Class 12 Physics Chapter 11 Important Questions (2026-27)

Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter: important questions with answers on the photoelectric effect for CBSE 2026-27.

Important Questions for Class 12 Physics Chapter 11 – Dual Nature of Radiation and Matter

Looking for Class 12 Physics Chapter 11 important questions? This page covers Dual Nature of Radiation and Matter with exam-focused questions and step-by-step answers, arranged by marks: 1 mark, 2–3 marks and 5 marks. Written for the CBSE 2026–27 syllabus, it focuses on the photoelectric effect, Einstein's equation and de Broglie wavelength. Attempt them after finishing the NCERT chapter.

Very Short Answer Questions (1 Mark)

Photoelectric Effect Basics

Question 1

The stopping potential in an experiment on the photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted?

Answer: \(K_{max} = eV_0 = e \times 1.5\ \text{V} =\) 1.5 eV

Question 2

The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?

Answer: \(V_0 = \dfrac{K_{max}}{e} =\) 3 V

Question 3

Define the term 'stopping potential' in relation to the photoelectric effect.

Answer: The stopping potential is the value of the retarding potential at which the photoelectric current becomes zero for a given frequency of incident radiation.

Question 4

Define 'intensity' of radiation in the photon picture of light.

Answer: Intensity is the number of photoelectrons emitted per second (or the number of photons incident per unit area per unit time).

Question 5

Define intensity of radiation on the basis of the photon picture of light. Write its SI unit.

Answer: It is the number of photoelectrons emitted per second per unit area. SI unit: m⁻² s⁻¹.

Question 6

Why is photoelectric emission not possible at all frequencies?

Answer: Photoelectric emission is possible only if the incident photon energy (\(h\nu\)) is greater than the work function of the metal (\(\phi_0 = h\nu_0\)). So the frequency must be greater than the threshold frequency \(\nu_0\).

Question 7

Name the phenomenon which shows the quantum nature of electromagnetic radiation.

Answer: The photoelectric effect.

Question 8

State one factor which determines the intensity of light in the photon picture of light.

Answer: The number of photons incident per unit area per unit time (equivalently, the number of photoelectrons emitted per second).

Question 9

State one reason to explain why the wave theory of light does not support the photoelectric effect.

Answer: The kinetic energy of photoelectrons does not depend on the intensity of the incident light — which contradicts wave theory.

Question 10

If the distance between the source of light and the cathode of a photocell is doubled, how does it affect the stopping potential applied to the photocell?

Answer: The stopping potential remains unchanged — it depends only on the frequency of incident light, not on intensity or distance.

De Broglie Hypothesis

Question 11

State the de Broglie hypothesis.

Answer: According to de Broglie, a particle of mass m moving with velocity v is associated with a matter wave of wavelength \(\lambda = \dfrac{h}{mv} = \dfrac{h}{p}\).

Question 12

An electron and an alpha particle have the same de Broglie wavelength associated with them. How are their kinetic energies related to each other?

Answer: For the same \(\lambda\), \(p = \dfrac{h}{\lambda}\) is the same. Kinetic energy \(K = \dfrac{p^2}{2m} \propto \dfrac{1}{m}\).

\(\dfrac{K_e}{K_\alpha} = \dfrac{m_\alpha}{m_e} = \dfrac{4m_p}{m_p} =\) 4 : 1

Question 13

A proton and an electron have the same kinetic energy. Which one has greater de Broglie wavelength and why?

Answer: \(\lambda = \dfrac{h}{\sqrt{2mK}}\). For the same K, \(\lambda \propto \dfrac{1}{\sqrt{m}}\). Since \(m_e < m_p\), the electron has the greater de Broglie wavelength.

Question 14

A proton and an electron have the same kinetic energy. Which one has smaller de Broglie wavelength and why?

Answer: The proton (greater mass, smaller wavelength).

Question 15

Write the expression for the de Broglie wavelength associated with a charged particle having charge q and mass m, when it is accelerated by a potential V.

Answer: \(\lambda = \dfrac{h}{\sqrt{2mqV}}\)

Question 16

Find the ratio of de Broglie wavelengths associated with two electrons accelerated through 25 V and 36 V.

