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PhysicsCh 11 Dual Nature of Radiation and Matter

Dual Nature of Radiation and Matter: Class 12 Physics · Chapter 11

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

Published · Updated

  • 85 min
  • 15 practice questions
  • 29 MCQs

What you will learn in Dual Nature of Radiation and Matter

  • Describe how cathode-ray experiments revealed the electron and measured its specific charge, charge and mass
  • Define the work function and the electron volt, and compare thermionic, field and photoelectric emission
  • Describe the observations of Hertz, Hallwachs and Lenard and the apparatus used to study photoelectric emission
  • Interpret graphs of photocurrent against intensity and collector potential, and of stopping potential against frequency
  • Explain why the wave picture of light fails for the photoelectric effect and how Einstein's light quanta succeed
  • Use Einstein's photoelectric equation to find threshold frequency, stopping potential, maximum kinetic energy and Planck's constant
  • State the properties of photons and calculate photon energy, momentum and the number of photons per second in a beam
  • Apply the de Broglie relation to particles and explain why matter waves show up only on the atomic scale

Original companion text aligned to NCERT Physics Class 12, Chapter 11, pp. 274–289. Not affiliated with or endorsed by NCERT.

11.1 Introduction

By now the case for light as a wave looks closed. Two slits give bright and dark fringes, a narrow gap spreads a beam, and a Polaroid can cut a reflection's glare, as you saw in Wave Optics. Maxwell's equations even predict the speed of the wave. Yet the light meter of a camera, the photomultiplier tube that counts single flashes in a particle detector, and the night-vision tube all work on an effect that the wave picture cannot explain. This chapter follows that puzzle to its answer, and then turns the answer round: if light can act like a stream of particles, can particles act like waves?

The puzzle grew out of experiments in the last quarter of the nineteenth century. In 1887 Heinrich Hertz made and detected electromagnetic waves, a triumph for Maxwell's wave theory. At the same time, physicists were passing electricity through gases at very low pressure, and these experiments kept producing surprises: X-rays (Wilhelm Röntgen, 1895) and, above all, the electron.

Seal two metal electrodes into a glass tube, pump out nearly all the air and connect a few thousand volts across them. Once the pressure is down to about a thousandth of a millimetre of mercury, roughly a millionth of an atmosphere, the glass at the far end from the negative electrode starts to glow. Its colour depends on the glass; ordinary soda glass glows green. Something is streaming out of the cathode and striking the glass. These cathode rays (कैथोड किरणें) were studied closely by William Crookes in the 1870s, and in 1879 he argued that they were streams of fast, negatively charged particles.

In 1897 J. J. Thomson tested the idea with a beam of cathode rays, an electric field between two plates and a magnetic field at right angles to both the beam and the electric field (Fig. 11.1). Each field alone bends the beam; set up together and adjusted, they can cancel. The cancellation gives the speed of the particles, and the bending by one field alone then gives their charge-to-mass ratio, or specific charge e/m. The particles turned out to move at about one-tenth to one-fifth of the speed of light, c=3×108 m/s.

3/3A magnetic field into the page pushes them down. When eE=evB the beam is straight again, so v=E/B.

Fig. 11.1Balancing an electric and a magnetic force on the beam measures the speed of the particles.

A long evacuated tube lies across the figure, with the cathode at the left end, an anode with a slit, a pair of horizontal plates (positive on top, negative below) and a screen at the right end. Step 1: the electron beam runs straight from the cathode to a glowing spot at the centre of the screen. Step 2: with the plates charged, the beam curves upwards towards the positive plate and the spot moves up; an arrow shows the upward electric force. Step 3: a magnetic field directed into the page is added over the plates, shown by crossed-circle symbols; a second, downward magnetic force balances the electric force and the beam runs straight again, which happens when v equals E divided by B.

Two findings made this particle special. Its e/m came out the same whatever metal the cathode was made of and whatever gas was left in the tube. And negative particles knocked out of a metal by ultraviolet light, or boiled out of a white-hot filament, had exactly the same e/m. One kind of particle, produced in very different ways from very different materials, had to be a common ingredient of all matter. Thomson called it the electron; the presently accepted value of its specific charge is e/m=1.76×1011 C/kg. He received the Nobel Prize in Physics in 1906.

The charge itself came from R. A. Millikan's oil-drop experiment, published in 1913. Millikan timed tiny charged oil drops as they rose and fell between two charged plates, and found that the charge on every drop was a whole-number multiple of one smallest charge, e=1.602×10−19 C. Charge is quantised, as you assumed in Chapter 1 of Part I. Dividing e by e/m then gave the electron's mass, about 9.11×10−31 kg, nearly two thousand times lighter than the lightest atom.

Example

Weighing the electron with two fields

In a cathode-ray tube the deflecting plates are 1.5 cm apart with 450 V across them. A magnetic field of 0.75 mT, perpendicular to both the beam and the electric field, makes the beam go straight. With the electric field switched off, the same magnetic field bends the beam into an arc of radius 30.3 cm. Find (a) the speed of the electrons, (b) their specific charge and (c) their mass, taking e=1.602×10−19 C.

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

Timeline

From cathode rays to matter waves

  1. 1879

    Crookes: cathode rays are charged particles

  2. 1887

    Hertz: ultraviolet light helps sparks jump

  3. 1888

    Hallwachs: ultraviolet light discharges zinc

  4. 1897

    Thomson measures e/m: the electron

  5. 1900

    Planck: radiation energy in steps of hν

    Introduced to explain the spectrum of radiation from hot bodies.

  6. 1902

    Lenard: electron energy ignores intensity

  7. 1905

    Einstein: light quanta explain photoemission

  8. 1913

    Millikan's oil drops fix e

  9. 1916

    Millikan confirms Einstein's equation

  10. 1923

    Compton: X-ray photons bounce off electrons

  11. 1924

    de Broglie: particles have waves

  12. 1927

    Electron diffraction observed

    C. J. Davisson and L. H. Germer in the USA; G. P. Thomson in Britain.

So, by the early 1900s, the electron was a particle with a definite charge and mass, and light was a wave. The rest of this chapter shows that each of these statements is only half the story: radiation and matter both have a dual nature (द्वैत प्रकृति).

Quick check
1 of 3
A cathode-ray beam passes undeflected through crossed electric and magnetic fields. What does this tell you about the particles?
Show answer

Answer (C) Their speed equals E/B

The two forces are equal and opposite, eE=evB, so v=E/B. Finding e/m needs a second measurement, such as the radius of the path in the magnetic field alone. An uncharged beam would also go straight, but it would not have been bent by either field on its own.

11.2 Electron emission

A copper wire is crowded with free electrons; that is why it conducts so well. Yet a wire lying on a table does not spray electrons into the room. Deep inside the metal, each free electron has positive ions on every side and their pulls cancel. At the surface that balance breaks. An electron that strays outside leaves the metal slightly positive, and the attraction of the ions it left behind drags it back.

