Q131 / 150 · MHT CET 2026 (15 April Shift 2) · Physics
When a metallic surface is illuminated with a radiation of wavelength '$\lambda$', the stopping potential is 'V'. If the same surface is illuminated with radiation of wavelength $6\lambda$, the stopping potential is $\left(\dfrac{V}{12}\right)$. The threshold wavelength for the surface is
- A$5\lambda$
- B$10\lambda$
- Correct answerC$11\lambda$
- D$12\lambda$
Solution
Using Einstein's photoelectric equation, $eV = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}$
For the first case:
$eV = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}$
For the second case:
$\dfrac{eV}{12} = \dfrac{hc}{6\lambda} - \dfrac{hc}{\lambda_0}$
Multiplying the second equation by $12$ gives:
$eV = \dfrac{2hc}{\lambda} - \dfrac{12hc}{\lambda_0}$
Equating the two expressions for $eV$:
$\dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0} = \dfrac{2hc}{\lambda} - \dfrac{12hc}{\lambda_0}$
$\dfrac{11hc}{\lambda_0} = \dfrac{hc}{\lambda}$
$\lambda_0 = 11\lambda$
Answer: $11\lambda$
For the first case:
$eV = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}$
For the second case:
$\dfrac{eV}{12} = \dfrac{hc}{6\lambda} - \dfrac{hc}{\lambda_0}$
Multiplying the second equation by $12$ gives:
$eV = \dfrac{2hc}{\lambda} - \dfrac{12hc}{\lambda_0}$
Equating the two expressions for $eV$:
$\dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0} = \dfrac{2hc}{\lambda} - \dfrac{12hc}{\lambda_0}$
$\dfrac{11hc}{\lambda_0} = \dfrac{hc}{\lambda}$
$\lambda_0 = 11\lambda$
Answer: $11\lambda$
Hence, the correct answer is option: 'C'.