Answer: \(\lambda \propto \dfrac{1}{\sqrt{V}}\). \(\dfrac{\lambda_1}{\lambda_2} = \sqrt{\dfrac{V_2}{V_1}} = \sqrt{\dfrac{36}{25}} =\) 6 : 5

Question 17

An electron is revolving around the nucleus with a constant speed of \(2.2 \times 10^8\ \text{m/s}\). Find the de Broglie wavelength associated with it.

Answer: \(\lambda = \dfrac{h}{mv} = \dfrac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.2 \times 10^8} \approx\) \(3.3 \times 10^{-12}\ \text{m}\)

Graphs and Interpretation

Question 18

The figure shows a plot of three curves a, b, c showing the variation of photocurrent vs. collector plate potential for three different intensities \(I_1, I_2, I_3\) having frequencies \(\nu_1, \nu_2, \nu_3\) respectively. Point out the two curves for which the incident radiations have the same frequency but different intensities.

Answer: The stopping potential is the same for the same frequency. Curves a and b have the same frequency but different intensities (\(I_2 > I_3\)).

Question 19

Show graphically the variation of the de Broglie wavelength (\(\lambda\)) with the potential (V) through which an electron is accelerated from rest.

Answer: \(\lambda \propto \dfrac{1}{\sqrt{V}}\) — a decreasing curve (rectangular hyperbola of the form \(\lambda \propto V^{-1/2}\)).

Question 20

Two lines A and B show the variation of de Broglie wavelength \(\lambda\) versus \(1/\sqrt{V}\) for two particles carrying the same charge. Which one represents a particle of smaller mass?

Answer: \(\lambda = \dfrac{h}{\sqrt{2mq}} \cdot \dfrac{1}{\sqrt{V}}\). The slope is proportional to \(1/\sqrt{m}\). Line with the greater slope corresponds to the smaller mass.

Question 21

The given graph shows the variation of photoelectric current (I) vs. applied voltage (V) for two different photosensitive materials and two different intensities of incident radiation. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation.

Answer: The pairs (2, 4) and (1, 3) have the same intensity but different materials.

Question 22

Show on a plot the nature of variation of photoelectric current with the intensity of radiation incident on a photosensitive surface.

Answer: Photocurrent vs intensity is a straight line through the origin (directly proportional).

Question 23

Figure shows a plot of \(1/\sqrt{V}\) (V = accelerating potential) vs the de Broglie wavelength \(\lambda\) for two particles with same charge q but different masses \(m_1, m_2\). Which line (A or B) represents a particle of larger mass?

Answer: Line B represents the particle of larger mass (slope ∝ 1/√m, so larger slope ⇒ smaller mass; line B has smaller slope).

Question 24

The graph shows variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has a higher value of work function? Justify.

Answer: Metal A, because it has the higher threshold frequency.

Question 25

The graph shows variation of stopping potential \(V_0\) vs frequency \(\nu\) for two photosensitive metals A and B. Which has higher threshold frequency and why?

Answer: Metal A — its intercept on the frequency axis is higher.

Question 26

Plot a graph showing the variation of de Broglie wavelength of an electron as a function of its kinetic energy.

Answer: \(\lambda \propto \dfrac{1}{\sqrt{K}}\) — a decreasing curve (rectangular hyperbola of the form \(\lambda \propto K^{-1/2}\)).

Short Answer Questions (2–3 Marks)

De Broglie Wavelength Comparisons

Question 27

An electron is accelerated through a potential difference of 100 volts. What is the de Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value correspond?

Answer: \(\lambda = \dfrac{12.27}{\sqrt{V}}\ \text{Å} = \dfrac{12.27}{\sqrt{100}} =\) 0.123 nm

This corresponds to the X-ray region of the electromagnetic spectrum.

Question 28

An electron is accelerated through a potential difference of 64 volts. What is the de Broglie wavelength? To which part of the EM spectrum does this correspond?

Answer: \(\lambda = \dfrac{12.27}{\sqrt{64}} \approx\) 1.53 Å ≈ 0.153 nm. This corresponds to the X-ray region.

