NEET Physics: Modern Physics and Electronics · Chapter 1

Dual Nature of Radiation and Matter

What to remember

Photoelectric emission needs light of frequency above the threshold frequency ν₀; Kmax = hν − φ₀ and eV₀ = Kmax. Intensity changes the number of photoelectrons, never Kmax.
A photon has energy E = hν = hc/λ and momentum p = h/λ = E/c; light shows particle nature in emission, absorption and collision.
Moving matter has a wave nature: λ = h/p = h/√(2mK); for an electron accelerated through V volts, λ ≈ 1.227/√V nm.
Graph of stopping potential against frequency is a straight line with slope h/e and a frequency intercept ν₀ = φ₀/h, the same slope for every metal.

Electron emission and work function

Electrons in a metal are free to move inside it, but they cannot leave the surface freely. A minimum energy is needed to take an electron out of the metal. This minimum energy is called the work function, written φ₀. It depends on the metal and on the state of its surface. It is measured in electron-volt (eV).

1 eV is the energy gained by one electron when it moves through a potential difference of 1 V. 1 eV = 1.6 × 10⁻¹⁹ J.

There are three ways to give an electron the energy to leave a metal.

Process	Energy source	Typical condition
Thermionic emission	Heat	Metal heated to a high temperature
Field emission	Very strong electric field	Field of about 10⁸ V m⁻¹ pulls electrons out
Photoelectric emission	Light of suitable frequency	Ultraviolet or visible light falls on the metal

Approximate work functions of some metals:

Metal	Work function φ₀ (eV)
Caesium	2.14
Sodium	2.75
Aluminium	4.28
Copper	4.65
Platinum	5.65

Alkali metals such as caesium and sodium have low work functions, so they respond even to visible light. Metals with high work functions need ultraviolet light.

Photoelectric effect: observations

Hertz observed that a spark between two electrodes at high voltage became easier when ultraviolet light fell on the negative electrode. Hallwachs and Lenard studied this further. A negatively charged zinc plate lost its charge when ultraviolet light fell on it. The light was knocking electrons out of the plate. The emission of electrons from a metal surface when light of suitable frequency falls on it is called the photoelectric effect. The emitted electrons are called photoelectrons.

Experimental set-up (in words). Two metal plates are placed in an evacuated glass tube. Light falls on the emitter plate. The other plate is the collector. A variable voltage source is connected between them, and a microammeter measures the current. The voltage can be made positive or negative for the collector with respect to the emitter.

Effect of intensity. At a fixed frequency and a fixed collector voltage, the photocurrent is directly proportional to the intensity of the light. More intensity means more photons per second, so more photoelectrons per second.

Effect of collector potential. When the collector is made more positive, the current rises and then becomes constant. This constant value is the saturation current. It is reached when all emitted electrons reach the collector. When the collector is made negative with respect to the emitter, the current falls. At one particular negative value the current becomes zero. This value is the stopping potential V₀. The graph of current against collector potential rises from the negative side, becomes zero at −V₀ and flattens at the saturation value. If the intensity is doubled, the saturation current doubles, but V₀ stays the same.

The maximum kinetic energy of photoelectrons is

Kmax = eV₀.

Effect of frequency. For a given metal, V₀ increases linearly with the frequency of the light. Kmax therefore increases linearly with frequency. It does not depend on intensity. The graph of V₀ against ν is a straight line:

Feature of the V₀ – ν graph	Meaning
Slope	h/e, the same for all metals
Intercept on the frequency axis	Threshold frequency ν₀ = φ₀/h
Intercept on the V₀ axis	−φ₀/e
Different metals	Parallel straight lines with different ν₀

Threshold frequency. For every metal there is a minimum frequency ν₀. Below ν₀ no photoelectrons are emitted, however large the intensity or the time of exposure. Above ν₀ emission starts at once. The delay is about 10⁻⁹ s or less, and does not depend on intensity.

Why wave theory fails. A wave picture would predict that a more intense light gives electrons more energy, that any frequency should work if the intensity is high enough, and that low-intensity light should cause a time lag while energy collects. None of these is seen. The observations show that Kmax depends on frequency, that a threshold frequency exists, and that emission is instantaneous.

