1. The work function of a metal is $W$ . If light of frequency $\nu$ is incident on the surface ( $\nu > W/h$ ), the maximum kinetic energy of the emitted photoelectrons is:

(a) $h\nu + W$
(b) $h\nu - W$
(c) $h\nu$
(d) $W - h\nu$

Answer: (b)
According to Einstein’s photoelectric equation: $K_{max} = h\nu - \phi_0$ (where $\phi_0$ is the work function $W$ ).

2. Which of the following graphs correctly represents the variation of stopping potential ( $V_0$ ) with frequency ( $\nu$ ) of incident radiation?

(a) Straight line passing through origin
(b) Straight line with negative y-intercept
(c) Straight line with positive y-intercept
(d) Hyperbola

Answer: (b)
From $V_0 = (h/e)\nu - (\phi_0/e)$ , the graph is a straight line with slope $h/e$ and negative y-intercept $-\phi_0/e$ .

3. A proton and an alpha particle have the same kinetic energy. The ratio of their de Broglie wavelengths ( $\lambda_p : \lambda_\alpha$ ) is:

(a) $1:1$
(b) $1:2$
(c) $2:1$
(d) $4:1$

Answer: (c)
$\lambda = h / \sqrt{2mK}$ . Since $K$ is same, $\lambda \propto 1/\sqrt{m}$ .
Mass of alpha is $4m_p$ . So $\lambda_p/\lambda_\alpha = \sqrt{m_\alpha/m_p} = \sqrt{4} = 2$ .

4. In the Davisson-Germer experiment, the diffraction of electrons confirms:

(a) Particle nature of electrons
(b) Wave nature of electrons
(c) Quantum nature of light
(d) None of the above

Answer: (b)
The observation of diffraction patterns (similar to X-rays) for electrons confirmed de Broglie’s hypothesis of matter waves.

5. The momentum of a photon of wavelength $\lambda$ is:

(a) $h\lambda$
(b) $h/\lambda$
(c) $hc/\lambda$
(d) $h/c\lambda$

Answer: (b)
From de Broglie’s relation (and particle theory), $p = E/c = (hc/\lambda)/c = h/\lambda$ .

6. If the intensity of incident light is doubled in a photoelectric experiment (keeping frequency constant), the stopping potential will:

(a) Double
(b) Halve
(c) Remain unchanged
(d) Become four times

Answer: (c)
Stopping potential depends only on the frequency of incident light and the material, not on the intensity.

7. The slope of the graph between stopping potential ( $V_0$ ) and frequency ( $\nu$ ) gives the value of:

(a) $h$
(b) $e$
(c) $h/e$
(d) $e/h$

Answer: (c)
Comparing Einstein’s equation with $y = mx + c$ , the slope $m = h/e$ .

8. An electron is accelerated through a potential difference of 100 V. Its de Broglie wavelength is approximately:

(a) 12.27 nm
(b) 1.227 nm
(c) 0.123 nm
(d) 0.012 nm

Answer: (c)
$\lambda = 1.227 / \sqrt{V}$ nm $= 1.227 / \sqrt{100} = 1.227 / 10 = 0.1227$ nm $\approx 1.23$ Å.

9. Light of frequency $1.5\nu_0$ (where $\nu_0$ is threshold frequency) is incident on a photosensitive material. If the frequency is halved and intensity is doubled, the photocurrent becomes:

(a) Doubled
(b) Halved
(c) Zero
(d) Quadrupled

Answer: (c)
New frequency $= 1.5\nu_0 / 2 = 0.75\nu_0$ . Since $\nu < \nu_0$ , no photoelectric emission occurs regardless of intensity.

10. Which one of the following has the largest de Broglie wavelength if all have the same velocity?

(a) Electron
(b) Proton
(c) Alpha particle
(d) Neutron

Answer: (a)
$\lambda = h/mv$ . For same velocity, $\lambda \propto 1/m$ . Since electron has the smallest mass, it has the largest wavelength.

Part 2: Assertion-Reason Questions

(A) Both A & R are true, R explains A.
(B) Both A & R are true, R does NOT explain A.
(C) A is true, R is false.
(D) A is false, R is true.

1. Assertion (A): Photoelectric effect demonstrates the particle nature of light.
Reason (R): The number of photoelectrons is proportional to the frequency of incident light.

