Q183 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Physics
When light of wavelength '$\lambda$' is incident on a photosensitive surface, the stopping potential is '$V$'. When light of wavelength '$2\lambda$' is incident on the same surface, the stopping potential is $V/3$. Threshold wavelength for the surface is
- A$2\lambda$
- B$3\lambda$
- Correct answerC$4\lambda$
- D$6\lambda$
Solution
Using Einstein's photoelectric equation:
$eV = \dfrac{hc}{\lambda} - \phi$
$\dfrac{eV}{3} = \dfrac{hc}{2\lambda} - \phi$
Multiplying the second equation by $3$:
$eV = \dfrac{3hc}{2\lambda} - 3\phi$
Equating the two expressions for $eV$:
$\dfrac{hc}{\lambda} - \phi = \dfrac{3hc}{2\lambda} - 3\phi$
$2\phi = \dfrac{3hc}{2\lambda} - \dfrac{hc}{\lambda}$
$2\phi = \dfrac{hc}{2\lambda}$
$\phi = \dfrac{hc}{4\lambda}$
The work function is given by $\phi = \dfrac{hc}{\lambda_0}$, where $\lambda_0$ is the threshold wavelength.
$\dfrac{hc}{\lambda_0} = \dfrac{hc}{4\lambda}$
$\lambda_0 = 4\lambda$
Answer: $4\lambda$
$eV = \dfrac{hc}{\lambda} - \phi$
$\dfrac{eV}{3} = \dfrac{hc}{2\lambda} - \phi$
Multiplying the second equation by $3$:
$eV = \dfrac{3hc}{2\lambda} - 3\phi$
Equating the two expressions for $eV$:
$\dfrac{hc}{\lambda} - \phi = \dfrac{3hc}{2\lambda} - 3\phi$
$2\phi = \dfrac{3hc}{2\lambda} - \dfrac{hc}{\lambda}$
$2\phi = \dfrac{hc}{2\lambda}$
$\phi = \dfrac{hc}{4\lambda}$
The work function is given by $\phi = \dfrac{hc}{\lambda_0}$, where $\lambda_0$ is the threshold wavelength.
$\dfrac{hc}{\lambda_0} = \dfrac{hc}{4\lambda}$
$\lambda_0 = 4\lambda$
Answer: $4\lambda$
Hence, the correct answer is option: 'C'.