Question 29

An alpha particle and a proton are accelerated from rest by the same potential. Find the ratio of their de Broglie wavelengths.

Answer: \(\lambda = \dfrac{h}{\sqrt{2mqV}}\). \(\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\dfrac{4m_p \cdot 2e}{m_p \cdot e}} = \sqrt{8} =\) 2√2 : 1

Question 30

A proton and a deuteron are accelerated through the same potential. Which has (a) greater de Broglie wavelength and (b) less momentum? Give reasons.

Answer: For both, qV is the same, so \(p = \sqrt{2mqV}\).

(a) Since \(\lambda = h/\sqrt{2mqV}\), the proton (smaller mass) has greater de Broglie wavelength.

(b) Since \(p \propto \sqrt{m}\), the proton has less momentum.

Question 31

A proton and an alpha particle are accelerated through the same potential. Which has (i) greater de Broglie wavelength and (ii) less kinetic energy? Give reasons.

Answer: (i) Proton (smaller mass → larger λ).

(ii) \(K = qV\). The proton has smaller charge (e) vs the α-particle (2e), so the proton has less kinetic energy.

Question 32

A deuteron and an alpha particle are accelerated with the same accelerating potential. Which has (1) greater de Broglie wavelength and (2) less kinetic energy? Explain.

Answer: (1) Deuteron (mass 2m_p vs. α's 4m_p — smaller mass → larger λ).

(2) \(K = qV\): deuteron has charge e, α has 2e, so the deuteron has less kinetic energy.

Question 33

A proton and an alpha particle have the same de Broglie wavelength. Determine the ratio of (i) their accelerating potentials and (ii) their speeds.

Answer: Same λ → same momentum \(p = h/\lambda\).

(i) \(V = \dfrac{p^2}{2mq}\). \(\dfrac{V_p}{V_\alpha} = \dfrac{m_\alpha q_\alpha}{m_p q_p} = \dfrac{4 \cdot 2}{1 \cdot 1} =\) 8 : 1

(ii) \(v = p/m\). \(\dfrac{v_p}{v_\alpha} = \dfrac{m_\alpha}{m_p} =\) 4 : 1

Question 34

An electron and a proton are accelerated through the same potential. Which has (i) greater de Broglie wavelength and (ii) less momentum? Justify.

Answer: (i) Since λ ∝ 1/√m, the electron has greater de Broglie wavelength.

(ii) \(p = \sqrt{2mqV}\). The electron (smaller mass) has less momentum.

Question 35

A photon and a proton have the same de Broglie wavelength. Show, by actual calculation, which has more total energy.

Answer: Photon energy: \(E = hc/\lambda\).

Proton energy: \(E' = \dfrac{p^2}{2m} = \dfrac{h^2}{2m\lambda^2}\).

Ratio: \(\dfrac{E}{E'} = \dfrac{hc/\lambda}{h^2/(2m\lambda^2)} = \dfrac{2mc\lambda}{h}\). For typical values, \(\dfrac{2mc\lambda}{h} \gg 1\), so the photon has much more energy.

Question 36

An electron and a photon each have a wavelength of 1.50 nm. Find (i) their momenta, (ii) the energy of the photon, (iii) the kinetic energy of the electron.

Answer: (i) \(p = h/\lambda = \dfrac{6.63 \times 10^{-34}}{1.5 \times 10^{-9}} =\) \(4.42 \times 10^{-25}\ \text{kg·m/s}\)

(ii) \(E = pc = 4.42 \times 10^{-25} \times 3 \times 10^8 =\) \(1.33 \times 10^{-16}\ \text{J} \approx 830\ \text{eV}\)

(iii) \(K = \dfrac{p^2}{2m} = \dfrac{(4.42 \times 10^{-25})^2}{2 \times 9.1 \times 10^{-31}} =\) \(1.07 \times 10^{-19}\ \text{J} \approx 0.67\ \text{eV}\)

Question 37

An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. Taking other factors to be the same, how does the resolving power of an electron microscope compare with that of an optical microscope using yellow light?

