CHAPTER 11: DUAL NATURE OF RADIATION AND MATTER
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: Q1–14 are MCQs and Q15–16 are Assertion-Reasoning questions of 1 mark each.
- Section B: Q17–21 are Very Short Answer questions of 2 marks each.
- Section C: Q22–28 are Short Answer questions of 3 marks each.
- Section D: Q29–30 are Case Study-based questions of 4 marks each.
- Section E: Q31–33 are Long Answer questions of 5 marks each.
- Physical Constants: $\displaystyle {h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}}$, $\displaystyle {c = 3 \times 10^8\text{ m/s}}$, $\displaystyle {e = 1.6 \times 10^{-19}\text{ C}}$, $\displaystyle {m_e = 9.11 \times 10^{-31}\text{ kg}}$, $\displaystyle {m_p = 1.67 \times 10^{-27}\text{ kg}}$, $\displaystyle {hc = 12400\text{ eV}\cdot\text{\AA}}$.
SECTION A (16 Marks)
Q.1. The work function of a metal is $\displaystyle {4.2\text{ eV}}$. Its threshold wavelength is nearly:
Explanation: $\displaystyle {\lambda_0 = {\frac{{hc}}{{\phi_0}}} = {\frac{{12400\text{ eV}\cdot\text{\AA}}}{{4.2\text{ eV}}}} \approx 2952.4\text{ \AA} \approx 2955\text{ \AA}}$.
Q.2. Photoelectric current is directly proportional to the:
Explanation: Higher intensity implies a greater number of incident photons per second, which ejects a proportionally higher number of photoelectrons per second above threshold frequency.
Q.3. The slope of the stopping potential ($\displaystyle {V_0}$) versus frequency ($\displaystyle {\nu}$) graph gives:
Explanation: From Einstein's equation, $\displaystyle {e V_0 = h \nu - \phi_0 \implies V_0 = {\left({\frac{{h}}{{e}}}\right)} \nu - {\frac{{\phi_0}}{{e}}}}$. Comparing with $\displaystyle {y = m x + c}$, the slope is $\displaystyle {m = h/e}$.
Q.4. The de Broglie wavelength of a particle of momentum $\displaystyle {p}$ is given by:
Explanation: According to de Broglie's matter wave hypothesis, wavelength is inversely proportional to momentum: $\displaystyle {\lambda = h/p = h/(m v)}$.
Q.5. An electron and a proton are accelerated through the same potential difference $\displaystyle {V}$. The ratio $\displaystyle {\lambda_e / \lambda_p}$ of their de Broglie wavelengths is:
Explanation: For charge $\displaystyle {q}$ accelerated through $\displaystyle {V}$, $\displaystyle {\lambda = {\frac{{h}}{{\sqrt{2 m q V}}}}}$. Since $\displaystyle {q_e = q_p = e}$, $\displaystyle {\lambda \propto 1/\sqrt{m} \implies \lambda_e / \lambda_p = \sqrt{m_p / m_e}}$.
Q.6. Einstein's photoelectric equation expressing conservation of energy is:
Explanation: Incident photon energy $\displaystyle {h\nu}$ is spent in two parts: work function $\displaystyle {\phi_0}$ (to free the electron) and maximum kinetic energy $\displaystyle {KE_{\max}}$ of the emitted electron.
Q.7. Which of the following elementary particles has zero rest mass?
Explanation: A photon is a quantum of electromagnetic energy and moves always at speed $\displaystyle {c}$ in vacuum; its rest mass $\displaystyle {m_0 = 0}$.
Q.8. The Davisson–Germer experiment historically demonstrated:
Explanation: Observing electron diffraction from a nickel crystal target verified de Broglie's matter wave hypothesis quantitatively.
Q.9. The relativistic/quantum momentum $\displaystyle {p}$ of a photon of energy $\displaystyle {E}$ is:
Explanation: From mass-energy-momentum relation $\displaystyle {E = p c \implies p = E/c = h\nu / c = h / \lambda}$.
Q.10. If frequency of incident light is increased above threshold while keeping intensity constant, the number of photoelectrons emitted per second:
Explanation: Number of emitted photoelectrons per second depends strictly on the rate of arriving photons (intensity), not on their frequency once $\displaystyle {\nu \ge \nu_0}$.
