Question

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. \(3\lambda\)
• C. \(4\lambda\)
• D. \(6\lambda\)

Hint:

Write Einstein's photoelectric equation for both wavelengths and eliminate the stopping potential to find the work function.

Answer 1

The Correct Option is C

Step 1: Eliminate eV
From case 1, $eV = \dfrac{hc}{\lambda} - \phi_{0}$. Put this in case 2: $\dfrac{hc}{2\lambda} = \phi_{0} + \dfrac{1}{3}\left(\dfrac{hc}{\lambda} - \phi_{0}\right)$.

Step 2: Simplify
$\dfrac{hc}{2\lambda} - \dfrac{hc}{3\lambda} = \dfrac{2\phi_{0}}{3}$, so $\dfrac{hc}{6\lambda} = \dfrac{2\phi_{0}}{3}$ and $\phi_{0} = \dfrac{hc}{4\lambda}$.

Step 3: Result
Comparing with $\phi_{0} = hc/\lambda_{0}$ gives $\lambda_{0} = 4\lambda$.

Final Answer:
The threshold wavelength is 4 lambda. \[ \boxed{\text{(C) }4\lambda} \]