To leave for good, an electron must be given a certain minimum amount of energy. This least energy needed to free an electron from a metal surface is the work function (कार्य फलन) of the metal, written ϕ0. Its value depends on the metal and also on the condition of the surface: a thin oxide layer, a film of adsorbed gas or a trace of another metal can change it noticeably.

Definition
Work function

The minimum energy ϕ0 that must be supplied to an electron to free it from the surface of a metal.

ExampleAbout 2.1 eV for caesium and about 4.5 eV for tungsten.

Work functions are a few times 10−19 J, clumsy numbers to carry around. Atomic and nuclear physics therefore uses a smaller unit, the electron volt: the kinetic energy an electron gains when it is accelerated from rest through a potential difference of 1 V. Since the energy gained is qV,

1 eV=(1.602×10−19 C)(1 V)=1.602×10−19 J

On this scale work functions lie between about 2 eV for caesium and nearly 6 eV for platinum. An electron accelerated through 1000 V gains 1000 eV, or 1 keV, whatever the size of the apparatus.

How can a metal's electrons be given that energy? There are three practical ways, compared in Fig. 11.2.

  1. Heating. At high temperature some free electrons gain enough thermal energy to cross the surface. This thermionic emission supplies the electrons in X-ray tubes and electron guns, from a glowing tungsten filament.
  2. A very strong electric field. A field of order 108 V/m at a metal surface can pull electrons straight out. This is field emission; the very sharp tips used as electron sources in some electron microscopes work this way.
  3. Light. Light of a suitable frequency falling on a metal can free electrons from it. This is photoelectric emission, and the electrons it frees are called photoelectrons (प्रकाशिक इलेक्ट्रॉन).

4/4Light of suitable frequency frees electrons: photoelectric emission.

Fig. 11.2To escape, the most energetic electrons must still be given an energy ϕ0, by heat, a strong field or light.

On the left, an energy diagram: electrons fill the lower part of a well that represents the metal, and a dashed line at the top marks an electron at rest outside the metal. A double arrow between the top of the filled part and the dashed line is labelled phi zero, the work function. On the right, a panel changes with each step. Step 1: an electron just outside a metal block is pulled back by the positive ions. Step 2: flames heat the metal, labelled thermionic emission, and one electron in the diagram rises out of the well. Step 3: a positively charged plate faces the metal with field lines between them, labelled field emission, and the electron again escapes. Step 4: a wavy light ray strikes the metal, labelled photoelectric emission, and the electron escapes.

Quick check
1 of 3
An electron starts from rest and is accelerated through a potential difference of 250 V. What kinetic energy does it gain, in joules?
J
Show answer

Answer 4.005e-17 J

It gains 250 eV=250×1.602×10−19 J=4.0×10−17 J.

11.3 Photoelectric effect

Of the three ways of freeing electrons, the third turned out to hold the deepest lesson. It was first noticed by accident.

11.3.1 Hertz’s observations

In 1887 Heinrich Hertz (1857–1894) was sending electromagnetic waves across his laboratory. His detector was a loop of wire with a tiny gap, and a received wave showed itself as a spark across the gap. Hertz noticed that the detector sparked more readily when ultraviolet light, from his transmitter's spark or from an arc lamp, fell on the metal near the gap. Shielding the gap from the light made the sparks weaker. We now know what was happening. The ultraviolet light handed certain surface electrons the energy they needed to break away from the metal's ions, and these free charges helped each spark to start.

11.3.2 Hallwachs’ and Lenard’s observations

Wilhelm Hallwachs followed the clue in 1888 with a clean zinc plate fixed to the top of a gold-leaf electroscope (Fig. 11.3). He found three things. A negatively charged plate lost its charge when ultraviolet light fell on it. An uncharged plate became positively charged under the same light. And a plate that was already positive became more positive. Each result says the same thing: ultraviolet light drives negative charge out of zinc.

2/3Electrons leave the plate; the leaves fall together as the charge is lost.

Fig. 11.3Ultraviolet light drives electrons out of zinc: a negative plate discharges and a neutral one turns positive.

A zinc plate sits on the rod of a gold-leaf electroscope inside a glass case, and an ultraviolet lamp is at the top left. Step 1: the lamp is off, the plate is marked with minus signs and the two leaves stand apart, also marked minus. Step 2: ultraviolet rays reach the plate, small electrons fly off it with arrows labelled electrons leave, and the leaves fall together. Step 3: a plate that started uncharged is under the same light; it is now marked with plus signs and the leaves, marked plus, stand apart again.

Philipp Lenard (1862–1947) took the measurements inside a vacuum. He sealed two metal plates in an evacuated glass tube and let ultraviolet light fall on one of them, the emitter, while the other, the collector, was kept positive. A current flowed round the circuit for exactly as long as the light was on. Electrons released at the emitter were drawn across the vacuum to the positive collector, so light falling on a metal could drive a current through an external circuit. Their later measurements asked three questions of this photocurrent: how does it respond to the collector's potential, to the light's frequency, and to its brightness?

One result stood out. For light below a certain frequency no electrons came out at all, however long the plate was lit. This minimum frequency, the threshold frequency (देहली आवृत्ति), is a property of the emitter, and it explains why metals behave so differently. For zinc, cadmium or magnesium it lies in the ultraviolet, so visible light does nothing. For the alkali metals (lithium, sodium, potassium, rubidium, caesium) it lies in the visible, so even ordinary lamplight frees electrons.

Definition
Photoelectric effect

The emission of electrons from a surface, usually a metal, when light of frequency above a threshold value falls on it. Surfaces that show it are called photosensitive; the electrons released are photoelectrons.

Quick check
1 of 3
A negatively charged zinc plate stands on an electroscope, with its leaves apart. Ultraviolet light is shone on the plate. What happens to the leaves?
Show answer

Answer (D) They close, because the plate loses electrons

Ultraviolet light ejects electrons from zinc, so the negative charge drains away and the leaves fall. Light carries no charge of its own; it frees electrons that are already in the metal.

11.4 Experimental study of photoelectric effect

To turn these observations into laws, we need a tube in which every variable can be set and measured. Fig. 11.4 shows one design. The glass bulb is evacuated, so that the electrons do not collide with gas molecules on their way. Inside is a photosensitive emitter C, here a curved sheet, and a collector A, here a rod. The light enters through a quartz window, because ordinary glass absorbs ultraviolet light.

Each quantity we want to vary has its own control. Colour, and so frequency: swap the filter in front of the lamp, or the lamp itself. Brightness: slide the lamp along a bench, since intensity falls off with distance. Potential: a dc supply with a reversing switch (often called a commutator) holds the collector at a chosen potential V relative to the emitter, either positive or negative. Two meters complete the circuit: a voltmeter for V, and a microammeter for the photocurrent, which is rarely more than a few microamperes.

