The surface of a certain metal is first illuminated with light of wavelength, λ1 = 350 nm, and then by light of wavelength, λ2 = 540 nm. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2. The work function of the metal (in eV ) is close to (Energy of photon = \( \displaystyle\frac{1240}{\lambda(\mathrm{nm})} \) eV)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.88
By Einstein’s photoelectric equation, the maximum kinetic energy of the photoelectrons is the photon energy minus the work function \( \phi_0 \):
\( \displaystyle K_{\mathrm{max}} = \frac{1}{2}mv_{\mathrm{max}}^{2} = E - \phi_0 \)
\( \displaystyle E_1 = \frac{1240}{350}\,\mathrm{eV} \approx 3.543\,\mathrm{eV} \)
\( \displaystyle E_2 = \frac{1240}{540}\,\mathrm{eV} \approx 2.296\,\mathrm{eV} \)
The shorter wavelength (350 nm) gives the faster electrons, so \( v_1 = 2v_2 \). Kinetic energy varies as \( v^{2} \), so doubling the speed means four times the kinetic energy.
\( \displaystyle E_1 - \phi_0 = 4(E_2 - \phi_0) \)
\( \displaystyle 3\phi_0 = 4E_2 - E_1 = 9.185\,\mathrm{eV} - 3.543\,\mathrm{eV} = 5.642\,\mathrm{eV} \)
\( \displaystyle \phi_0 \approx 1.88\,\mathrm{eV} \)
Check: \( K_1 \approx 1.66 \) eV and \( K_2 \approx 0.42 \) eV, whose ratio is 4.