Q.11. For photoelectric emission to occur from a photosensitive surface, the frequency $\displaystyle {\nu}$ of incident light must be:
Explanation: Emission requires photon energy $\displaystyle {h\nu \ge \phi_0 = h\nu_0 \implies \nu \ge \nu_0}$.
Q.12. A photocell operates fundamentally on the principle of:
Q.13. The de Broglie wavelength associated with a ball of mass $\displaystyle {1\text{ kg}}$ moving at speed $\displaystyle {1\text{ m/s}}$ is of the order of:
Explanation: $\displaystyle {\lambda = {\frac{{h}}{{m v}}} = {\frac{{6.63 \times 10^{-34}}}{{1 \times 1}}} = 6.63 \times 10^{-34}\text{ m} \sim 10^{-34}\text{ m}}$.
Q.14. Two metals A and B have threshold frequencies $\displaystyle {\nu_0}$ and $\displaystyle {2\nu_0}$ respectively. The work function of A compared to B is:
Explanation: $\displaystyle {\phi_A = h \nu_0}$ and $\displaystyle {\phi_B = h (2\nu_0) = 2 \phi_A \implies \phi_A / \phi_B = 1/2}$.
Directions for Q.15 & Q.16:
(a) Both Assertion and Reason are true and Reason is correct explanation of A.
(b) Both Assertion and Reason are true but Reason is NOT correct explanation of A.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.
Q.15. Assertion (A): Photoelectric effect cannot be explained on the basis of the wave theory of light.
Reason (R): According to wave theory, energy is distributed continuously over the wavefront and does not depend on frequency.
Explanation: Continuous wave energy prediction fails to account for instantaneous emission and the existence of a threshold frequency.
Q.16. Assertion (A): The de Broglie wavelength of a moving electron decreases as its speed increases.
Reason (R): The de Broglie wavelength is directly proportional to the momentum of the particle.
Explanation: Assertion is true ($\lambda \propto 1/v$), but Reason is false because de Broglie wavelength is inversely proportional to momentum ($\lambda = h/p$).
SECTION B (10 Marks)
Q.17. State two important features of the photon picture of electromagnetic radiation.
1. Radiation interacts with matter as discrete packets of energy called photons, each possessing energy $\displaystyle {E = h\nu}$ and momentum $\displaystyle {p = h/\lambda}$.
2. Photons are electrically neutral and unaffected by electric or magnetic fields; in photon-particle collisions, total energy and momentum are conserved.
Q.18. The threshold wavelength for photoelectric emission from a metal is $\displaystyle {5200\text{ \AA}}$. Calculate the work function of the metal in $\displaystyle {\text{eV}}$.
Q.19. Why does photoelectric current increase with an increase in intensity of incident radiation, while stopping potential remains unchanged? Explain.
Q.20. Derive the expression for the de Broglie wavelength of an electron accelerated from rest through a potential difference $\displaystyle {V}$.
Momentum: $\displaystyle {p = \sqrt{2 m_e K} = \sqrt{2 m_e e V}}$.
de Broglie Wavelength: $\displaystyle {\lambda = {\frac{{h}}{{p}}} = {\frac{{h}}{{\sqrt{2 m_e e V}}}}}$.
Substituting standard numerical constants ($\displaystyle {h, m_e, e}$): $\displaystyle {\lambda = {\frac{{12.27}}{{\sqrt{V}}}}\text{ \AA} = {\frac{{1.227}}{{\sqrt{V}}}}\text{ nm}}$.
Q.21. Sketch a graph showing the variation of stopping potential ($\displaystyle {V_0}$) with frequency ($\displaystyle {\nu}$) of incident radiation for two photosensitive materials M1 and M2 ($\displaystyle {\phi_1 < \phi_2}$). What physical quantity does the slope represent?
SECTION C (21 Marks)
Q.22. State Einstein's photoelectric equation. Using it, explain: (i) why maximum kinetic energy is independent of intensity, and (ii) why there exists a threshold frequency for a given metal.
(i) Each photon interacts with a single electron transferring fixed energy $\displaystyle {h\nu}$. Increasing intensity increases photon quantity but not energy per photon, leaving $\displaystyle {KE_{\max}}$ unchanged.
(ii) For emission, $\displaystyle {KE_{\max} \ge 0 \implies h\nu \ge \phi_0 \implies \nu \ge \phi_0/h}$. If $\displaystyle {\nu < \nu_0 = \phi_0/h}$, emission cannot occur regardless of light intensity.
