Two radiations of photon energies 1 eV and 2.5 eV successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1 : 2
By Einstein’s photoelectric equation, the maximum kinetic energy of the emitted electrons is \( K_{\mathrm{max}}=h\nu-\phi_0 \). Since \( K_{\mathrm{max}}=\frac{1}{2}mv_{\mathrm{max}}^{2} \), the maximum speed is proportional to \( \sqrt{K_{\mathrm{max}}} \).
For the two radiations, with \( \phi_0=0.5\,\mathrm{eV} \):
\( \begin{aligned} K_1 &= 1\,\mathrm{eV}-0.5\,\mathrm{eV} \\ &= 0.5\,\mathrm{eV} \end{aligned} \)
\( \begin{aligned} K_2 &= 2.5\,\mathrm{eV}-0.5\,\mathrm{eV} \\ &= 2\,\mathrm{eV} \end{aligned} \)
The ratio of the maximum speeds is:
\( \begin{aligned} \frac{v_1}{v_2} &= \sqrt{\frac{K_1}{K_2}} \\ &= \sqrt{\frac{0.5\,\mathrm{eV}}{2\,\mathrm{eV}}} \\ &= \frac{1}{2} \end{aligned} \)
Therefore, \( v_1:v_2 \) = 1 : 2.