisation energyआयनन ऊर्जाaayanan oorja: The least energy that frees an electron from an atom in its ground state; $13.6 e V$ for hydrogen.First seen in the chapter →

nucleusनाभिकnabhik: The tiny, dense, positive centre of an atom, holding all its positive charge and nearly all its mass.First seen in the chapter →

principal quantum numberमुख्य क्वांटम संख्याmukhya quantum sankhya: The whole number $n$ labelling a Bohr orbit; $L = n h / 2 π$ and $E_{n} = - 13.6 e V / n^{2}$ .First seen in the chapter →

stationary stateस्थायी अवस्थाsthayi avastha: An allowed state of an atom in which the electron does not radiate and the energy is constant.First seen in the chapter →

α-particleऐल्फ़ा कणalpha kan: A helium nucleus (two protons, two neutrons), charge $+ 2 e$ , emitted by some radioactive nuclei.First seen in the chapter →

12.1 Introduction

12.2 Alpha-particle Scattering and Rutherford's Nuclear Model of Atom

12.2.1 Alpha-particle trajectory

12.2.2 Electron orbits

12.3 Atomic Spectra

12.4 Bohr Model of the Hydrogen Atom

12.4.1 Energy levels

12.5 The Line Spectra of the Hydrogen Atom

12.6 De Broglie's Explanation of Bohr's Second Postulate of Quantisation

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