Q.23. Light of wavelength $\displaystyle {4000\text{ \AA}}$ falls on a metal surface of work function $\displaystyle {2.0\text{ eV}}$. Calculate: (i) threshold frequency, and (ii) maximum kinetic energy of emitted photoelectrons.
(i) $\displaystyle {\nu_0 = {\frac{{\phi_0}}{{h}}} = {\frac{{2.0 \times 1.6 \times 10^{-19}\text{ J}}}{{6.63 \times 10^{-34}\text{ J}\cdot\text{s}}}} \approx 4.83 \times 10^{14}\text{ Hz}}$.
(ii) Incident photon energy $\displaystyle {E = {\frac{{hc}}{{\lambda}}} = {\frac{{12400}}{{4000}}} = 3.1\text{ eV}}$.
$\displaystyle {KE_{\max} = E - \phi_0 = 3.1\text{ eV} - 2.0\text{ eV} = 1.1\text{ eV} = 1.76 \times 10^{-19}\text{ J}}$.
Q.24. Describe briefly, with a labeled diagram, the Davisson–Germer experiment. State its historical significance in quantum physics.
Q.25. An electron and a photon each possess energy $\displaystyle {100\text{ eV}}$. Calculate and compare their associated de Broglie wavelengths.
Photon ($\displaystyle {E = 100\text{ eV}}$): $\displaystyle {\lambda_p = {\frac{{hc}}{{E}}} = {\frac{{12400\text{ eV}\cdot\text{\AA}}}{{100\text{ eV}}}} = 124\text{ \AA}}$.
Electron ($\displaystyle {V = 100\text{ V}}$): $\displaystyle {\lambda_e = {\frac{{12.27}}{{\sqrt{100}}}}\text{ \AA} = 1.227\text{ \AA}}$.
Comparison: $\displaystyle {\lambda_p / \lambda_e = 124 / 1.227 \approx 101}$. The photon's wavelength is approximately 100 times larger than the electron's for the same energy.
Q.26. The graph below shows variation of photoelectric current with collector plate potential for two intensities $\displaystyle {I_1}$ and $\displaystyle {I_2}$ ($\displaystyle {I_2 > I_1}$) at constant frequency. Explain why saturation currents differ while stopping potential is identical.
Q.27. Starting from the photon relation $\displaystyle {E = h\nu}$, derive the expression for photon momentum and show its consistency with de Broglie relation $\displaystyle {\lambda = h/p}$.
Q.28. A metal surface is illuminated successively by light of wavelengths $\displaystyle {300\text{ nm}}$ and $\displaystyle {400\text{ nm}}$. The maximum speeds of emitted photoelectrons are $\displaystyle {v_1}$ and $\displaystyle {v_2}$ respectively, with $\displaystyle {v_1 = 2 v_2}$. Calculate the work function of the metal.
Since $\displaystyle {v_1 = 2 v_2}$, $\displaystyle {KE_1 = 4 KE_2}$.
$\displaystyle {E_1 - \phi_0 = 4(E_2 - \phi_0) \implies 4.133 - \phi_0 = 4(3.10) - 4\phi_0 = 12.40 - 4\phi_0}$.
$\displaystyle {3\phi_0 = 12.40 - 4.133 = 8.267 \implies \phi_0 \approx 2.76\text{ eV}}$.
SECTION D: CASE STUDY QUESTIONS (8 Marks)
Q.29. Case Study 1: Photoelectric Effect & Stopping Potential Analysis.
In an experiment measuring stopping potential $\displaystyle {V_0}$ against frequency $\displaystyle {\nu}$ for a photosensitive metal, recorded values were:
$\displaystyle {\nu \,(\times 10^{14}\text{ Hz}): 5.0 \quad 6.0 \quad 7.0 \quad 8.0}$
$\displaystyle {V_0 \,(\text{V}): 0.00 \quad 0.41 \quad 0.83 \quad 1.24}$
(i) Threshold frequency $\displaystyle {\nu_0}$ of metal: (a) $\displaystyle {5.0 \times 10^{14}\text{ Hz}}$ (b) $\displaystyle {6.0 \times 10^{14}\text{ Hz}}$
(ii) Slope $\displaystyle {h/e}$ calculated from data: (a) $\displaystyle {4.13 \times 10^{-15}\text{ V}\cdot\text{s}}$ (b) $\displaystyle {1.6 \times 10^{-19}\text{ V}\cdot\text{s}}$
(iii) Work function $\displaystyle {\phi_0}$ in eV: (a) $\displaystyle {2.07\text{ eV}}$ (b) $\displaystyle {4.14\text{ eV}}$
(iv) If frequency increases beyond $\displaystyle {8 \times 10^{14}\text{ Hz}}$ at constant intensity: (a) $\displaystyle {V_0}$ increases linearly, current constant (b) Both increase
(i) (a) $\displaystyle {5.0 \times 10^{14}\text{ Hz}}$ (where $\displaystyle {V_0 = 0}$).
