Radiation of frequency \( 2v_{0} \) is incident on a metal with threshold frequency \( v_{0} \). The correct statement of the following is ___________________.
Fill in the blank with the correct answer from the options given below.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Maximum kinetic energy of photoelectrons emitted can be \( hv_{0} \)
By Einstein's photoelectric equation, the maximum kinetic energy is the photon energy minus the work function \( \phi_0 = hv_{0} \):
\( \displaystyle K_{\max} = h(2v_{0}) - hv_{0} = hv_{0} \)
Electrons coming from deeper inside the metal lose some energy on the way out, so the emitted electrons have kinetic energies from zero up to this maximum.
Option (1) No photoelectrons will be emitted — Incorrect. The frequency \( 2v_0 \) is above the threshold \( v_0 \), so photoelectrons are emitted.
Option (2) All photoelectrons emitted will have kinetic energy equal to \( hv_{0} \) — Incorrect. Only the fastest electrons have \( hv_0 \); the others have less, so not all have the same energy.
Option (3) Maximum kinetic energy of photoelectrons emitted can be \( hv_{0} \) — Correct. The maximum kinetic energy is \( 2hv_0 - hv_0 = hv_0 \).
Option (4) Maximum kinetic energy of photoelectrons emitted will be \( 2hv_{0} \) — Incorrect. \( 2hv_0 \) is the photon energy; the work function \( hv_0 \) has to be subtracted.
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
\( \displaystyle K_{max} = h\nu - \phi_0 \qquad (11.2) \)”
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.”