2/3The positive collector attracts the electrons: a photocurrent flows.

Fig. 11.4A phototube: light frees electrons from the emitter, and the collector's potential decides how many of them arrive.NCERT Fig. 11.1 · p. 277

A round evacuated glass bulb holds a curved emitter sheet labelled emitter C on the left and a straight rod labelled collector A near the centre. Light from a lamp on the right passes through a filter and a quartz window in the bulb and falls on the emitter. Below, the emitter and collector are joined through a microammeter and an adjustable supply, with a voltmeter across the supply. Step 1: light falls on the emitter and electrons just leave its surface; the meter reads zero. Step 2: the supply makes the collector positive, arrows show electrons crossing to the collector and the microammeter needle swings. Step 3: the supply is reversed so the collector is negative; most electrons curve back to the emitter, only the fastest one (dashed) reaches the collector, and the needle barely moves.

Four things can now be varied one at a time: the brightness of the light, its frequency, the collector's potential and the emitter's material. The next three subsections take them in turn, holding everything else fixed.

11.4.1 Effect of intensity of light on photocurrent

Keep the collector positive, keep the colour of the light fixed and change only the intensity. A simple way is to move the lamp: intensity falls off as the inverse square of distance, so halving the distance makes the light four times as intense. The microammeter then reads four times as much (Fig. 11.5). Repeat for several distances and the readings fall on a straight line through the origin: the photocurrent is directly proportional to the intensity.

Every electron that reaches the collector carries the same charge e, so the current counts electrons: n electrons per second give I=ne. A current proportional to intensity therefore tells us that the rate at which photoelectrons are released rises in proportion to the intensity of the light.

3/3All the points lie on a straight line through the origin: current ∝ intensity.

Fig. 11.5Halving the distance quadruples the intensity, and the photocurrent quadruples too.NCERT Fig. 11.2 · p. 278

On the left, a phototube and a lamp of fixed colour. On the right, a graph of photocurrent against intensity. Step 1: the lamp is at distance d and the reading is plotted as a point at intensity 1 and current 1. Step 2: the lamp moves to half the distance, the intensity becomes 4 and a second point appears at current 4. Step 3: a straight line through the origin passes through both points, labelled straight line.

11.4.2 Effect of potential on photoelectric current

Now fix both the frequency and the intensity of the light and vary the collector's potential V. Start with the collector positive and raise V step by step. The photocurrent grows, because a stronger attraction gathers electrons that would otherwise have strayed, and then it levels off. Beyond a certain potential, raising V further changes nothing: every electron emitted is already being collected. This largest current is the saturation current (संतृप्त धारा).

Even at V=0 some current flows, because the electrons leave the emitter with kinetic energy of their own and many reach the collector unaided. To find out how much energy they have, reverse the supply so that the collector is negative with respect to the emitter. The field now pushes electrons back towards the emitter, and only those with enough kinetic energy can climb the potential difference to reach the collector. As the collector is made more negative, the current falls quickly, and at one sharply defined value of the retarding potential it becomes zero.

Definition
Stopping potential

For light of a given frequency, the magnitude V0 of the negative (retarding) potential of the collector at which the photocurrent just becomes zero. It is also called the cut-off potential.

Here is how to read it. An electron that travels from emitter to collector against a retarding potential difference V loses kinetic energy eV. The photoelectrons do not all leave with the same energy, so a small retarding potential stops only the slowest. The current vanishes when the retarding potential is just enough to stop the fastest electrons, those with the maximum kinetic energy Kmax⁡. Therefore

Kmax⁡=eV0
(11.1)
where
Kmax⁡
the largest kinetic energy of the photoelectrons
V0
the stopping potential

A stopping potential of 0.8 V means that the fastest photoelectrons leave with 0.8 eV: the electron volt makes this conversion free. Now repeat the whole run with brighter light of the same frequency (Fig. 11.6). The saturation current rises in proportion to the intensity, since more electrons are emitted each second. But the curve reaches zero at exactly the same stopping potential. For a given frequency, the stopping potential does not depend on the intensity: brighter light frees more electrons, not faster ones. The maximum kinetic energy depends on the light's frequency and on the emitter, not on how bright the light is.

3/3Doubling the intensity doubles the saturation current; the cut-off stays at −0.8 V.

Fig. 11.6Brighter light raises the saturation current but leaves the stopping potential unchanged.NCERT Fig. 11.3 · p. 278

Axes show photocurrent against collector potential, from about minus 2 to plus 5 volts. Step 1: for positive potentials the current rises and levels off at a value labelled saturation. Step 2: the curve is continued to negative potentials, where the current falls and reaches zero at minus 0.8 volt, marked with a dot and the label minus 0.8 V. Step 3: a dashed curve for light twice as bright levels off at twice the saturation current but reaches zero at the same minus 0.8 volt.

Example

From stopping potential to speed

Ultraviolet light falls on a clean sodium surface. The photocurrent just stops when the collector is 1.20 V below the emitter. Find the maximum kinetic energy of the photoelectrons, in eV and in joules, and their greatest speed. (me=9.11×10−31 kg)

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

Quick check
1 of 3
Light of fixed frequency falls on an emitter. Its intensity is doubled. What happens?
Show answer

Answer (B) The saturation current doubles; the stopping potential is unchanged

More intense light frees more electrons per second, so the saturation current doubles. The fastest electrons are no faster, so V0 (and Kmax⁡=eV0) stays the same. The idea that V0 should grow with brightness is the wave-picture expectation that experiment rules out.

11.4.3 Effect of frequency of incident radiation on stopping potential

Finally, change the frequency of the light. To compare like with like, adjust the intensity at each frequency until the saturation current is the same. The current–potential curves then differ in only one place: where they meet the axis (Fig. 11.7). The higher the frequency, the larger the stopping potential, so the more negative the collector must be made to stop the current. Higher-frequency light gives faster photoelectrons.

3/3Still higher frequency: the cut-off is at −2.7 V.

Fig. 11.7Light of higher frequency (shorter wavelength) needs a larger retarding potential to stop the current.NCERT Fig. 11.4 · p. 279

Axes show photocurrent against collector potential. Step 1: a curve for 400 nanometre light rises to a saturation level and falls to zero at minus 0.8 volt. Step 2: a dashed curve for 300 nanometre light is added; it has the same saturation level but reaches zero at minus 1.8 volts. Step 3: a dotted curve for 250 nanometre light is added; same saturation level, zero at minus 2.7 volts. A label reads same saturation.

Now plot the stopping potential against the frequency for one emitter (Fig. 11.8). The points fall on a straight line, and the line meets the frequency axis at a definite frequency ν0 where the stopping potential is zero. Below ν0 there is nothing to stop, because no electrons are emitted. Repeat for another metal and you get another straight line, parallel to the first but crossing the axis at a different frequency.