Answer: \(\lambda = \dfrac{12.27}{\sqrt{50000}}\ \text{Å} \approx 5.5 \times 10^{-12}\ \text{m}\).

For yellow light: \(\lambda = 5.9 \times 10^{-7}\ \text{m}\).

Resolving power ∝ 1/λ: RP of electron microscope is about \(10^5\) times greater.

Question 38

A proton and an alpha particle are accelerated through the same potential difference. Find the ratio of their de Broglie wavelengths.

Answer: Same as Question 29: \(\lambda_p : \lambda_\alpha =\) 2√2 : 1

Question 39

Calculate the de Broglie wavelength of the electron orbiting in the n = 2 state of hydrogen atom.

Answer: From Bohr quantization: \(mvr = \dfrac{nh}{2\pi} \Rightarrow \lambda = \dfrac{h}{mv} = \dfrac{2\pi r}{n}\).

For n = 2, \(r = 4 \times 0.53 = 2.12\ \text{Å}\).

\(\lambda = \dfrac{2\pi \times 2.12}{2} =\) 6.66 Å

Question 40

Calculate the kinetic energy of an electron having a de Broglie wavelength of 1 Å.

Answer: \(K = \dfrac{h^2}{2m\lambda^2} = \dfrac{(6.63 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (10^{-10})^2}\)

\(= \dfrac{43.96 \times 10^{-68}}{1.82 \times 10^{-50}} \approx 2.4 \times 10^{-17}\ \text{J} \approx\) 150 eV

Photoelectric Equation and Its Features

Question 41

Write Einstein's photoelectric equation. State clearly the three salient features observed in the photoelectric effect which can be explained on the basis of this equation.

Answer: \(K_{max} = h\nu - \phi_0\)

Features explained:
(i) K_max depends linearly on ν and not on intensity.
(ii) There is a threshold frequency ν₀ = φ₀/h below which no emission occurs.
(iii) Photoelectric current is proportional to intensity (number of photons).

Question 42

Two monochromatic radiations of frequencies \(\nu_1\) and \(\nu_2\) (\(\nu_1 > \nu_2\)), having the same intensity, are incident on a photosensitive surface. Explain, giving reason, in which case (i) more electrons will be emitted and (ii) the maximum kinetic energy of emitted photoelectrons will be more.

Answer: (i) Same intensity means same number of photons per second ⇒ same number of photoelectrons per second. So same number in both cases.

(ii) \(K_{max} = h\nu - \phi_0\). Larger ν → larger K_max, so ν₁ gives more kinetic energy.

Question 43

Write the three basic properties of photons which are used to obtain Einstein's photoelectric equation. Use this equation to draw a plot of maximum kinetic energy of the electrons emitted versus frequency of incident radiation.

Answer: Properties: (i) Each photon has energy E = hν, (ii) photon energy depends only on frequency, not intensity, (iii) each photon interacts with one electron (one-to-one).

Plot: A straight line with slope h and intercept −φ₀ on the y-axis; it crosses the x-axis at threshold frequency ν₀.

Question 44

(i) Define the term 'threshold frequency' as used in the photoelectric effect. (ii) Plot a graph showing the variation of photoelectric current as a function of anode potential for two light beams having the same frequency but different intensities \(I_1\) and \(I_2\) (\(I_1 > I_2\)).

Answer: (i) The minimum frequency ν₀ which incident light must have to eject photoelectrons from a metal surface is the threshold frequency.

(ii) Both curves have the same stopping potential (same frequency); the I₁ curve has a higher saturation current than I₂.

Question 45

Using the graph of stopping potential vs. incident frequency, calculate Planck's constant.

Answer: Slope of the V₀ vs ν graph \(= \dfrac{h}{e}\), so \(h = e \times \text{slope}\). (Numerical value depends on the given graph.)

Question 46

Draw a graph showing the variation of stopping potential with frequency of incident radiation for two photosensitive materials with work functions \(W_1\) and \(W_2\) (\(W_1 > W_2\)). On what factors do (i) the slope and (ii) the intercept of the lines depend?

