In hydrogen atom light corresponding to the transition n = 4 to n = 2 falls on cesium metal. The work function of the metal is 1.9 eV. What is the maximum kinetic energy of the photoelectrons emitted?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.65 eV
The photon emitted in the \( n = 4 \to n = 2 \) transition has energy equal to the difference of the two Bohr levels, \( E_n = -\frac{13.6}{n^{2}}\,\mathrm{eV} \). By Einstein’s photoelectric equation the maximum kinetic energy of the photoelectrons is this photon energy minus the work function.
\( \displaystyle E_{\mathrm{photon}} = E_4 - E_2 = 13.6\left(\frac{1}{2^{2}} - \frac{1}{4^{2}}\right) = 13.6\times\frac{3}{16} = 2.55\,\mathrm{eV} \)
\( K_{\max} = E_{\mathrm{photon}} - \phi_0 = 2.55\,\mathrm{eV} - 1.9\,\mathrm{eV} = 0.65\,\mathrm{eV} \)