3/4Zinc gives a parallel line with a higher threshold.

Fig. 11.8Stopping potential rises linearly with frequency, along lines of the same slope for every metal.NCERT Fig. 11.5 · p. 279

Axes show stopping potential in volts, from minus 4 to plus 4, against frequency in units of ten to the fourteen hertz, from 0 to 15. Step 1: five measured points for potassium. Step 2: a straight line through them meets the frequency axis at about 5.6, the threshold frequency, marked nu zero. Step 3: a dashed line for zinc, parallel to the first, meets the axis at about 10.4. Step 4: both lines are extended backwards as dotted lines to zero frequency, where they cut the potential axis at minus 2.3 and minus 4.3 volts, labelled minus phi zero over e.

Definition
Threshold frequency

The minimum frequency ν0 of light that can eject photoelectrons from a given surface. Below it there is no emission, no matter how intense the light.

Not every surface is equally willing. Copper gives photoelectrons under ultraviolet light but none at all under green or red light, however bright, while selenium responds more readily than either copper or zinc.

One more fact completes the picture. Whenever the frequency is above threshold, emission begins the moment the light is switched on, with no delay that any experiment has detected, even when the light is extremely dim. Measurements put any delay below about 10−9 s.

Together these experiments give four laws of photoelectric emission, for a given emitter:

  1. Above the threshold frequency, the photocurrent is directly proportional to the intensity of the light (Fig. 11.5).
  2. The saturation current is proportional to the intensity, but the stopping potential, and so Kmax⁡, does not depend on the intensity (Fig. 11.6).
  3. There is a threshold frequency below which no photoelectrons are emitted, however intense the light. Above it, the stopping potential, and so Kmax⁡, increases linearly with frequency (Fig. 11.8).
  4. Emission is instantaneous: it starts within about 10−9 s of the light arriving, even in very dim light.
InteractivePhotoelectric lab

A simulated phototube with an emitter on the left and a collector on the right, lit by a lamp whose colour follows the chosen wavelength. Controls choose the emitter metal (caesium, potassium, zinc or copper), the wavelength from 200 to 700 nanometres, the brightness as photons per second, and the collector potential from minus 5 to plus 5 volts. A Play button sets electrons moving across the tube. Readouts give the photon energy, the frequency, the maximum kinetic energy, the stopping potential and the photocurrent, and a status line explains what is happening. A graph shows photocurrent against collector potential for the present settings, and a second graph collects recorded stopping potentials against frequency and reports the slope and the value of Planck's constant it implies.

Fig. 11.9Choose a metal, set the wavelength, brightness and collector potential, and record stopping potentials to measure Planck's constant.
Quick check
1 of 3
For a given emitter, which change increases the stopping potential?
Show answer

Answer (C) Using light of higher frequency

The stopping potential depends only on the frequency of the light and on the emitter. Brightness and plate area change how many electrons are emitted, not how fast the fastest ones are.

11.5 Photoelectric effect and wave theory of light

By 1900 the wave theory of light seemed beyond doubt. It explained interference, diffraction and polarisation, and Maxwell's theory identified light as an electromagnetic wave. In that picture the energy of a light beam is spread smoothly over each wavefront and flows continuously, like water from a hose rather than like a stream of pebbles. Let us test this picture against the four laws.

A brighter wave has stronger electric and magnetic fields, so it should shake each surface electron harder and give it more energy. The maximum kinetic energy of the photoelectrons ought to grow with intensity. Since the energy simply accumulates, any frequency should work if the light is bright enough or shines for long enough, so there should be no threshold. Both predictions contradict the experiments (laws 2 and 3).

Compare

Wave picture against experiment

What the wave picture predicts

  • Brighter light gives faster electrons
  • Any frequency works if the light is bright enough
  • Dim light: a long wait before emission begins
  • Frequency plays no special role

What experiments show

  • Kmax⁡ does not depend on intensity
  • Below ν0, no emission at any intensity
  • Emission within about 10−9 s, however dim
  • Kmax⁡ rises linearly with frequency

The timing is just as damaging. A wave delivers its energy over the whole illuminated area, shared among a vast number of electrons, so each electron's share arrives slowly. The example below estimates the wait. It comes out at hours for dim light, against the observed 10−9 s. In short, the wave picture cannot account for even the simplest facts of photoemission (Fig. 11.10).

Example

How long would a wave make us wait?

Dim light of intensity 1.0×10−3 W/m2 falls on potassium, whose work function is 2.3 eV. Suppose, as the wave picture allows, that an electron soaks up all the energy falling on a patch the size of an atom, a circle of radius 0.10 nm. How long must it wait to collect 2.3 eV?

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

3/4One photon gives all its energy hν to one electron, which leaves at once.

Fig. 11.10In the wave picture every electron collects energy slowly; in the photon picture one photon gives one electron all its energy at once.

The figure has two rows. Top row, the wave picture: wavefronts arrive over a metal surface on which five cups stand for five electrons, each with a low level of energy. Step 2: the cups have filled to three quarters, labelled dim light: hours. Bottom row, the photon picture. Step 3: a single wavy packet strikes the metal and one electron leaves at once, labelled at once. Step 4: a second packet arrives and a second electron leaves, showing that more photons mean more electrons.

Quick check
1 of 3
Which observation is consistent with the wave picture of light?
Show answer

Answer (A) The photocurrent is proportional to the intensity

A more intense wave brings more energy and could free more electrons, so the current rising with intensity fits both pictures. The other three observations are exactly the ones the wave picture cannot explain.

11.6 Einstein’s photoelectric equation: energy quantum of radiation

In 1905 Albert Einstein proposed a radical way out. In 1900 Max Planck had found that he could explain the radiation from hot bodies only by assuming that energy passed between a hot body and its radiation in steps of hν, where ν is the frequency and h is a new constant of nature, now called Planck's constant. Einstein went further. He suggested that the energy of radiation itself is made of separate lumps, or quanta, each of energy hν. Light is not absorbed as a smooth flow: an electron in the metal takes in one whole quantum, or nothing.

DerivationEinstein’s photoelectric equation

An electron near the surface absorbs one quantum and so gains energy hν. To leave the metal it must spend at least the work function ϕ0. Energy is conserved, so whatever is left over appears as kinetic energy outside the metal.

The electrons that need only ϕ0 to escape are the least tightly bound ones, and they come out with the most kinetic energy:

Einstein's photoelectric equation
Kmax⁡=hν−ϕ0
(11.2)

Electrons from lower energy levels, or that lose energy in collisions on their way to the surface, use up more than ϕ0 and emerge with less than Kmax⁡. This is why the photoelectrons have a spread of energies up to a sharp maximum.

Fig. 11.11 shows the energy budget for potassium, whose work function is 2.30 eV. In this picture, for light of one fixed frequency, intensity just counts quanta: how many arrive on each square metre every second. Making the light brighter sends more quanta; it does not make any one quantum bigger.