Answer: Both lines are straight and parallel (slope = h/e in both cases); the line for W₁ has a larger x-intercept (ν₀₁ > ν₀₂).

(i) Slope depends only on h/e — independent of material.
(ii) x-intercept = ν₀ = W/h — depends on the material's work function.

Question 47

Answer: (i) Threshold frequency of material with W₁ is higher than that of W₂.

(ii) Both lines have the same slope (h/e) — independent of material.

Einstein's Equation and Its Derivations

Question 48

Write Einstein's photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation. Write the three salient features observed in the photoelectric effect which can be explained using this equation.

Answer: \(h\nu = \phi_0 + K_{max}\) (or \(K_{max} = h\nu - \phi_0\)).

Derivation: A photon of energy hν is absorbed by an electron. Part of the energy (φ₀) is used to escape the metal; the rest appears as the electron's maximum kinetic energy. Conservation of energy gives hν = φ₀ + K_max.

Three features explained:
(i) K_max linear in ν, independent of intensity.
(ii) Threshold frequency ν₀ = φ₀/h exists.
(iii) Photoelectric current ∝ intensity.

Question 49

Define the terms (i) 'cut-off voltage' and (ii) 'threshold frequency' in relation to the photoelectric effect. Using Einstein's equation, show how these can be determined from a suitable plot.

Answer: (i) Cut-off (stopping) voltage: The retarding potential at which photocurrent becomes zero.

(ii) Threshold frequency: The minimum frequency below which no photoelectric emission occurs.

From \(V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}\), the V₀ vs ν plot is a straight line. Slope = h/e → gives h. x-intercept = ν₀ (threshold frequency). y-intercept = −φ₀/e → gives φ₀.

Question 50

A photon and a proton have the same de Broglie wavelength. Show, by actual calculations, which has more total energy.

Answer: Same as Question 35.

Wave Theory vs. Photon Picture

Question 51

Write Einstein's photoelectric equation and point out any two characteristic properties of photons on which this equation is based. Briefly explain the three observed features which can be explained by this equation.

Answer: \(K_{max} = h\nu - \phi_0\).

Two properties: (i) Photoelectric emission only if hν ≥ φ₀. (ii) A photon gives all its energy to one electron.

Three features: (a) threshold frequency, (b) instantaneous emission, (c) K_max depends on ν not intensity.

Question 52

State three important properties of photons which describe the particle picture of electromagnetic radiation. Use Einstein's equation to define (i) stopping potential and (ii) threshold frequency.

Answer: Three properties: (i) Light consists of photons with energy E = hν, momentum p = h/λ. (ii) Intensity is the number of photons per second per unit area. (iii) One photon interacts with one electron.

(i) Stopping potential: Minimum negative potential V₀ at which photocurrent is zero. \(eV_0 = h\nu - \phi_0\).

(ii) Threshold frequency: Minimum frequency ν₀ = φ₀/h needed to eject photoelectrons.

Question 53

(a) Why cannot the photoelectric effect be explained on the basis of wave nature of light? Give reasons. (b) Write the basic features of the photon picture of electromagnetic radiation on which Einstein's photoelectric equation is based.

Answer: (a) Wave theory fails because:

(i) K_max should increase with intensity — but it does not.
(ii) Threshold frequency should not exist — but it does.
(iii) Photoelectric emission should have a time lag — but it is instantaneous.

(b) Photon picture features: (i) photons with E = hν, (ii) intensity = number of photons/second, (iii) one-to-one collision between photon and electron, (iv) energy conservation hν = φ₀ + K_max.

Question 54

(a) Describe briefly three experimentally observed features in the photoelectric effect. (b) Discuss briefly how wave theory of light cannot explain these features.

Answer: (a) Features: (i) photocurrent ∝ intensity (same frequency), (ii) stopping potential independent of intensity but linear in frequency, (iii) threshold frequency exists, (iv) emission is instantaneous.

(b) Wave theory predicts: K_max should increase with intensity, no threshold frequency should exist, and there should be a time lag. All these contradict observations.

