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: Understanding the Concept
Einstein's equation is \(\dfrac{hc}{\lambda} = \phi_{0} + eV\), where \(\phi_{0} = \dfrac{hc}{\lambda_{0}}\) is the work function.

Step 2: Write both cases
Case 1: \(\dfrac{hc}{\lambda} = \phi_{0} + eV\).
Case 2: \(\dfrac{hc}{2\lambda} = \phi_{0} + \dfrac{eV}{3}\).

Step 3: Calculation
Multiply case 2 by 3: \(\dfrac{3hc}{2\lambda} = 3\phi_{0} + eV\).
Subtract case 1: \(\dfrac{hc}{2\lambda} = 2\phi_{0}\), so \(\phi_{0} = \dfrac{hc}{4\lambda}\).

Step 4: Threshold wavelength
Since \(\phi_{0} = \dfrac{hc}{\lambda_{0}}\), we get \(\lambda_{0} = 4\lambda\).

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