Einstein's equation and the photon

Light travels as small packets of energy called photons (quanta). Each photon has energy E = hν, where h = 6.63 × 10⁻³⁴ J s is Planck's constant. The energy of a photon depends only on the frequency. Intensity is the number of photons crossing unit area per second, multiplied by the energy of each photon.

Properties of a photon

Energy E = hν = hc/λ.
Momentum p = E/c = h/λ.
Rest mass is zero. It travels at the speed of light c = 3 × 10⁸ m s⁻¹ in vacuum.
It is electrically neutral and is not deflected by electric or magnetic fields.
In a collision with an electron, energy and momentum are conserved, as for particles.
Photon energy does not change with the intensity of the light.

A convenient value: hc ≈ 1240 eV nm. So the energy of a photon in eV is 1240 divided by its wavelength in nm.

Einstein's photoelectric equation. One photon gives all its energy to one electron. Part of this energy, φ₀, is used to leave the metal. The rest becomes kinetic energy. For the fastest electrons (those that lose no energy inside the metal):

hν = φ₀ + Kmax, that is, Kmax = hν − φ₀.

Using Kmax = eV₀ we get eV₀ = hν − φ₀. At the threshold, Kmax = 0, so hν₀ = φ₀ and ν₀ = φ₀/h. The threshold wavelength is λ₀ = hc/φ₀.

The equation explains all observations:

Observation	Explanation
Kmax depends on frequency only	Each electron takes one photon; photon energy is hν
Threshold frequency	If hν < φ₀, no electron can leave
Current proportional to intensity	More photons per second, more electrons per second
No time lag	Energy is transferred in a single event
Slope of V₀ – ν line is h/e	V₀ = (h/e)ν − φ₀/e

Photon flux. The number of photons per second in a beam of power P is N = P/(hν) = Pλ/(hc). It is often called photon flux when counted per unit area per second. A source of 1 W at 500 nm emits about 2.5 × 10¹⁸ photons per second.

Wave nature of matter

Light shows both wave and particle behaviour. De Broglie proposed that moving particles of matter also have a wave nature. A particle of momentum p has an associated wavelength

λ = h/p = h/(mv).

For a particle of kinetic energy K, p = √(2mK), so λ = h/√(2mK).

For a particle of charge q accelerated from rest through a potential difference V, K = qV and λ = h/√(2mqV).

For an electron accelerated through V volts, with h = 6.63 × 10⁻³⁴ J s, m = 9.1 × 10⁻³¹ kg and e = 1.6 × 10⁻¹⁹ C:

λ ≈ 1.227/√V nm.

So for V = 100 V, λ ≈ 0.1227 nm, which is comparable to the spacing between atoms in crystals. Heavier particles have a shorter wavelength for the same speed. A cricket ball of mass 0.1 kg moving at 10 m s⁻¹ has λ = 6.63 × 10⁻³⁴ m, far too small to observe. This is why wave nature is seen only for very small particles such as electrons. Experiments in which electrons are diffracted by crystals confirmed that matter has a wave nature.

Case	Wavelength relation
Any particle of momentum p	λ = h/p
Particle of kinetic energy K	λ = h/√(2mK)
Particle of charge q through potential V	λ = h/√(2mqV)
Electron through V volts	λ ≈ 1.227/√V nm
Photon	λ = h/p = c/ν

For the same accelerating potential, λ ∝ 1/√(mq). A proton and an alpha particle (mass 4 times, charge 2 times) give λp : λα = 2√2 : 1.

Worked examples

Example 1 (photon energy and threshold). A metal has φ₀ = 2.0 eV. Find the threshold wavelength and Kmax for light of 400 nm. Use hc = 1240 eV nm.

Threshold: λ₀ = 1240/2.0 = 620 nm.

Photon energy at 400 nm: 1240/400 = 3.1 eV.

Kmax = 3.1 − 2.0 = 1.1 eV, so V₀ = 1.1 V.

Example 2 (photon flux). Monochromatic light of frequency 5 × 10¹⁴ Hz has power 3.315 W. Find photons per second. Use h = 6.63 × 10⁻³⁴ J s.

Energy of one photon = 6.63 × 10⁻³⁴ × 5 × 10¹⁴ = 3.315 × 10⁻¹⁹ J.

N = 3.315 / 3.315 × 10⁻¹⁹ = 1 × 10¹⁹ photons per second.