Answer: (C)
Assertion is True. Reason is False. The number of photoelectrons depends on **intensity**, not frequency.

2. Assertion (A): The de Broglie wavelength of a moving cricket ball is not observable.
Reason (R): The mass of the cricket ball is large, making $\lambda$ extremely small ( $h/mv$ ).

Answer: (A)
Correct. The wavelength is so small (order of $10^{-34}$ m) that diffraction effects are negligible.

3. Assertion (A): Stopping potential depends on the distance of the source from the metal surface.
Reason (R): Changing distance changes intensity, and stopping potential depends on intensity.

Answer: (D)
Both are False. Stopping potential depends on **frequency**, not intensity (distance).

4. Assertion (A): Alkali metals are suitable for photoelectric emission with visible light.
Reason (R): Alkali metals have low work functions.

Answer: (A)
Correct. Low work function means lower threshold frequency, falling within the visible range.

5. Assertion (A): Mass of a moving photon is $h\nu/c^2$ .
Reason (R): Rest mass of a photon is zero.

Answer: (B)
Both statements are true. From $E = mc^2 = h\nu$ , effective mass is $h\nu/c^2$ . However, R is not the explanation for the derivation of relativistic mass.

6. Assertion (A): An electron microscope has higher resolution than an optical microscope.
Reason (R): The wavelength of electrons is much shorter than visible light.

Answer: (A)
Correct. Resolution is inversely proportional to wavelength. Smaller $\lambda$ of electrons allows resolving smaller details.

7. Assertion (A): Kinetic energy of photoelectrons emitted is variable, ranging from zero to a maximum.
Reason (R): Electrons lose variable amounts of energy in collisions while coming out of the metal surface.

Answer: (A)
Correct. Electrons deeper inside the metal lose energy before reaching the surface, while surface electrons possess maximum KE.

8. Assertion (A): If the frequency of incident light is doubled, the stopping potential becomes more than double.
Reason (R): $eV_0 = h\nu - \phi_0$ .

Answer: (A)
Let initial $eV_1 = h\nu - \phi_0$ . New $eV_2 = h(2\nu) - \phi_0 = 2(h\nu) - \phi_0 = 2(eV_1 + \phi_0) - \phi_0 = 2eV_1 + \phi_0$ . Thus $V_2 > 2V_1$ .

9. Assertion (A): A photocell converts light energy into electrical energy.
Reason (R): It works on the principle of thermoelectric emission.

Answer: (C)
Assertion is True. Reason is False. It works on the **photoelectric** effect, not thermoelectric.

10. Assertion (A): Matter waves travel faster than the speed of light.
Reason (R): Phase velocity of matter waves is $c^2/v$ , which is greater than $c$ .

Answer: (A)
Correct. The phase velocity ( $v_p = E/p = c^2/v$ ) is always $> c$ because particle velocity $v < c$ . Note: Group velocity is equal to particle velocity $v$ .

Part 3: Important Derivations & Theory

1. Write Einstein’s Photoelectric Equation and explain the terms. How does it explain the existence of threshold frequency?

Equation: $K_{max} = h\nu - \phi_0$
Or $eV_0 = h\nu - \phi_0$
Where $h\nu$ = Energy of incident photon, $\phi_0$ = Work function, $K_{max}$ = Max Kinetic Energy.
Explanation for Threshold Frequency:
For emission to happen, $K_{max}$ must be non-negative ( $K_{max} \ge 0$ ).
Therefore, $h\nu - \phi_0 \ge 0 \Rightarrow h\nu \ge \phi_0 \Rightarrow \nu \ge \phi_0/h$ .
This minimum frequency $\nu_0 = \phi_0/h$ is the threshold frequency.

2. Derive the expression for the de Broglie wavelength of an electron accelerated through a potential V.

Step 1: Kinetic energy gained by electron $= eV$ .
Step 2: Relation between KE ( $K$ ) and momentum ( $p$ ): $K = p^2 / 2m \Rightarrow p = \sqrt{2mK} = \sqrt{2meV}$ .
Step 3: de Broglie hypothesis: $\lambda = h/p$ .
Substituting $p$ : $\lambda = \frac{h}{\sqrt{2meV}}$ .
Putting constants ( $h, m, e$ ): $\lambda \approx \frac{1.227}{\sqrt{V}}$ nm.

