Every Book Back multiple-choice question from Dual Nature of Radiation and Matter (12th Standard Physics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
The wavelength \(\lambda_e\) of an electron and \(\lambda_p\) of a photon of same energy E are related by
- A. \(\lambda_p \propto \lambda_e\)
- B. \(\lambda_p \propto \sqrt{\lambda_e}\)
- C. \(\lambda_p \propto \frac{1}{\sqrt{\lambda_e}}\)
- D. \(\lambda_p \propto \lambda_e^2\)Correct
Explanation. Energy of a photon is inversely proportional to its wavelength, while for an electron, energy is inversely proportional to the square of its de Broglie wavelength. Therefore, the photon wavelength varies with the square of the electron wavelength.
Q2
In an electron microscope, the electrons are accelerated by a voltage of 14 kV. If the voltage is changed to 224 kV, then the de Broglie wavelength associated with the electrons would
- A. increase by 2 times
- B. decrease by 2 times
- C. decrease by 4 timesCorrect
- D. increase by 4 times
Explanation. The de Broglie wavelength is inversely proportional to the square root of the accelerating potential. Since the voltage increases from 14 kV to 224 kV (a factor of 16), the wavelength decreases by the square root of 16, which is 4.
Q3
The wave associated with a moving particle of mass \(3 \times 10^{-6}\) g has the same wavelength as an electron moving with a velocity \(6 \times 10^6\) m s\(^{-1}\). The velocity of the particle is
- A. \(1.82 \times 10^{-18}\) m s\(^{-1}\)
- B. \(9 \times 10^{-2}\) m s\(^{-1}\)
- C. \(3 \times 10^{-31}\) m s\(^{-1}\)
- D. \(1.82 \times 10^{-15}\) m s\(^{-1}\)Correct
Explanation. Since wavelengths are equal, their momenta must be equal (\(mv = m_e v_e\)). By converting units to SI and solving for the particle velocity using the known mass and velocity of the electron, we get \(1.82 \times 10^{-15}\) m s\(^{-1}\).
Q4
When a metallic surface is illuminated with radiation of wavelength \(\lambda\), the stopping potential is V. If the same surface is illuminated with radiation of wavelength \(2\lambda\), the stopping potential is \(V/4\). The threshold wavelength for the metallic surface is
- A. \(4\lambda\)
- B. \(5\lambda\)
- C. \(\frac{5}{2}\lambda\)
- D. \(3\lambda\)Correct
Explanation. Using Einstein's photoelectric equation for both cases, we set up two equations relating energy, work function, and stopping potential. Solving these simultaneous equations for the threshold wavelength yields a value of three times the initial incident wavelength.
Q5
If a light of wavelength 330 nm is incident on a metal with work function 3.55 eV, the electrons are emitted. Then the wavelength of the wave associated with the emitted electron is (Take h = \(6.6 \times 10^{-34}\) Js)
- A. \(< 2.75 \times 10^{-9}\) m
- B. \(\ge 2.75 \times 10^{-9}\) mCorrect
- C. \(\le 2.75 \times 10^{-12}\) m
- D. \(< 2.75 \times 10^{-10}\) m
Explanation. The incident photon energy is calculated and the work function is subtracted to find the maximum kinetic energy. This maximum kinetic energy defines the minimum possible de Broglie wavelength. The wavelength of emitted electrons will therefore be greater than or equal to this minimum.
Q6
A photoelectric surface is illuminated successively by monochromatic light of wavelength \(\lambda\) and \(\lambda/2\). If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the material is
- A. \(\frac{hc}{\lambda}\)
- B. \(\frac{2hc}{\lambda}\)
- C. \(\frac{hc}{3\lambda}\)
- D. \(\frac{hc}{2\lambda}\)Correct
Explanation. Einstein's equation states \(K = hc/\lambda - \phi\). Given \(K_2 = 3K_1\) for wavelengths \(\lambda\) and \(\lambda/2\), substituting these into the equation and solving for the work function \(\phi\) results in \(hc/2\lambda\).
Q7
In photoelectric emission, a radiation whose frequency is 4 times threshold frequency of a certain metal is incident on the metal. Then the maximum possible velocity of the emitted electron will be
- A. \(\sqrt{\frac{hv_0}{m}}\)
- B. \(\sqrt{\frac{6hv_0}{m}}\)Correct
- C. \(\sqrt{\frac{2hv_0}{m}}\)
- D. \(\sqrt{\frac{hv_0}{2m}}\)
Explanation. Max kinetic energy is \(h(4v_0) - hv_0 = 3hv_0\). Setting this equal to \(1/2 mv^2\) and solving for velocity \(v\) gives the square root of \(6hv_0/m\).