(ii) (a) $\displaystyle {\text{Slope} = \Delta V_0 / \Delta \nu = 1.24 / (3.0 \times 10^{14}) = 4.13 \times 10^{-15}\text{ V}\cdot\text{s}}$.
(iii) (a) $\displaystyle {\phi_0 = (h/e)\nu_0 = 4.13 \times 10^{-15} \times 5.0 \times 10^{14} = 2.07\text{ eV}}$.
(iv) (a) Stopping potential increases linearly while saturation current remains constant.
Q.30. Case Study 2: Matter Waves & de Broglie Wavelength.
Louis de Broglie proposed that moving particles exhibit wave properties with wavelength $\displaystyle {\lambda = h/p}$.
(i) Expression for de Broglie wavelength of particle of mass $\displaystyle {m}$ and kinetic energy $\displaystyle {K}$: (a) $\displaystyle {h/\sqrt{2mK}}$ (b) $\displaystyle {\sqrt{2mK}/h}$
(ii) If electron and proton have equal kinetic energy, longer wavelength belongs to: (a) Electron (b) Proton
(iii) de Broglie wavelength of electron accelerated through $\displaystyle {150\text{ V}}$: (a) $\displaystyle {1.00\text{ \AA}}$ (b) $\displaystyle {12.27\text{ \AA}}$
(i) (a) $\displaystyle {\lambda = h/\sqrt{2mK}}$.
(ii) (a) Electron ($\displaystyle {\lambda \propto 1/\sqrt{m}}$, smaller mass gives longer wavelength).
(iii) (a) $\displaystyle {\lambda = 12.27 / \sqrt{150} \approx 1.00\text{ \AA}}$.
SECTION E (15 Marks)
Q.31. (a) State the laws of photoelectric emission. Explain how Einstein's equation accounts for each law.
(b) Draw graphs showing variation of: (i) photoelectric current with intensity at constant frequency, and (ii) stopping potential with frequency.
Derive Einstein's photoelectric equation from photon-electron interaction and sketch the corresponding experimental graphs.
Laws & Explanation:
1. Photocurrent $\displaystyle {\propto}$ Intensity: More photons per second eject more electrons per second.
2. Existence of Threshold Frequency: Emission requires $\displaystyle {h\nu \ge \phi_0 \implies \nu \ge \nu_0 = \phi_0/h}$.
3. $\displaystyle {KE_{\max} \propto \nu}$, independent of intensity: Energy per photon is $\displaystyle {h\nu}$, independent of photon count.
4. Instantaneous emission ($\displaystyle {\sim 10^{-9}\text{ s}}$): Single photon-electron collision transfers energy instantaneously.
Graphs: (i) Linear line through origin for current vs intensity. (ii) Straight line with positive slope $\displaystyle {h/e}$ cutting frequency axis at $\displaystyle {\nu_0}$.
Q.32. Work function of caesium is $\displaystyle {2.14\text{ eV}}$.
(i) Find threshold frequency.
(ii) If light of frequency $\displaystyle {6 \times 10^{14}\text{ Hz}}$ is incident, calculate stopping potential and maximum speed of emitted photoelectrons.
(i) Derive $\displaystyle {\lambda = h/\sqrt{2meV}}$. (ii) Calculate wavelength for $\displaystyle {100\text{ V}}$ electron. (iii) Compare with proton accelerated through $\displaystyle {100\text{ V}}$.
(i) $\displaystyle {\nu_0 = \phi_0 / h = (2.14 \times 1.6 \times 10^{-19}) / (6.63 \times 10^{-34}) = 5.16 \times 10^{14}\text{ Hz}}$.
(ii) Incident photon energy $\displaystyle {E = h\nu = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 3.978 \times 10^{-19}\text{ J} = 2.486\text{ eV}}$.
$\displaystyle {KE_{\max} = 2.486 - 2.14 = 0.346\text{ eV} \implies \text{Stopping Potential } V_0 = 0.346\text{ V}}$.