3/3A 405 nm photon brings 3.06 eV: the electron leaves with up to 0.76 eV.

Fig. 11.11A photon frees an electron only if hν≥ϕ0; whatever is left over is the electron's kinetic energy.

A band at the bottom stands for the most energetic electrons in potassium, and a dashed line 2.30 electronvolts above it is labelled escape level, with a double arrow marked phi zero. Step 1: a vermillion arrow for a 650 nanometre photon of 1.91 electronvolts rises from the band but stops below the escape level. Step 2: a green arrow for a 539 nanometre photon of 2.30 electronvolts just reaches the escape level. Step 3: a purple arrow for a 405 nanometre photon of 3.06 electronvolts passes the escape level; the part above it, 0.76 electronvolt, is shaded and labelled K max.

Now check the equation against each of the four laws.

  1. Frequency, not intensity, sets Kmax⁡. Each emission is one quantum absorbed by one electron, so by Eq. (11.2) Kmax⁡ depends on ν and on the metal, rising linearly with ν. How many quanta arrive per second, which is what intensity measures, does not enter.
  2. There is a threshold. Kinetic energy cannot be negative, so emission needs hν≥ϕ0, that is ν≥ν0 with ν0 given by Eq. (11.3) below. A metal with a larger work function needs a higher threshold frequency. Below ν0 no single quantum is big enough, and the quanta cannot pool their energy, so turning up the brightness, or waiting, achieves nothing.
  3. Current is proportional to intensity. More quanta per second means more electrons absorbing one each, so more photoelectrons per second (for ν>ν0).
  4. No delay. Absorbing a quantum is a single, instantaneous event. Dim light means fewer such events per second, not slower ones, so the first electrons appear at once.
threshold frequency
ν0=ϕ0h
(11.3)

Combining Eq. (11.2) with Kmax⁡=eV0 from Eq. (11.1) gives a prediction that can be tested with a voltmeter, valid for ν≥ν0:

V0=he ν−ϕ0e
(11.4)
where
h/e
the slope, the same for every metal
ϕ0/e
the intercept, which depends on the metal

This is a strong prediction. A graph of V0 against ν must be a straight line, its slope must be h/e for every metal, it must cross the frequency axis at ν0, and extended back to ν=0 it must cut the potential axis at −ϕ0/e. Look again at the last step of Fig. 11.8: that is exactly what is observed.

The sternest test came from someone who expected the equation to fail. Robert Millikan spent ten years, 1906 to 1916, on it, building an apparatus that shaved fresh, clean surfaces of alkali metals inside a vacuum. To his surprise the graphs came out straight and parallel, exactly as Eq. (11.4) demands. Their slope multiplied by e gave h; for sodium it matched the constant Planck had extracted from hot-body radiation, an entirely different phenomenon. Today's value is

h=6.626×10−34 J s

Two independent routes to the same h, and work functions that matched those from other methods, left little room for doubt. Einstein's quanta were accepted, though many physicists stayed uneasy about what they meant for the wave theory.

Example

Measuring h with a mercury lamp

In a school laboratory, filters pick out two lines of a mercury lamp to light a phototube. Light of wavelength 365 nm gives a stopping potential of 1.40 V; light of 546 nm gives 0.27 V. Find Planck's constant, the work function of the emitter and its threshold wavelength. Take c=3.00×108 m/s and e=1.602×10−19 C.

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

Quick check
1 of 3
Photons of energy 4.5 eV fall on a surface whose work function is 2.0 eV. What is the stopping potential, in volts?
V
Show answer

Answer 2.5 V

Kmax⁡=4.5−2.0=2.5 eV, and V0=Kmax⁡/e=2.5 V.

11.7 Particle nature of light: the photon

Energy delivered in lumps of hν does not by itself make light a stream of particles. What would? A cricket ball carries momentum as well as energy and hands both over when it hits something. Maxwell's theory already says that an electromagnetic wave carrying energy E also carries momentum E/c (Part I, Chapter 8). Einstein applied this to a single quantum: each one must carry momentum hν/c. In terms of wavelength,

E=hν=hcλ,p=Ec=hνc=hλ

With a definite energy and a definite momentum, each quantum behaves like a particle; the name photon for it was coined in 1926. The clinching evidence came in the early 1920s from Arthur Compton, who scattered X-rays off electrons. The scattered X-rays had longer wavelengths, by exactly the amount expected if each photon had collided with an electron like one billiard ball with another, sharing energy and momentum. Einstein received the 1921 Nobel Prize, Millikan the 1923 prize (for the electron's charge and for his photoelectric work), and Compton the 1927 prize.

Here is what the photon picture says, property by property.

  • Energy and momentum. A photon of frequency ν carries E=hν and p=hν/c=h/λ, and in vacuum it always moves at c.
  • Identical photons. Light of a single frequency consists of identical photons. Brightness tells you how many cross each square metre per second, not how big each one is.
  • No charge. Photons are electrically neutral, so neither electric nor magnetic fields can bend their paths.
  • Collisions. When a photon meets an electron or another particle, energy and momentum are both conserved, but the photon count is not: photons can be absorbed, and new ones created.
  • Where it applies. These particle properties show up when light is emitted, absorbed or scattered by matter; while it travels, interference still needs the wave description.

2/3Bright red light: many more photons, but each is the same red photon.

Fig. 11.12Brightness sets how many photons arrive each second; frequency sets how much energy each one carries.

A lamp on the left sends wave packets towards a detector on the right. Step 1: dim red light, only three packets, labelled few photons. Step 2: bright red light, nine packets of exactly the same size and wavelength, labelled many photons. Step 3: dim blue light, three packets with a visibly shorter wavelength, labelled few, bigger photons. A formula below reads E equals h nu, p equals h over lambda.

Example 11.1NCERT Example 11.1 · p. 284

Photons from a water purifier

The ultraviolet lamp in a home water purifier gives out 2.5 W of radiation at a wavelength of 254 nm. (a) What is the energy of one photon, in joules and in electron volts? (b) How many photons does the lamp emit each second? (c) What is the momentum of each photon?

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

Example 11.2NCERT Example 11.2 · p. 284

Will red light do?

The work function of potassium is 2.30 eV. (a) Find its threshold frequency and threshold wavelength. (b) Light from a 405 nm violet laser falls on it. Find the maximum kinetic energy of the photoelectrons, the stopping potential and the greatest speed of the electrons. (c) Would light of 650 nm eject any electrons? (h=6.626×10−34 J s, c=3.00×108 m/s, me=9.11×10−31 kg)

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

Quick check
1 of 4
What is the energy, in electron volts, of a photon of orange light of wavelength 620 nm? (hc=1240 eV nm)
eV
Show answer

Answer 2 eV

E=hc/λ=1240/620=2.00 eV.

