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:

Updated On: Nov 20, 2024
  • 5\( \lambda \)
  • 3\( \lambda \)
  • 9\( \lambda \)
  • 4.5\( \lambda \)
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The Correct Option is C

Solution and Explanation

\[ E = \phi + K_{\text{max}} \]

\[ \phi = \frac{hc}{\lambda_0} \]

\[ K_{\text{max}} = eV_0 \]

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

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

On solving (i) & (ii),

\[ \lambda_0 = 9\lambda \]

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