$\displaystyle {v_{\max} = \sqrt{2 KE_{\max} / m_e} = \sqrt{(2 \times 0.346 \times 1.6 \times 10^{-19}) / (9.11 \times 10^{-31})} \approx 3.49 \times 10^5\text{ m/s}}$.
OR Part: (ii) $\displaystyle {\lambda_e = 12.27/\sqrt{100} = 1.227\text{ \AA}}$. (iii) $\displaystyle {\lambda_p = \lambda_e \sqrt{m_e/m_p} = 1.227 / \sqrt{1836} \approx 0.0286\text{ \AA}}$.
Q.33. What is meant by wave–particle duality? Describe the Davisson–Germer experiment with a diagram and explain how it confirms electron wave nature.
A monochromatic beam of wavelength $\displaystyle {663\text{ nm}}$ and power $\displaystyle {3\text{ mW}}$ is incident on a metal surface of work function $\displaystyle {1.5\text{ eV}}$. Assuming 1% photon efficiency, find photoelectrons emitted per second and maximum kinetic energy.
OR Part Calculations:
Photon Energy: $\displaystyle {E = hc/\lambda = (6.63 \times 10^{-34} \times 3 \times 10^8) / (663 \times 10^{-9}) = 3 \times 10^{-19}\text{ J} = 1.875\text{ eV}}$.
Photons per second: $\displaystyle {N_p = P / E = 3 \times 10^{-3} / (3 \times 10^{-19}) = 10^{16}\text{ s}^{-1}}$.
Photoelectrons emitted per second (1% efficiency): $\displaystyle {N_e = 0.01 \times 10^{16} = 10^{14}\text{ s}^{-1}}$.
Maximum Kinetic Energy: $\displaystyle {KE_{\max} = E - \phi_0 = 1.875\text{ eV} - 1.5\text{ eV} = 0.375\text{ eV} = 6 \times 10^{-20}\text{ J}}$.
ASSIGNMENT – 2 (MCQ PRACTICE)
1. Work function of a metal surface is defined as the minimum energy required to:
2. If accelerating potential of an electron is increased by 4 times, its de Broglie wavelength changes by a factor of:
3. Which particle has the largest de Broglie wavelength if all are moving with the same kinetic energy?
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): In photoelectric emission, all emitted electrons do not possess the same kinetic energy.
Reason (R): Electrons emitted from deeper atomic layers lose varying amounts of energy in collisions before escaping the surface.
2. Assertion (A): Stopping potential depends only on frequency of incident light and not on its intensity.
Reason (R): Maximum kinetic energy $\displaystyle {KE_{\max} = h\nu - \phi_0}$ depends on photon energy, not on photon quantity.
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study: Millikan's Photoelectric Verification
Millikan verified Einstein's equation by measuring stopping potential $\displaystyle {V_0}$ against frequency $\displaystyle {\nu}$ for various metals.
1. The y-intercept of the $\displaystyle {V_0}$ vs $\displaystyle {\nu}$ graph represents: (a) $\displaystyle {-\phi_0/e}$ (b) $\displaystyle {h/e}$
2. The slope of the $\displaystyle {V_0}$ vs $\displaystyle {\nu}$ graph equals: (a) $\displaystyle {h/e}$ (b) $\displaystyle {\phi_0}$
3. If work function is $\displaystyle {2.5\text{ eV}}$, threshold wavelength is: (a) $\displaystyle {4960\text{ \AA}}$ (b) $\displaystyle {2480\text{ \AA}}$
ASSIGNMENT – 5 (FORMULAE & CONCEPTUAL QUESTIONS)
CORE FORMULAE SUMMARY
General: $\displaystyle {\lambda = {\frac{{h}}{{p}}} = {\frac{{h}}{{m v}}} = {\frac{{h}}{{\sqrt{2 m K}}}}}$
Accelerated Electron: $\displaystyle {\lambda_e = {\frac{{h}}{{\sqrt{2 m_e e V}}}} = {\frac{{12.27}}{{\sqrt{V}}}}\text{ \AA} = {\frac{{1.227}}{{\sqrt{V}}}}\text{ nm}}$
CONCEPTUAL SHORT QUESTIONS
1. Why is wave theory incapable of explaining the existence of threshold frequency in photoelectric emission?
2. Why is the wave nature of a moving cricket ball not observable in daily life?
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