11.8 Wave nature of matter

Look back over this chapter and Wave Optics, and light seems to change character with the experiment. Fringes and diffraction patterns can only be built from waves; the photoelectric effect and Compton scattering, in which light hands over energy and momentum, can only be explained with photons. Often one device needs both descriptions at once. In a phone camera the lens forms the image, and how sharp that image can be is limited by diffraction, a wave effect. Each pixel of the sensor, though, works on the photoelectric idea: arriving photons free electrons inside the silicon, at most one electron per photon, and the camera counts them (Fig. 11.13).

2/2In each pixel, an arriving photon frees an electron; the camera reads out the count.

Fig. 11.13A camera uses both pictures: the lens forms the image as a wave, and the sensor counts photons through the electrons they free.

On the left, a camera body with a lens at the front and a sensor chip at the back. Step 1: three rays enter from the left and the lens bends them so that they meet at a point on the sensor. Step 2: a magnified circle of the sensor shows a grid of nine pixels; two wave packets, photons, arrive at two pixels, and each of those pixels now holds a freed electron, drawn as a dot, ready to be counted.

Louis de Broglie turned the argument round in 1924. Light, long called a wave, had turned out to carry momentum in particle-sized parcels; perhaps electrons, long called particles, carry a wave. His reason was a belief in symmetry. Matter and radiation are the two great ingredients of the universe, and it seemed unlikely that only one of them should lead a double life. He proposed that any particle with momentum p is accompanied by a wave of wavelength

de Broglie relation
λ=hp=hmv
(11.5)
where
m
the mass of the particle
v
its speed

Equation (11.5) is the de Broglie relation, and λ is the particle's de Broglie wavelength (दे ब्रॉग्ली तरंगदैर्ध्य), the wavelength of its matter wave. Read the two sides separately and the duality is in plain view: λ is something only a wave can have, p something we picture as belonging to a moving lump, and the tiny constant h converts one into the other.

For electrons and atoms, the relation was in 1924 a bold guess waiting for experiment. For light it can be checked at once. A photon of frequency ν has momentum

p=hνc
(11.6)

so de Broglie's recipe gives it the wavelength

λ=hp=cν
(11.7)

nothing other than the familiar wavelength of the light itself. The new rule contains the old one.

Why do we never notice the waves of moving objects? Because h is so small, the wavelength is minute unless the mass and speed are tiny too. A raindrop of mass 4 mg falling at 9 m/s has λ=h/mv≈1.8×10−29 m, some 1014 times smaller than a proton. No slit or grating could ever reveal it. An electron is a different story: its mass is so small that its de Broglie wavelength is comparable with the spacing of atoms in a crystal, and then the wave behaviour is real and measurable (Fig. 11.14). Note too that λ depends only on momentum, not on the charge or the kind of particle.

4/4That is the size of atoms, so crystals can diffract electrons.

Fig. 11.14Matter waves are far too short to notice for everyday objects; for electrons they are about the size of atoms.

A horizontal logarithmic axis of wavelength runs from ten to the minus 38 to ten to the minus 4 metres, with tick labels every six powers of ten. Step 1: markers for a sprinter, near ten to the minus 36 metres, and a falling raindrop, near ten to the minus 29 metres. Step 2: a pollen grain drifting slowly, near ten to the minus 19 metres. Step 3: an electron accelerated through 150 volts, at ten to the minus 10 metres. Step 4: a shaded band for the size of atoms covers the electron's marker, and a narrow band for visible light lies further right.

DerivationWavelength of an electron accelerated through a potential difference

An electron starting from rest and accelerated through a potential difference V gains kinetic energy K=eV. Since K=p2/2m, its momentum is p=2meV, and Eq. (11.5) gives

λ=h2meV

Putting in h, m and e for an electron, with V in volts,

λ=1.227V nm

For V=150 V this gives λ=0.100 nm, about the distance between neighbouring atoms in a crystal. (The formula ignores relativity, which is fine up to a few thousand volts.)

Example 11.3NCERT Example 11.3 · p. 286

Three de Broglie wavelengths

Find the de Broglie wavelength of (a) an electron moving at 1.2×107 m/s, (b) an oxygen molecule (m=5.3×10−26 kg) moving at 480 m/s, a typical speed in air, and (c) a 70 kg sprinter running at 10 m/s. (h=6.626×10−34 J s, me=9.11×10−31 kg)

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

Example

Electrons for a diffraction experiment

Electrons are to be diffracted by a crystal whose atomic planes are 0.20 nm apart, so their de Broglie wavelength should also be about 0.20 nm. Through what potential difference should they be accelerated from rest?

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

Quick check
1 of 4
An electron is accelerated from rest through 600 V. What is its de Broglie wavelength, in nanometres?
nm
Show answer

Answer 0.0501 nm

λ=1.227/600 nm=0.050 nm.
Recap

Dual Nature of Radiation and Matter: key points and summary

  1. Cathode rays are streams of electrons. Crossed fields gave their specific charge, e/m=1.76×1011 C/kg, the same for every source; Millikan's oil drops gave e=1.602×10−19 C (Section 11.1).
  2. The work function ϕ0 is the least energy that frees an electron from a metal surface. Heating, a strong field or light can supply it; 1 eV=1.602×10−19 J (Section 11.2).
  3. In the photoelectric effect, light of high enough frequency ejects electrons from a surface. Some metals (zinc, copper) need ultraviolet light; the alkali metals respond to visible light (Section 11.3).
  4. Photocurrent is proportional to the intensity of the light, because the number of electrons emitted per second is (Section 11.4.1).
  5. Raising the collector's potential increases the current up to a saturation value. A retarding potential reduces it to zero at the stopping potential V0, with eV0=Kmax⁡; V0 does not depend on intensity (Section 11.4.2).
  6. V0 rises linearly with frequency; below the threshold frequency ν0 there is no emission at any intensity, and above it emission is instantaneous (Section 11.4.3).
  7. The wave picture of light predicts that Kmax⁡ grows with intensity, that there is no threshold, and that dim light causes a delay. All three are wrong (Section 11.5).
  8. Einstein: light energy comes in quanta hν, and one electron absorbs one quantum, so Kmax⁡=hν−ϕ0 and ν0=ϕ0/h (Section 11.6).
  9. Hence V0=(h/e)ν−ϕ0/e: a straight line of slope h/e for every metal. Millikan's measurements confirmed it and gave h=6.626×10−34 J s.
  10. A photon has energy E=hν and momentum p=hν/c=h/λ, carries no charge and travels at c. Intensity sets the number of photons, not their energy (Section 11.7).
  11. Neither picture of light is complete on its own: fringes need waves, photoemission needs photons, and a device such as a camera uses both (Section 11.8).
  12. De Broglie: a particle of momentum p has a wavelength λ=h/p=h/mv, independent of its charge. It is measurable for electrons (about the spacing of atoms) and far too small for everyday objects (Section 11.8).
Table 11.3Quantities met in this chapter
QuantitySymbolSI unitDimensionsWhere it appears
Planck's constanthJ s[ML2T−1]E=hν, p=h/λ
Work functionϕ0J (often eV)[ML2T−2]Kmax⁡=hν−ϕ0
Threshold frequencyν0Hz[T−1]ν0=ϕ0/h
Stopping potentialV0V[ML2T−3A−1]eV0=Kmax⁡
Photon momentumpkg m/s[MLT−1]p=hν/c
de Broglie wavelengthλm[L]λ=h/p
Flash cards · 9

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Practice · 15 questions

Practice questions with answers (NCERT “Exercises”)

  1. NumericalEasyNCERT Exercises 11.1

    A dental X-ray machine accelerates electrons through 65 kV before they strike a tungsten target. An electron that is stopped in a single collision can give all its kinetic energy to one X-ray photon.

