Question

When a metallic surface is illuminated with a radiation of wavelength '\(λ\)', the stopping potential is 'V'. If the same surface is illuminated with radiation of wavelength \(6λ\), the stopping potential is \((\frac{V}{12})\). The threshold wavelength for the surface is

• A. \(5λ\)
• B. \(10λ\)
• C. \(11λ\)
• D. \(12λ\)

Hint:

Use eV = hc/lambda - phi for both wavelengths and eliminate eV.

Answer 1

The Correct Option is C

Step 1: Subtract equations
$eV - \dfrac{eV}{12} = \dfrac{hc}{\lambda} - \dfrac{hc}{6\lambda}$, so $\dfrac{11eV}{12} = \dfrac{5hc}{6\lambda}$ and $eV = \dfrac{10hc}{11\lambda}$.

Step 2: Work function
$\phi = \dfrac{hc}{\lambda} - eV = \dfrac{hc}{11\lambda}$.

Step 3: Threshold
$\lambda_0 = hc/\phi = 11\lambda$. Option (C).

Final Answer:
Option (C). \[ \boxed{11\lambda} \]