Question 55

Explain the following, giving reasons: (a) Photoelectric current in a photocell increases with the increase in intensity of incident radiation. (b) The stopping potential varies linearly with frequency of incident radiation for a given photosensitive surface with the slope remaining the same for different surfaces. (c) Maximum kinetic energy of photoelectrons is independent of intensity of incident radiation.

Answer: (a) More photons per second → more photoelectrons per second → larger current.

(b) \(V_0 = \dfrac{h}{e}\nu - \dfrac{\phi_0}{e}\) is linear in ν with slope h/e — a universal constant, independent of material.

Question 56

State the two characteristic properties of the photon picture of radiation on which Einstein's photoelectric equation is based. Briefly explain the three observed features explained by this equation.

Answer: Properties: (i) Photons of energy hν interact with single electrons, (ii) Energy conservation hν = φ₀ + K_max.

Features: (a) threshold frequency, (b) instantaneous emission, (c) linear dependence of K_max on ν.

Photon Numbers and Energy

Question 57

Monochromatic light of frequency \(6.0 \times 10^{14}\ \text{Hz}\) is produced by a laser. The power emitted is \(2.0 \times 10^{-3}\ \text{W}\). Estimate the number of photons emitted per second on average by the source.

Answer: Energy per photon \(= h\nu = 6.63 \times 10^{-34} \times 6.0 \times 10^{14} = 3.98 \times 10^{-19}\ \text{J}\).

Number per second \(= \dfrac{2.0 \times 10^{-3}}{3.98 \times 10^{-19}} \approx\) \(5.0 \times 10^{15}\) photons/s

Question 58

The work function of a metal X equals \(3 \times 10^{-19}\ \text{J}\). Calculate the number N of photons of light of wavelength 26.52 nm whose total energy equals W.

Answer: Energy per photon \(= \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{26.52 \times 10^{-9}} \approx 7.5 \times 10^{-18}\ \text{J}\).

Number of photons: \(N = \dfrac{3 \times 10^{-19}}{7.5 \times 10^{-18}} =\) 0.04 ≈ 4 × 10⁻² (i.e., about 0.04 photons).

Miscellaneous

Question 59

State two properties of photons. For a monochromatic radiation incident on a photosensitive surface, why do all photoelectrons not come out with the same energy?

Answer: Properties: (i) Photons are electrically neutral, (ii) each photon has energy hν.

Photoelectrons have different energies because different electrons need different amounts of energy to escape from the surface — some are more tightly bound than others (the emitted electron loses some energy in collisions while escaping).

Question 60

An electron is accelerated through a potential difference of 100 V. What is the de Broglie wavelength? To which part of the EM spectrum does this value of wavelength correspond?

Answer: Same as Question 27: 0.123 nm, X-ray region.

Question 61

Using de Broglie's hypothesis, explain with the help of a suitable diagram Bohr's second postulate of quantization of energy levels in a hydrogen atom.

Answer: For a stable orbit, the electron's matter wave must form a standing wave around the orbit (constructive interference) — circumference must be an integer multiple of the wavelength:

\(2\pi r = n\lambda = \dfrac{nh}{mv}\)

So \(mvr = \dfrac{nh}{2\pi}\) — Bohr's quantization condition.

Question 62

Calculate the de Broglie wavelength of the electron orbiting in the n = 2 state of hydrogen atom.

Answer: Same as Question 39: 6.66 Å.

Question 63

Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, \(\nu_1 > \nu_2\), of incident radiation having the same intensity. In which case will the stopping potential be higher? Justify.

Answer: Same saturation current (same intensity); the ν₁ curve has a more negative stopping potential.

Stopping potential is higher for ν₁, because \(V_0 = \dfrac{h\nu - \phi_0}{e}\) increases with ν.

Question 64

A photon and a proton have the same de Broglie wavelength. Show, by actual calculations, which has more total energy.

Answer: Same as Question 35. The photon has more energy.

Long Answer Questions (5 Marks)

Wave Theory vs. Photon Picture

Question 65

(a) Write three observed features of the photoelectric effect which cannot be explained by wave theory. Explain how Einstein's photoelectric equation is used to describe these features satisfactorily. (b) Figure shows a plot of stopping potential \(V_0\) with frequency ν of incident radiation for two photosensitive materials M₁ and M₂. Explain (i) why the slope of both lines is the same, (ii) for which material the emitted electrons have greater kinetic energy for the same frequency.