    1. What is the highest frequency in the X-rays produced?

      Hz
      Show solution
      hνmax⁡=eV  ⇒  νmax⁡=eVh=(1.602×10−19 C)(6.5×104 V)6.626×10−34 J s=1.57×1019 Hz
    2. What is the shortest wavelength, in nanometres?

      nm
      Show solution
      λmin⁡=cνmax⁡=3.00×108 m/s1.57×1019 Hz=1.91×10−11 m=0.0191 nm

      Equivalently, λmin⁡=1240 eV nm/65 000 eV=0.0191 nm.

  2. NumericalMediumNCERT Exercises 11.2

    Lithium has a work function of 2.90 eV. Ultraviolet light of wavelength 350 nm falls on a lithium surface.

    1. Find the photoelectrons' maximum kinetic energy, in eV.

      eV
      Show solution
      hν=hcλ=1240 eV nm350 nm=3.55 eV
      Kmax⁡=hν−ϕ0=3.55 eV−2.90 eV=0.65 eV
    2. What is the stopping potential?

      V
      Show solution

      eV0=Kmax⁡, so with Kmax⁡ in eV the stopping potential in volts has the same number: V0=0.65 V.

    3. What is the greatest speed of the photoelectrons?

      m/s
      Show solution
      vmax⁡=2Kmax⁡me=2×0.645×1.602×10−19 J9.11×10−31 kg=4.8×105 m/s
  3. NumericalEasyNCERT Exercises 11.3

    In a photoelectric experiment the current falls to zero when the collector is 0.85 V negative with respect to the emitter.

    1. Find the photoelectrons' maximum kinetic energy, in joules.

      J
      Show solution
      Kmax⁡=eV0=(1.602×10−19 C)(0.85 V)=1.36×10−19 J=0.85 eV
    2. How fast are the fastest photoelectrons moving?

      m/s
      Show solution
      vmax⁡=2×1.36×10−19 J9.11×10−31 kg=5.5×105 m/s
  4. NumericalMediumNCERT Exercises 11.4

    A laser diode in an optical mouse emits 4.5 mW of infrared light of wavelength 780 nm.

    1. Find the energy (in joules and eV) and the momentum of each photon.

      Show solution

      AnswerE=2.55×10−19 J=1.59 eV; p=8.49×10−28 kg m/s.

      E=hcλ=(6.626×10−34)(3.00×108)780×10−9 J=2.55×10−19 J=1.59 eV
      p=hλ=6.626×10−34 J s780×10−9 m=8.49×10−28 kg m/s
    2. How many photons leave the diode each second, on average?

      photons per second
      Show solution
      N=PE=4.5×10−3 W2.55×10−19 J=1.8×1016 per second
    3. How fast would a neutron (m=1.675×10−27 kg) have to move to have the same momentum as one of these photons?

      m/s
      Show solution
      v=pm=8.49×10−28 kg m/s1.675×10−27 kg=0.51 m/s

      Slower than walking pace: a single photon's momentum is tiny.

  5. NumericalMediumNCERT Exercises 11.5

    A student measures stopping potentials for one emitter: 0.17 V at 5.5×1014 Hz, 0.79 V at 7.0×1014 Hz, 1.41 V at 8.5×1014 Hz and 2.03 V at 10.0×1014 Hz.

    1. What is the slope of the graph of V0 against ν, in units of 10−15 V s?

      × 10⁻¹⁵ V s
      Show solution

      The points rise by 0.62 V for every 1.5×1014 Hz, so they lie on a straight line. Using the end points:

      slope=2.03 V−0.17 V10.0×1014 Hz−5.5×1014 Hz=4.13×10−15 V s
    2. What value of Planck's constant does this give?

      J s
      Show solution
      h=e×slope=(1.602×10−19 C)(4.13×10−15 V s)=6.62×10−34 J s
    3. Find the threshold frequency and the work function of the emitter.

      Show solution

      Answerν0≈5.1×1014 Hz; ϕ0≈2.1 eV.

      The line falls to V0=0 at 0.17 V below the first point:

      ν0=5.5×1014 Hz−0.17 V4.13×10−15 V s=5.1×1014 Hz
      ϕ0=hν0=(6.62×10−34)(5.09×1014)1.602×10−19 eV=2.1 eV
  6. NumericalEasyNCERT Exercises 11.6

    The threshold wavelength of a photosensitive surface is 620 nm. Violet light of wavelength 410 nm falls on it. What stopping potential is needed?

    V
    Show solution
    ϕ0=hcλ0=1240 eV nm620 nm=2.00 eV,hν=1240 eV nm410 nm=3.02 eV
    eV0=hν−ϕ0=3.02 eV−2.00 eV=1.02 eV  ⇒  V0≈1.0 V
  7. Short answerEasyNCERT Exercises 11.7

    Tungsten has a work function of 4.5 eV. A lamp gives ultraviolet light of wavelength 280 nm. Will it eject photoelectrons from a tungsten surface? Explain.

    Show solution

    AnswerNo: each photon has only 4.43 eV, just under the 4.5 eV work function.

    hν=hcλ=1240 eV nm280 nm=4.43 eV<4.5 eV

    No single photon has enough energy, so no electrons are emitted, however intense the lamp. The threshold wavelength for tungsten is 1240/4.5=276 nm; only shorter wavelengths work.

  8. NumericalMediumNCERT Exercises 11.8

    Light of wavelength 365 nm from an ultraviolet LED falls on a metal, and the fastest photoelectrons leave at 4.4×105 m/s. Find the threshold frequency of the metal.

    Hz
    Show solution
    ν=cλ=3.00×108 m/s365×10−9 m=8.22×1014 Hz
    Kmax⁡=12mev2=12(9.11×10−31 kg)(4.4×105 m/s)2=8.82×10−20 J

    By Eq. (11.2), hν0=ϕ0=hν−Kmax⁡, so

    ν0=ν−Kmax⁡h=8.22×1014 Hz−8.82×10−20 J6.626×10−34 J s=6.9×1014 Hz
  9. NumericalEasyNCERT Exercises 11.9

    A 445 nm blue laser shines on the emitter of a phototube, and the stopping potential is 0.62 V. What is the work function of the emitter?

    eV
    Show solution
    hν=1240 eV nm445 nm=2.79 eV
    ϕ0=hν−eV0=2.79 eV−0.62 eV=2.17 eV

    This is close to the work function of caesium.

