A metal surface of work function \( \phi_0 \) is illuminated with monochromatic light in which each photon has an energy \( 3\phi_0 \). Consider the following statements:
(A) The maximum kinetic energy of the emitted photoelectrons is \( 2\phi_0 \).
(B) The threshold wavelength of the metal is one-third of the wavelength of the incident light.
(C) If the wavelength of the incident light is halved, the maximum kinetic energy of the photoelectrons becomes \( 5\phi_0 \).
(D) If the intensity of the incident light is doubled, the maximum kinetic energy of the photoelectrons becomes \( 4\phi_0 \).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
Each photon gives all its energy to one electron. By Einstein’s photoelectric equation the fastest electrons leave with \( K_{max}=E-\phi_0 \), where \( E=hc/\lambda \) is the photon energy. The threshold wavelength is the wavelength whose photons carry just \( \phi_0 \), that is \( \lambda_0=hc/\phi_0 \).
(A) The maximum kinetic energy of the emitted photoelectrons is \( 2\phi_0 \). — True. With \( E=3\phi_0 \):
\( \begin{aligned} K_{max} &= 3\phi_0-\phi_0 \\ &= 2\phi_0 \end{aligned} \)
(B) The threshold wavelength of the metal is one-third of the wavelength of the incident light. — False. The incident light has \( \lambda=hc/3\phi_0 \), while \( \lambda_0=hc/\phi_0 \). So \( \lambda_0=3\lambda \). The threshold wavelength is three times the incident wavelength, as it must be for emission to take place.
(C) If the wavelength of the incident light is halved, the maximum kinetic energy of the photoelectrons becomes \( 5\phi_0 \). — True. Halving the wavelength doubles the photon energy to \( 6\phi_0 \). The work function stays the same:
\( \begin{aligned} K_{max} &= 6\phi_0-\phi_0 \\ &= 5\phi_0 \end{aligned} \)
(D) If the intensity of the incident light is doubled, the maximum kinetic energy of the photoelectrons becomes \( 4\phi_0 \). — False. A more intense beam of the same wavelength has more photons per second, but each photon still carries \( 3\phi_0 \). More electrons come out, and the maximum kinetic energy stays \( 2\phi_0 \).
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function φ0), the electron is emitted with maximum kinetic energy”
\( K_{max}=h\nu-\phi_0 \)\( \qquad (11.2) \)
“More tightly bound electrons will emerge with kinetic energies less than the maximum value. Note that the intensity of light of a given frequency is determined by the number of photons incident per second. Increasing the intensity will increase the number of emitted electrons per second. However, the maximum kinetic energy of the emitted photoelectrons is determined by the energy of each photon.”
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… Thus, there exists a threshold frequency ν0 (= φ0/h) for the metal surface, below which no photoelectric emission is possible, no matter how intense the incident radiation may be or how long it falls on the surface.”