The work function of cesium is 2.27 eV. The cut off voltage which stops the emission of electron from cesium when irradiated with light of wavelength 400 nm is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.84 V
By Einstein's photoelectric equation, the maximum kinetic energy of the emitted electrons is the photon energy minus the work function, and the cut-off (stopping) potential \( V_0 \) is the retarding potential that just stops them, \( eV_0 = K_{\max} \).
Photon energy, with \( hc \approx 1240\,\mathrm{eV\,nm} \):
\( \displaystyle E = \frac{hc}{\lambda} = \frac{1240\,\mathrm{eV\,nm}}{400\,\mathrm{nm}} = 3.1\,\mathrm{eV} \)
\( eV_0 = E - \phi = 3.1\,\mathrm{eV} - 2.27\,\mathrm{eV} = 0.83\,\mathrm{eV} \)
\( V_0 \approx 0.84\,\mathrm{V} \)
The last digit depends on the value of \( hc \) used (with \( h = 6.626\times10^{-34}\,\mathrm{J\,s} \) one gets \( E = 3.106\,\mathrm{eV} \) and \( V_0 = 0.836\,\mathrm{V} \)); Option (1) is the nearest.