  10. NumericalEasyNCERT Exercises 11.10

    Find the de Broglie wavelength of each of these.

    1. A 1500 kg car moving at 20 m/s.

      m
      Show solution
      λ=hmv=6.626×10−34 J s(1500 kg)(20 m/s)=2.2×10−38 m
    2. A mosquito of mass 2.5 mg flying at 0.50 m/s.

      m
      Show solution
      λ=6.626×10−34 J s(2.5×10−6 kg)(0.50 m/s)=5.3×10−28 m
    3. A smoke particle of mass 1.0×10−15 kg drifting at 1.0 mm/s.

      m
      Show solution
      λ=6.626×10−34 J s(1.0×10−15 kg)(1.0×10−3 m/s)=6.6×10−16 m

      Even this, the largest of the three, is smaller than an atomic nucleus, far too small to observe.

  11. ProofMediumNCERT Exercises 11.11

    Starting from E=hν for a photon and the fact that light carries momentum p=E/c, show that h/p for a photon equals the wavelength of the light. Then use the result to find the momentum of a photon of wavelength 600 nm.

    Show solution

    Answerh/p=λ; p=1.10×10−27 kg m/s at 600 nm.

    p=Ec=hνc  ⇒  hp=cν=λ

    So the de Broglie wavelength of a photon is the wavelength of the radiation it belongs to.

    p=hλ=6.626×10−34 J s600×10−9 m=1.10×10−27 kg m/s
  12. Quick answerEasy

    Light of a single frequency, above the threshold, falls on a metal. Which change will increase the number of photoelectrons emitted per second?

    Show answer

    Answer (A) Increasing the intensity of the light

    More intense light brings more photons per second, so more electrons are freed. A more negative collector does not change emission (it only turns electrons back), and the other two reduce the number emitted.
  13. Quick answerMedium

    Photons of energy 1.0 eV have a wavelength of nm (to the nearest 10 nm).

    Show answer

    Answer 1240

    λ=hc/E=1240 eV nm/1.0 eV=1240 nm, in the infrared.
  14. Quick answerMedium

    An electron and a proton have the same kinetic energy. About how many times longer is the electron's de Broglie wavelength?

    times
    Show answer

    Answer 42.85 times

    λe/λp=mp/me=1836≈43.
  15. Quick answerEasy

    Match each observation with the idea that explains it.

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

    Show answer
    Answer
    • Threshold frequency : one photon must supply at least ϕ0
    • No time lag : a photon is absorbed in a single event
    • Current proportional to intensity : more photons free more electrons
    • Electron diffraction by crystals : electrons have a wavelength h/p
    Each photoelectric law follows from one-photon, one-electron absorption; electron diffraction is evidence for de Broglie's matter waves.

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 11.1Practice Q1
NCERT Exercises 11.2Practice Q2
NCERT Exercises 11.3Practice Q3
NCERT Exercises 11.4Practice Q4
NCERT Exercises 11.5Practice Q5
NCERT Exercises 11.6Practice Q6
NCERT Exercises 11.7Practice Q7
NCERT Exercises 11.8Practice Q8
NCERT Exercises 11.9Practice Q9
NCERT Exercises 11.10Practice Q10
NCERT Exercises 11.11Practice Q11

Quick check: MCQs on Dual Nature of Radiation and Matter

  1. 11.1 Introduction3 questions
  2. 11.2 Electron emission3 questions
  3. 11.3.2 Hallwachs’ and Lenard’s observations3 questions
  4. 11.4.2 Effect of potential on photoelectric current3 questions
  5. 11.4.3 Effect of frequency of incident radiation on stopping potential3 questions
  6. 11.5 Photoelectric effect and wave theory of light3 questions
  7. 11.6 Einstein’s photoelectric equation: energy quantum of radiation3 questions
  8. 11.7 Particle nature of light: the photon4 questions
  9. 11.8 Wave nature of matter4 questions
Glossary · 15

Key terms (English–Hindi)

cathode raysकैथोड किरणेंkaithod kiranen
Streams of electrons that leave the cathode of a gas-discharge tube at very low pressure.First seen in the chapter →
de Broglie wavelengthदे ब्रॉग्ली तरंगदैर्ध्यde broglie tarangdairghya
The wavelength λ=h/p of the wave associated with a moving particle of momentum p.First seen in the chapter →
dual natureद्वैत प्रकृतिdvait prakriti
The fact that radiation and matter both show wave-like and particle-like behaviour, depending on the experiment.First seen in the chapter →
electron volt (eV)इलेक्ट्रॉन वोल्टelectron volt
The energy an electron gains when accelerated through 1 V: 1 eV=1.602×10−19 J.First seen in the chapter →
field emissionक्षेत्र उत्सर्जनkshetra utsarjan
Release of electrons from a metal by a very strong electric field, of order 108 V/m.First seen in the chapter →
photoelectric effectप्रकाश-विद्युत प्रभावprakash-vidyut prabhav
Emission of electrons from a surface when light of high enough frequency falls on it.
photoelectronप्रकाशिक इलेक्ट्रॉनprakashik electron
An electron set free from a surface by light.First seen in the chapter →
photonफ़ोटॉनphoton
A quantum of electromagnetic radiation, with energy hν and momentum h/λ, travelling at c.First seen in the chapter →
Planck's constantप्लांक नियतांकplank niyatank
The constant h=6.626×10−34 J s linking a quantum's energy to its frequency.First seen in the chapter →
saturation currentसंतृप्त धाराsantript dhara
The largest photocurrent at a given light setting, reached when every emitted electron reaches the collector.First seen in the chapter →
specific chargeविशिष्ट आवेशvishisht avesh
The charge of a particle divided by its mass; for the electron e/m=1.76×1011 C/kg.First seen in the chapter →
stopping potentialनिरोधी विभवnirodhi vibhav
The retarding potential V0 of the collector at which the photocurrent just becomes zero; eV0=Kmax⁡.
thermionic emissionतापायनिक उत्सर्जनtapayanik utsarjan
Release of electrons from a metal heated strongly enough that some free electrons gain energy above the work function.First seen in the chapter →
threshold frequencyदेहली आवृत्तिdehali avritti
The lowest frequency ν0=ϕ0/h that can eject electrons from a given surface, however intense the light.First seen in the chapter →
work functionकार्य फलनkarya phalan
The least energy needed to free an electron from the surface of a metal, ϕ0; it depends on the metal and its surface.First seen in the chapter →