Example 3 (de Broglie wavelength). Find the wavelength of an electron accelerated through 150 V.

λ = 1.227/√150 nm = 1.227/12.25 nm ≈ 0.100 nm.

Check: K = 150 eV = 150 × 1.6 × 10⁻¹⁹ = 2.4 × 10⁻¹⁷ J. p = √(2 × 9.1 × 10⁻³¹ × 2.4 × 10⁻¹⁷) = 6.6 × 10⁻²⁴ kg m s⁻¹. λ = 6.63 × 10⁻³⁴/6.6 × 10⁻²⁴ ≈ 1.0 × 10⁻¹⁰ m.

Example 4 (photon momentum). A photon has λ = 663 nm. p = 6.63 × 10⁻³⁴ / 6.63 × 10⁻⁷ = 1 × 10⁻²⁷ kg m s⁻¹. If 10²⁰ such photons per second are fully absorbed by a surface, the momentum given to it each second is 10²⁰ × 10⁻²⁷ = 10⁻⁷ kg m s⁻², that is, a force of 10⁻⁷ N.

Example 5 (two-step problem). A photon of 5 eV gives a stopping potential of 3 V. Then Kmax = 3 eV, φ₀ = 5 − 3 = 2 eV and the threshold wavelength is 1240/2 = 620 nm.

Formula summary

Quantity	Formula	Symbols and units
Photon energy	E = hν = hc/λ	h in J s, ν in Hz, λ in m
Photon momentum	p = h/λ = E/c	kg m s⁻¹
Einstein's equation	Kmax = hν − φ₀	φ₀ work function in J or eV
Stopping potential	eV₀ = Kmax	V₀ in volt
Threshold	ν₀ = φ₀/h; λ₀ = hc/φ₀	Hz; m
Photons per second	N = P/(hν) = Pλ/(hc)	P in W
de Broglie wavelength	λ = h/p = h/√(2mK)	m
Electron through V	λ ≈ 1.227/√V nm	V in volt
Energy conversion	1 eV = 1.6 × 10⁻¹⁹ J; hc ≈ 1240 eV nm

NEET traps

Intensity and frequency are different. Doubling the intensity doubles the saturation current but does not change Kmax or V₀.
Doubling the frequency does not double Kmax. Kmax = hν − φ₀, so it more than doubles when φ₀ is non-zero.
Stopping potential is the magnitude of the retarding voltage. The collector is made negative with respect to the emitter, not positive.
The slope of the V₀ – ν graph is h/e, not h. The slope of the Kmax – ν graph is h.
The slope does not depend on the metal. Only the intercepts change with the metal.
A photon is not deflected by fields, because it is neutral. Electrons emitted are deflected.
A photon has momentum even though its rest mass is zero.
Electron and photon of the same wavelength have the same momentum, but not the same energy.
For the same accelerating voltage a heavier or more highly charged particle has a shorter wavelength; do not use 1.227/√V nm for protons or alpha particles.
Work function is the minimum energy needed. Electrons from deeper inside the metal come out with less than Kmax.

One-liners

Work function φ₀ is the least energy needed to remove an electron from a metal surface.
1 eV = 1.6 × 10⁻¹⁹ J.
Thermionic emission uses heat; field emission uses a field of about 10⁸ V m⁻¹.
Photocurrent at saturation is proportional to light intensity.
Stopping potential is independent of intensity.
Stopping potential increases linearly with frequency.
No emission below the threshold frequency, whatever the intensity.
Photoelectric emission is instantaneous, about 10⁻⁹ s or less.
Photon energy E = hν, momentum p = h/λ.
hc ≈ 1240 eV nm.
Kmax = hν − φ₀ = eV₀.
The V₀ – ν line has slope h/e and frequency intercept φ₀/h.
λ = h/p applies to every moving particle.
Electron through V volts has λ ≈ 1.227/√V nm.
Electron diffraction by crystals confirmed the wave nature of matter.

