Question

When a metal surface is illuminated by light of wavelength $\lambda$, the stopping potential is 8V. When the same surface is illuminated by light of wavelength $3\lambda$, the stopping potential is 2V. The threshold wavelength for this surface is:

• A. 5$\lambda$
• B. 3$\lambda$
• C. 9$\lambda$
• D. 4.5$\lambda$

Answer 1

Answer: (C)

9$\lambda$

Answer 2

$E = \phi + K_{\text{max}}$

$\phi = \frac{hc}{\lambda_0}$

$K_{\text{max}} = eV_0$

$8e = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad \text{(i)}$

$2e = \frac{hc}{3\lambda} - \frac{hc}{\lambda_0} \quad \text{(ii)}$

On solving (i) & (ii),

$\lambda_0 = 9\lambda$