The work function for Tungsten and Sodium are 4.2 eV and 2.3 eV, respectively. If the threshold wavelength for Sodium is \( 5460\,\mathring{\mathrm{A}} \), then that for tungsten is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2990\,\mathring{\mathrm{A}} \)
At the threshold wavelength the photon energy just equals the work function, so \( \phi_0 = hc/\lambda_0 \) and the threshold wavelength is inversely proportional to the work function.
\( \displaystyle \lambda_0 = \frac{hc}{\phi_0} \;\Rightarrow\; \frac{\lambda_{W}}{\lambda_{Na}} = \frac{\phi_{Na}}{\phi_{W}} \)
\( \displaystyle \lambda_{W} = 5460 \times \frac{2.3}{4.2} = \frac{12558}{4.2} \approx 2990\,\mathring{\mathrm{A}} \)
Tungsten has the larger work function, so its threshold wavelength must be shorter than that of sodium; this alone rules out Options (2) and (3).