Light of frequency \(1.0\times10^{15}\ \mathrm{Hz}\) falls on a metal surface. The stopping potential of the emitted photoelectrons is 1.65 V. The threshold frequency of the metal is (Take \(h = 6.6\times10^{-34}\ \mathrm{J\,s}\) and \(e = 1.6\times10^{-19}\ \mathrm{C}\).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(6.0\times10^{14}\ \mathrm{Hz}\)
The stopping potential \( V_0 \) is just enough to stop the fastest photoelectrons. So \( eV_0 \) is their maximum kinetic energy. With the threshold frequency \( \nu_0 \), Einstein’s equation becomes \( eV_0=h(\nu-\nu_0) \).
\( \nu_0=\nu-\frac{eV_0}{h} \)
\( \begin{aligned} \frac{eV_0}{h} &= \frac{1.6\times10^{-19}\times1.65}{6.6\times10^{-34}} \\ &= 4.0\times10^{14}\ \mathrm{Hz} \end{aligned} \)
\( \begin{aligned} \nu_0 &= \nu-\frac{eV_0}{h} \\ &= 1.0\times10^{15}-4.0\times10^{14} \\ &= 6.0\times10^{14}\ \mathrm{Hz} \end{aligned} \)
From the above, \( \nu_0=6.0\times10^{14}\ \mathrm{Hz} \). Light of a lower frequency cannot eject electrons from this metal.
NCERT (2026-27):
Summary – “7. … The maximum kinetic energy (1/2)m v2max is equal to the photon energy (hν) minus the work function φ0 (= hν0) of the target metal:”
\( \displaystyle \frac{1}{2}mv_{\max}^{2}= \)\( V_0e=h\nu-\phi_0=h(\nu-\nu_0) \)