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: Understanding the Concept
Einstein equation: \(eV_0 = \dfrac{hc}{\lambda} - \phi\), where \(\phi = \dfrac{hc}{\lambda_0}\). Let \(E = \dfrac{hc}{\lambda}\).

Step 2: Two equations
1. \(eV = E - \phi\).
2. \(\dfrac{eV}{12} = \dfrac{E}{6} - \phi\).
Multiply (2) by 12: \(eV = 2E - 12\phi\). Equate with (1):
\[ E - \phi = 2E - 12\phi \Rightarrow 11\phi = E \Rightarrow \phi = \frac{E}{11} \]

Step 3: Threshold wavelength
\[ \frac{hc}{\lambda_0} = \frac{1}{11}\cdot\frac{hc}{\lambda} \Rightarrow \lambda_0 = 11\lambda \]

Final Answer:
The threshold wavelength is \(11\lambda\), option (C). \[ \boxed{11\lambda} \]