Practice questions

The photoelectric effect provides direct evidence for which nature of light?
1. Wave nature only
2. Interference nature
3. Transverse wave nature
4. Particle nature (photons)
Answer
D. Particle nature (photons)
Emission of electrons by single photons with energy hν can be explained only if light is treated as a stream of particles.
A photon has wavelength 400 nm. Taking hc = 1240 eV nm, its energy is
1. 0.31 eV
2. 6.2 eV
3. 3.1 eV
4. 1.55 eV
Answer
C. 3.1 eV
E = 1240/400 = 3.1 eV.
A metal has work function 2.0 eV. Taking hc = 1240 eV nm, its threshold wavelength is
1. 310 nm
2. 620 nm
3. 2480 nm
4. 1240 nm
Answer
B. 620 nm
λ₀ = hc/φ₀ = 1240/2.0 = 620 nm.
Light of photon energy 4.5 eV falls on a metal of work function 2.5 eV. The stopping potential is
1. 2.0 V
2. 2.5 V
3. 7.0 V
4. 4.5 V
Answer
A. 2.0 V
eV₀ = hν − φ₀ = 4.5 − 2.5 = 2.0 eV, so V₀ = 2.0 V.
In a photoelectric experiment the stopping potential depends on
1. the intensity of the light only
2. the distance between the electrodes
3. the frequency of the light and the metal used
4. the intensity of the light and the metal used
Answer
C. the frequency of the light and the metal used
eV₀ = hν − φ₀ contains only frequency and work function (the metal), not intensity.
An electron is accelerated through 150 V from rest. Using λ ≈ 1.227/√V nm, its de Broglie wavelength is about
1. 0.0818 nm
2. 0.150 nm
3. 1.227 nm
4. 0.100 nm
Answer
D. 0.100 nm
λ = 1.227/√150 = 1.227/12.25 ≈ 0.100 nm. Check with λ = h/√(2meV) gives the same value.
Take h = 6.63 × 10⁻³⁴ J s. The momentum of a photon of wavelength 663 nm is
1. 3.0 × 10⁻¹⁹ kg m s⁻¹
2. 1.0 × 10⁻²⁷ kg m s⁻¹
3. 9.0 × 10⁻¹¹ kg m s⁻¹
4. 1.0 × 10⁻³⁶ kg m s⁻¹
Answer
B. 1.0 × 10⁻²⁷ kg m s⁻¹
p = h/λ = 6.63 × 10⁻³⁴ / 6.63 × 10⁻⁷ = 1.0 × 10⁻²⁷. Check: E/c = 3 × 10⁻¹⁹/3 × 10⁸ = 10⁻²⁷.
Emission of electrons from a heated metal surface is called
1. thermionic emission
2. radioactive emission
3. photoelectric emission
4. field emission
Answer
A. thermionic emission
Heat gives electrons enough energy to leave the metal: thermionic emission.
A proton and an alpha particle are accelerated from rest through the same potential difference. The ratio of their de Broglie wavelengths λp : λα is
1. 2√2 : 1
2. 2 : 1
3. 4 : 1
4. √2 : 1
Answer
A. 2√2 : 1
λ = h/√(2mqV) ∝ 1/√(mq). For the alpha particle mq is 4 × 2 = 8 times that of the proton, so λp/λα = √8 = 2√2.
Light of frequency 2ν₀ (ν₀ is the threshold frequency, φ₀ the work function) falls on a metal. The maximum kinetic energy of photoelectrons is
1. zero
2. φ₀
3. 2φ₀
4. φ₀/2
Answer
B. φ₀
Kmax = h(2ν₀) − hν₀ = hν₀ = φ₀.
For a plot of stopping potential V₀ against frequency ν, the slope is equal to
1. e/h
2. φ₀/e
3. h
4. h/e
Answer
D. h/e
V₀ = (h/e)ν − φ₀/e, so the slope is h/e.
In a graph of Kmax against frequency, the intercept on the frequency axis equals
1. e/h
2. h/e
3. ν₀ = φ₀/h
4. φ₀
Answer
C. ν₀ = φ₀/h
Kmax = 0 at ν = ν₀, and hν₀ = φ₀.
Take h = 6.63 × 10⁻³⁴ J s. A beam of power 3.315 W has frequency 5 × 10¹⁴ Hz. The number of photons per second is
1. 1.0 × 10²⁰
2. 1.0 × 10¹⁸
3. 1.0 × 10¹⁹
4. 3.3 × 10⁻¹⁹
Answer
C. 1.0 × 10¹⁹
E = hν = 3.315 × 10⁻¹⁹ J. N = 3.315/3.315 × 10⁻¹⁹ = 1.0 × 10¹⁹ s⁻¹.
An electron has kinetic energy 25 eV. Using λ ≈ 1.227/√V nm (V in volt), its de Broglie wavelength is about
1. 0.0245 nm
2. 0.123 nm
3. 0.490 nm
4. 0.245 nm
Answer
D. 0.245 nm
A 25 eV electron has been accelerated through 25 V: λ = 1.227/5 = 0.245 nm.
An electron and a photon have the same wavelength. Which statement is correct?
1. They have the same momentum
2. They have the same speed
3. The photon has greater momentum