Q8
Two radiations with photon energies 0.9 eV and 3.3 eV respectively are falling on a metallic surface successively. If the work function of the metal is 0.6 eV, then the ratio of maximum speeds of emitted electrons in the two cases will be
- A. 1:4
- B. 1:3Correct
- C. 1:1
- D. 1:9
Explanation. Kinetic energies are \(0.9-0.6 = 0.3\) eV and \(3.3-0.6 = 2.7\) eV. Since velocity is proportional to the square root of kinetic energy, the ratio of speeds is \(\sqrt{0.3/2.7} = \sqrt{1/9}\), which is 1:3.
Q9
A light source of wavelength 520 nm emits \(1.04 \times 10^{15}\) photons per second while the second source of 460 nm produces \(1.38 \times 10^{15}\) photons per second. Then the ratio of power of second source to that of first source is
- A. 1.00
- B. 1.02
- C. 1.5Correct
- D. 0.98
Explanation. Power is the product of photon emission rate and the energy per photon (\(hc/\lambda\)). The ratio is calculated by comparing \((n_2 \cdot \lambda_1) / (n_1 \cdot \lambda_2)\). Using the given values, the resulting power ratio is 1.5.
Q10
The threshold wavelength for a surface \(\lambda_0\) is 6000 Å. The maximum kinetic energy of photoelectrons emitted by radiation of wavelength 4000 Å is
- A. 1.03 eVCorrect
- B. 2.06 eV
- C. 3.09 eV
- D. 0.52 eV
Explanation. Maximum kinetic energy is found using \(hc(1/\lambda - 1/\lambda_0)\). Substituting \(hc = 12400\) eVÅ and the given wavelengths, the result is approximately 1.03 eV.
Q11
When a photon of energy 7 eV is incident on a metal surface, the maximum kinetic energy of photoelectron is 4 eV. The stopping potential is
- A. 7 V
- B. 4 VCorrect
- C. 3 V
- D. 11 V
Explanation. The stopping potential is numerically equal to the maximum kinetic energy expressed in electron-volts. Since the max kinetic energy is 4 eV, the stopping potential required to halt these electrons is 4 V.
Q12
If the energy of the photon is increased by a factor of 4, then its momentum is increased by a factor of
- A. 4
- B. 2
- C. 4Correct
- D. 2
Explanation. For a photon, energy and momentum are directly proportional (\(E = pc\)). Thus, if energy increases by a factor of 4, the momentum must also increase by the same factor of 4.
Q13
Photons of wavelength \(\lambda\) are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius R by a perpendicular magnetic field having magnitude B. The work function of the metal is
- A. \(\frac{hc}{\lambda} - \frac{m_e c^2 + e^2 B^2 R^2}{2m_e}\)
- B. \(\frac{hc}{\lambda} + \frac{(eBR)^2}{2m_e}\)
- C. \(\frac{hc}{\lambda} - \frac{m_e c^2 - e^2 B^2 R^2}{2m_e}\)
- D. \(\frac{hc}{\lambda} - \frac{(eBR)^2}{2m_e}\)Correct
Explanation. The maximum kinetic energy \(K\) of an electron in a magnetic field is \((eBR)^2/2m\). The work function is the incident photon energy minus this kinetic energy, leading to the formula in the fourth option.
Q14
The work functions for metals A, B and C are 1.92 eV, 2.0 eV and 5.0 eV respectively. The metal/metals which will emit photoelectrons for a radiation of wavelength 4100 Å is/are
- A. A only
- B. both A and BCorrect
- C. all these metals
- D. none
Explanation. The energy of incident radiation is \(12400/4100 \approx 3.02\) eV. Photoemission occurs only if this energy exceeds the work function. This condition is met for metals A and B, but not for metal C.
Q15
Emission of electrons by the absorption of heat energy is called.........emission.
- A. photoelectric
- B. field
- C. thermionicCorrect
- D. secondary
Explanation. Thermionic emission is the process where free electrons are liberated from a metal surface by supplying thermal energy, allowing them to overcome the attractive pull of the atomic nuclei.