3. Draw a graph showing the variation of stopping potential with frequency of incident radiation for two photosensitive materials having work functions $W_1$ and $W_2$ ( $W_1 > W_2$ ).

Features:

Both lines will be parallel straight lines with positive slope ( $h/e$ ).
The line for $W_1$ will have a larger x-intercept (threshold frequency $\nu_{01}$ ) than $W_2$ ( $\nu_{02}$ ), since $\nu_0 \propto W$ .
Hence, the graph for Material 1 starts at a higher frequency on the x-axis compared to Material 2.

Part 4: Numericals

1. Light of wavelength 400 nm is incident on a metal surface with work function 2.14 eV. Calculate the (a) energy of the photon in eV, (b) kinetic energy of emitted electrons, and (c) stopping potential.

(a) $E = \frac{1240}{\lambda (\text{nm})}$ eV $= \frac{1240}{400} = \textbf{3.1 eV}$ .
(b) $K_{max} = E - \phi_0 = 3.1 - 2.14 = \textbf{0.96 eV}$ .
(c) Stopping potential $V_0 = K_{max}/e = \textbf{0.96 V}$ .

2. An electron, an alpha particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?

Formula: $\lambda = \frac{h}{\sqrt{2mK}}$ .
Since $K$ is constant, $\lambda \propto \frac{1}{\sqrt{m}}$ .
Mass order: $m_\alpha > m_p > m_e$ .
Therefore, wavelength order: $\lambda_\alpha < \lambda_p < \lambda_e$ .
The Alpha particle has the shortest wavelength.

3. The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

$K_{max} = e \times V_0$ .
$K_{max} = e \times 1.5$ V $= \textbf{1.5 eV}$ .
(In Joules: $1.5 \times 1.6 \times 10^{-19} = 2.4 \times 10^{-19}$ J).

Part 5: Case Study

Case Study: Solar Cells & Photoelectric Effect
The photoelectric effect is the phenomenon where light energy forces electrons to be released from a material’s surface. While this usually happens in a vacuum tube in labs, a similar principle governs solar cells (photovoltaic cells). In semiconductors, photons knock electrons from the valence band to the conduction band, creating an electron-hole pair that generates current. However, unlike the external photoelectric effect where electrons leave the surface, this is an internal photoelectric effect. The Einstein equation $h\nu = \phi + K$ is fundamental to understanding energy transfer in these quantum processes.

Basic Photoelectric Process — Schematic of electron emission by photons

Does the photoelectric effect support the wave nature or particle nature of light?
Why do we not observe photoelectric emission from a piece of wood?
If green light causes emission from a metal, will violet light definitely cause emission? Why?
Define threshold frequency in the context of this phenomenon.

1. It supports the particle nature of light (photons), as wave theory could not explain instantaneous emission and threshold frequency.
2. Wood is an insulator with extremely tightly bound electrons (very high work function). Visible light photons do not have enough energy to eject them.
3. Yes. Violet light has a higher frequency (and energy) than green light. If green light ( $E_g$ ) is sufficient ( $E_g > \phi_0$ ), then violet ( $E_v > E_g > \phi_0$ ) will definitely cause emission.
4. Threshold frequency is the minimum frequency of incident radiation required to just eject an electron from the metal surface with zero kinetic energy.

← Back to Chapter Notes Go to CBSE Hub →

Dual Nature of Radiation and Matter: Question Bank

Dual Nature of Radiation and Matter: Question Bank

Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions

Part 3: Important Derivations & Theory

Part 4: Numericals

Part 5: Case Study

Editorial Team

Leave a Reply Cancel reply

Part 1: Multiple Choice Questions (1 Mark)

Part 2: Assertion-Reason Questions

Part 3: Important Derivations & Theory

Part 4: Numericals

Part 5: Case Study

Editorial Team

You might also like

CBSE Class 12 Physics: Alternating Current Previous Year Questions (2019–2025)

Transformer Working Principle – Class 12 Physics Explained (With Diagram & Formula)

Power in AC Circuits: Why Average Power ≠ V₀I₀ (Real, Apparent & Power Factor)

Leave a Reply Cancel reply