Light of wavelength 310 nm falls on a metal surface. The fastest photoelectrons come out with a speed of \(8.0\times10^{5}\ \mathrm{m\,s^{-1}}\). The work function of the metal is (Take \(hc = 1240\ \mathrm{eV\,nm}\), mass of the electron \(= 9.0\times10^{-31}\ \mathrm{kg}\) and \(1\ \mathrm{eV} = 1.6\times10^{-19}\ \mathrm{J}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.2 eV
One photon gives all its energy \( h\nu \) to one electron. A part of this energy frees the electron from the metal. This part is the work function \( \phi_0 \). The fastest electron carries the rest as kinetic energy, so \( \phi_0=h\nu-K_{\max} \).
Given:
\( \lambda=310\ \mathrm{nm} \)
\( v_{\max}=8.0\times10^{5}\ \mathrm{m\,s^{-1}} \)
Energy of each photon:
\( \begin{aligned} h\nu &= \frac{hc}{\lambda} \\ &= \frac{1240\ \mathrm{eV\,nm}}{310\ \mathrm{nm}} \\ &= 4.0\ \mathrm{eV} \end{aligned} \)
Maximum kinetic energy of the photoelectrons:
\( \begin{aligned} K_{\max} &= \frac{1}{2}mv_{\max}^{2} \\ &= \frac{1}{2}\times9.0\times10^{-31}\times(8.0\times10^{5})^{2} \\ &= 2.88\times10^{-19}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} K_{\max} &= \frac{2.88\times10^{-19}}{1.6\times10^{-19}}\ \mathrm{eV} \\ &= 1.8\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} \phi_0 &= h\nu-K_{\max} \\ &= 4.0-1.8 \\ &= 2.2\ \mathrm{eV} \end{aligned} \)
Therefore, the work function of the metal is 2.2 eV.
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”
\( K_{\max}=h\nu-\phi_0 \)\( \qquad (11.2) \)