4. They have the same energy
Answer
A. They have the same momentum
p = h/λ for both, so equal λ means equal momentum. Their energies and speeds differ.
One electron-volt is equal to
1. 9.1 × 10⁻³¹ J
2. 1.6 × 10⁻¹⁹ J
3. 1.6 × 10¹⁹ J
4. 6.63 × 10⁻³⁴ J
Answer
B. 1.6 × 10⁻¹⁹ J
1 eV = e × 1 V = 1.6 × 10⁻¹⁹ J.
Given below are two statements: Statement I: The saturation photocurrent is proportional to the intensity of incident light for a fixed frequency above the threshold. Statement II: The stopping potential increases in proportion to the intensity of light. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
More photons per second give more electrons, so I is correct. The stopping potential depends on frequency, not intensity, so II is incorrect.
Given below are two statements: Statement I: No photoelectrons are emitted below the threshold frequency, however intense the light. Statement II: A wave picture of light predicts that such a threshold frequency should exist. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
I is correct. A wave picture predicts that enough intensity should eject electrons at any frequency, so II is incorrect.
Given below are two statements: Statement I: A photon has zero rest mass. Statement II: A photon of wavelength λ has energy hc/λ and momentum h/λ. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are standard properties of a photon.
Which of the following statements are correct? 1. The energy of a photon is hν. 2. The momentum of a photon is h/λ. 3. A photon is deflected by an electric field.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
A photon is neutral and is not deflected by an electric field; 1 and 2 are correct.
Which of the following statements are correct? 1. For equal speed, a heavier particle has a shorter de Broglie wavelength. 2. A moving neutral particle has no de Broglie wavelength. 3. The relation λ = h/p applies to every moving particle.
1. 2 and 3 only
2. 1 and 3 only
3. 1, 2 and 3
4. 1 and 2 only
Answer
B. 1 and 3 only
λ = h/mv applies to all moving particles, neutral or charged, and decreases as m increases.
Given below are two statements: Statement I: The maximum kinetic energy of photoelectrons increases linearly with the frequency of light. Statement II: The slope of the graph of Kmax against frequency equals Planck's constant h for every metal. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Kmax = hν − φ₀ is a straight line of slope h for any metal.
Given below are two statements: Statement I: The stopping potential of a given metal increases with the frequency of the light. Statement II: The stopping potential is the positive potential of the collector at which the photocurrent becomes zero. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
I is correct. The collector must be made negative with respect to the emitter, so II is incorrect.
Given below are two statements: Statement I: Field emission of electrons occurs when a very strong electric field, of about 10⁸ V m⁻¹, acts on a metal surface. Statement II: Thermionic emission occurs when a metal is heated to a high temperature. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both describe the processes correctly.
Given below are two statements: Statement I: A time lag of several seconds is seen for photoelectric emission when the light is of low intensity. Statement II: The wave theory of light explains the threshold frequency of photoelectric emission. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Emission is instantaneous (about 10⁻⁹ s or less) and the wave theory cannot explain the threshold frequency; both are incorrect.
