Class 12 Dual Nature of Radiation and Matter Mock Test (CBSE)

Topics in Dual Nature of Radiation and Matter

• Electron emission
• Photoelectric effect
• Experimental study of photoelectric effect
• Photoelectric effect and wave theory of light
• Einstein's photoelectric equation
• Particle nature of light: the photon
• Wave nature of matter
• de Broglie wavelength

Dual Nature of Radiation and Matter practice questions with answers

1. Q1Multiple choice·1 mark
   An electron and a proton are accelerated through the same potential difference $V$. The ratio of their de Broglie wavelengths $(\lambda_e / \lambda_p)$ is:

   1. (a)$\sqrt{\frac{m_p}{m_e}}$
   2. (b)$\frac{m_e}{m_p}$
   3. (c)$\sqrt{\frac{m_e}{m_p}}$
   4. (d)$1$
   Show answer and step marks

   1. Correct option: (a) $\sqrt{\frac{m_p}{m_e}}$+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   Which of the following statements is correct regarding matter waves?

   1. (a)They are electromagnetic waves produced by accelerated charges.
   2. (b)They travel with a speed equal to the velocity of light in vacuum under all conditions.
   3. (c)They can be produced only by uncharged particles like neutrons.
   4. (d)Their wavelength is inversely proportional to the momentum of the moving particle.
   Show answer and step marks

   1. Correct option: (d) Their wavelength is inversely proportional to the momentum of the particle.+1

   Answer: (d)
3. Q3Short answer·2 marks
   Define the work function of a metal. Name any two methods by which electrons can be emitted from a metal surface.
   Show answer and step marks

   1. Work function: the minimum energy needed by an electron to escape from the metal surface.+1
   2. First method, any one of: thermionic emission; field emission; photoelectric emission; or any other valid method (e.g. secondary emission).+0.5
   3. Second method, a different one from the same list.+0.5

   Answer: Work function: the minimum energy an electron needs to escape from the metal surface. Any two of: thermionic emission, field emission, photoelectric emission (or secondary emission).
4. Q4Short answer·2 marks
   State any two methods of electron emission from a metal surface and briefly explain the process for each.
   Show answer and step marks

   1. First method, any one of: thermionic emission; field emission; photoelectric emission; or any other valid method (e.g. secondary emission).+0.5
   2. Its process: thermionic, heating gives free electrons enough thermal energy to escape; field, a very strong electric field (about $10^8\text{ V m}^{-1}$) pulls electrons out; photoelectric, light of suitable frequency gives electrons the energy to escape; secondary, fast electrons striking the surface knock electrons out.+0.5
   3. Second method, a different one from the same list.+0.5
   4. Its process, as described above.+0.5

   Answer: Any two of: thermionic emission (heating supplies the energy), field emission (a very strong electric field pulls electrons out), photoelectric emission (light of suitable frequency supplies the energy), secondary emission (fast electrons knock electrons out).
5. Q5Short answer·3 marks
   In an experimental study of the photoelectric effect, a graph is plotted between the stopping potential ($V_0$) and the frequency ($\nu$) of the incident radiation. (a) How does the slope of this graph depend on the nature of the emitting metal surface? (b) What physical quantity does the intercept of this graph on the frequency axis represent? (c) If the frequency of the incident light is doubled, does the stopping potential double? Give reason.
   Show answer and step marks

   1. Using $eV_0 = h\nu - \phi_0$, the slope is $\frac{h}{e}$, which is a universal constant and independent of the nature of the metal.+1
   2. The intercept on the frequency axis represents the threshold frequency ($\nu_0$) of the metal.+1
   3. No, the stopping potential does not double because $V_0 = \frac{h\nu}{e} - \frac{\phi_0}{e}$; doubling $\nu$ makes the new stopping potential more than double due to the constant work function term.+1

   Answer: (a) Slope is independent of the material ($\frac{h}{e}$) (b) Threshold frequency (c) No, it becomes more than double.
6. Q6Short answer·3 marks
   Monochromatic light of frequency $1.0 \times 10^{15} \text{ Hz}$ is incident on a metal surface. The threshold frequency for the metal is $5.0 \times 10^{14} \text{ Hz}$. (a) Calculate the maximum kinetic energy of the emitted photoelectrons in Joules. (b) State the effect on the photoelectric current if the intensity of the incident light is increased while keeping the frequency constant.
   Show answer and step marks

   1. State Einstein's photoelectric equation for maximum kinetic energy: $K_{\max} = h(\nu - \nu_0)$.+1
   2. Substitute the given values: $K_{\max} = 6.63 \times 10^{-34} \text{ J s} \times (1.0 \times 10^{15} - 5.0 \times 10^{14}) \text{ Hz} = 3.315 \times 10^{-19} \text{ J}$.+1
   3. State that the photoelectric current increases linearly with an increase in the intensity of incident light.+1

   Answer: Maximum kinetic energy is $3.315 \times 10^{-19} \text{ J}$ and photoelectric current increases linearly with intensity.

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