Answer: (a) Same as Q51/Q53/Q54: (i) K_max independent of intensity, (ii) threshold frequency, (iii) instantaneous emission. Einstein's equation \(K_{max} = h\nu - \phi_0\) explains all three.

(b) (i) Slope = h/e — a universal constant independent of the material.
(ii) M₁ has the greater stopping potential at the same ν, so M₁ gives electrons with greater kinetic energy.

Question 66

(a) Describe briefly how the wave nature of moving electrons was established experimentally. (b) Estimate the ratio of de Broglie wavelengths associated with deuterons and alpha particles when they are accelerated from rest through the same accelerating potential V.

Answer: (a) Davisson–Germer experiment: A beam of accelerated electrons was scattered from a nickel crystal, and the intensity of scattered electrons was measured as a function of the angle of scattering for different accelerating voltages. A sharp peak at 54 V and 50° gave a wavelength of 0.165 nm, in excellent agreement with the de Broglie prediction \(\lambda = h/\sqrt{2meV}\) — thus confirming the wave nature of electrons.

(b) \(\lambda = \dfrac{h}{\sqrt{2mqV}}\). For deuteron (\(2m_p\), e) and α-particle (\(4m_p\), 2e): \(\dfrac{\lambda_d}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha q_\alpha}{m_d q_d}} = \sqrt{\dfrac{4 \cdot 2}{2 \cdot 1}} = \sqrt{4} =\) 2 : 1

Key Formulas to Remember

Concept	Formula
Photon energy	\(E = h\nu = \dfrac{hc}{\lambda}\)
Photon momentum	\(p = \dfrac{h}{\lambda} = \dfrac{E}{c}\)
Einstein's photoelectric equation	\(K_{max} = h\nu - \phi_0\)
Stopping potential	\(eV_0 = h\nu - \phi_0 = K_{max}\)
Threshold frequency	\(\nu_0 = \dfrac{\phi_0}{h}\)
de Broglie wavelength	\(\lambda = \dfrac{h}{p} = \dfrac{h}{mv}\)
de Broglie (KE form)	\(\lambda = \dfrac{h}{\sqrt{2mK}}\)
de Broglie (accelerated through V)	\(\lambda = \dfrac{h}{\sqrt{2mqV}}\)
de Broglie wavelength (electron, Å)	\(\lambda = \dfrac{12.27}{\sqrt{V}}\ \text{Å}\)
Bohr's quantization from de Broglie	\(2\pi r = n\lambda \Rightarrow mvr = \dfrac{nh}{2\pi}\)
Photon energy in eV	\(E(\text{eV}) = \dfrac{1240}{\lambda(\text{nm})}\)

FAQs

Are these questions enough for CBSE Class 12 Physics Chapter 11?

Yes. These questions cover every major topic — the photoelectric effect and Einstein's equation, the photon picture of light, de Broglie hypothesis and wavelength, Davisson–Germer experiment, and wave–particle duality — that CBSE has asked in previous board exams. Combine them with NCERT exercises and you'll be well prepared.

Which topics are most important for board exams?

Einstein's photoelectric equation and its three explaining features (threshold frequency, stopping potential, intensity dependence), derivation of de Broglie wavelength, comparisons between electrons/protons/α-particles at the same potential or same kinetic energy, the slope and intercept of the V₀–ν graph, and photon number calculations are consistently high-weightage topics.

How should I prepare for numericals in this chapter?

Practise de Broglie wavelength calculations (\(\lambda = 12.27/\sqrt{V}\ \text{Å}\) for electrons; general \(\lambda = h/\sqrt{2mqV}\) for other particles), photon energy and momentum calculations, photoelectric equation applications to find threshold frequency / work function / stopping potential, photon number per second from power, and ratios of de Broglie wavelengths for different particles. Be careful with unit conversions (eV ↔ J, nm ↔ m).