Given below are two statements: Statement I: Only charged particles in motion have an associated de Broglie wavelength. Statement II: The de Broglie wavelength of a particle is inversely proportional to its momentum. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Every moving particle has a wavelength, so I is incorrect. λ = h/p, so II is correct.
Assertion (A): If the intensity of light of fixed frequency is doubled, the stopping potential does not change. Reason (R): The stopping potential depends on the photon energy and the work function, not on the number of photons.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Doubling intensity doubles the number of photons only; eV₀ = hν − φ₀ is unchanged.
Assertion (A): No photoelectrons are emitted below the threshold frequency, even for very intense light. Reason (R): One electron absorbs one photon, and the photon energy depends on frequency only.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
If hν < φ₀ no single photon can free an electron, and electrons do not add energy from several photons.
Assertion (A): The wave nature of a moving cricket ball is not observed. Reason (R): Its de Broglie wavelength is extremely small compared with the size of the ball and everyday distances.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
λ = h/mv is of the order of 10⁻³⁴ m for a cricket ball, too small to detect.
Assertion (A): The maximum kinetic energy of photoelectrons increases with the frequency of light. Reason (R): The saturation photocurrent is proportional to the intensity of light.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but the second does not explain the first, which follows from Kmax = hν − φ₀.
Assertion (A): The stopping potential is higher for light of higher frequency. Reason (R): The energy of a photon decreases as the frequency of light increases.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
A is true because eV₀ = hν − φ₀. R is false since E = hν increases with frequency.
Assertion (A): Photoelectric emission shows a long time lag when the intensity of light is very low. Reason (R): Each electron receives the energy of one photon in a single event.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
Emission is instantaneous, so A is false. R is true.
Match List I with List II. List I: a. Photon energy b. Photon momentum c. de Broglie wavelength of an electron through V volt d. Threshold frequency List II: i. h/λ ii. hν iii. 1.227/√V nm iv. φ₀/h
1. a-i, b-ii, c-iii, d-iv
2. a-ii, b-i, c-iv, d-iii
3. a-iii, b-iv, c-i, d-ii
4. a-ii, b-i, c-iii, d-iv
Answer
D. a-ii, b-i, c-iii, d-iv
Energy hν; momentum h/λ; electron wavelength 1.227/√V nm; ν₀ = φ₀/h.
Match List I with List II. List I: a. Planck's constant b. Work function c. Photon momentum d. Stopping potential List II: i. kg m s⁻¹ ii. J s iii. eV iv. V
1. a-ii, b-iv, c-i, d-iii
2. a-ii, b-iii, c-i, d-iv
3. a-iii, b-ii, c-iv, d-i
4. a-i, b-iii, c-ii, d-iv
Answer
B. a-ii, b-iii, c-i, d-iv
h is in J s, φ₀ in eV, p in kg m s⁻¹, V₀ in volt.
Match List I with List II. List I: a. Wavelength 620 nm b. Wavelength 310 nm c. Wavelength 248 nm d. Wavelength 1240 nm List II: i. 5.0 eV ii. 1.0 eV iii. 2.0 eV iv. 4.0 eV
1. a-iii, b-iv, c-ii, d-i
2. a-iv, b-iii, c-i, d-ii
3. a-i, b-ii, c-iii, d-iv
4. a-iii, b-iv, c-i, d-ii
Answer
D. a-iii, b-iv, c-i, d-ii
E = 1240/λ: 620 nm → 2.0 eV; 310 nm → 4.0 eV; 248 nm → 5.0 eV; 1240 nm → 1.0 eV.
Match List I with List II. List I: a. Electron through 1 V b. Electron through 4 V c. Electron through 25 V d. Electron through 100 V List II: i. 0.1227 nm ii. 1.227 nm iii. 0.2454 nm iv. 0.6135 nm
1. a-ii, b-iii, c-iv, d-i
2. a-iv, b-ii, c-iii, d-i
3. a-i, b-iii, c-iv, d-ii
4. a-ii, b-iv, c-iii, d-i
Answer
D. a-ii, b-iv, c-iii, d-i
λ = 1.227/√V nm: 1.227, 0.6135, 0.2454 and 0.1227 nm.
Match List I with List II. List I: a. Hertz b. Hallwachs c. Einstein d. de Broglie List II: i. Photon explanation of photoelectric effect ii. Wave nature of matter iii. Sparking between electrodes eased by ultraviolet light on the negative electrode iv. A negatively charged zinc plate lost charge in ultraviolet light
1. a-iv, b-iii, c-i, d-ii
2. a-iii, b-iv, c-i, d-ii
3. a-i, b-iv, c-iii, d-ii
4. a-iii, b-iv, c-ii, d-i
Answer
B. a-iii, b-iv, c-i, d-ii
Hertz: easier sparking; Hallwachs: zinc plate discharge; Einstein: photons; de Broglie: matter waves.
Match List I with List II. List I: a. Saturation photocurrent b. Stopping potential c. Threshold frequency d. Maximum kinetic energy List II: i. φ₀/h ii. eV₀ iii. depends on intensity iv. depends on frequency
1. a-iii, b-i, c-iv, d-ii
2. a-iii, b-iv, c-ii, d-i
3. a-iii, b-iv, c-i, d-ii
4. a-iv, b-iii, c-i, d-ii
Answer
C. a-iii, b-iv, c-i, d-ii
Saturation current depends on intensity; V₀ on frequency; ν₀ = φ₀/h; Kmax = eV₀.
Light of wavelength 400 nm falls on a metal of work function 2.3 eV. Take hc = 1240 eV nm. The maximum kinetic energy of the photoelectrons is
1. 5.4 eV
2. 0.4 eV
3. 3.1 eV
4. 0.8 eV
Answer
D. 0.8 eV
Photon energy 1240/400 = 3.1 eV; Kmax = 3.1 − 2.3 = 0.8 eV.
A laser of power 1.5 mW emits light of wavelength 663 nm. Take h = 6.63 × 10⁻³⁴ J s and c = 3 × 10⁸ m s⁻¹. The number of photons emitted per second is
1. 5.0 × 10¹⁸
2. 5.0 × 10¹⁵
3. 5.0 × 10¹²
4. 1.5 × 10¹⁹
Answer
B. 5.0 × 10¹⁵
Photon energy = hc/λ = 3.0 × 10⁻¹⁹ J. N = 1.5 × 10⁻³/3.0 × 10⁻¹⁹ = 5.0 × 10¹⁵ s⁻¹.
Electrons are accelerated from rest through 400 V. Using λ ≈ 1.227/√V nm, the de Broglie wavelength is about
1. 0.123 nm
2. 0.0307 nm
3. 0.0614 nm
4. 0.614 nm
Answer
C. 0.0614 nm
λ = 1.227/20 = 0.0614 nm.
Photons of energy 5 eV give photoelectrons from a metal with a stopping potential of 3 V. Taking hc = 1240 eV nm, the threshold wavelength of the metal is
1. 413 nm
2. 620 nm
3. 1240 nm
4. 248 nm
Answer
B. 620 nm
φ₀ = 5 − 3 = 2 eV; λ₀ = 1240/2 = 620 nm.
Take h = 6.63 × 10⁻³⁴ J s. A ball of mass 0.1 kg moves at 10 m s⁻¹. Its de Broglie wavelength is
1. 6.63 × 10⁻³³ m
2. 6.63 × 10⁻³⁶ m
3. 6.63 × 10⁻³⁵ m
4. 6.63 × 10⁻³⁴ m
Answer
D. 6.63 × 10⁻³⁴ m
λ = h/mv = 6.63 × 10⁻³⁴/(0.1 × 10) = 6.63 × 10⁻³⁴ m.
A point source of monochromatic light of frequency above the threshold is placed near a photocell. The source is moved to twice the distance (intensity falls as the inverse square of distance). Then
1. the saturation current becomes half and the stopping potential becomes half
2. the saturation current is unchanged and the stopping potential becomes one-fourth
3. the saturation current becomes one-fourth and the stopping potential is unchanged
4. the saturation current becomes one-fourth and the stopping potential becomes half
Answer
C. the saturation current becomes one-fourth and the stopping potential is unchanged
Intensity becomes one-fourth, so the number of photoelectrons per second falls to one-fourth. Frequency is unchanged, so V₀ is unchanged.
Take h = 6.63 × 10⁻³⁴ J s. A surface fully absorbs 10²⁰ photons per second, each of wavelength 663 nm, falling normally on it. The force on the surface is
1. 1.0 × 10⁻²⁷ N
2. 2.0 × 10⁻⁷ N
3. 1.0 × 10⁻⁷ N
4. 3.0 × 10⁻¹¹ N
Answer
C. 1.0 × 10⁻⁷ N
Momentum of one photon = h/λ = 10⁻²⁷ kg m s⁻¹. Force = rate × momentum = 10²⁰ × 10⁻²⁷ = 10⁻⁷ N (a reflecting surface would feel twice this).

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