Question

If the frequency of incident photon on a metal surface is doubled, then stopping potential will become:

• A. doubled
• B. less than double
• C. more than double
• D. less than existing value

Hint:

Because of work function subtraction, doubling frequency increases stopping potential more than double.

Answer 1

The Correct Option is C

Concept: Photoelectric equation: \[ eV_0 = h\nu - \phi \]

Step 1:Initial stopping potential \[ V_0 = \frac{h\nu - \phi}{e} \]

Step 2:When frequency is doubled \[ V_0' = \frac{2h\nu - \phi}{e} \]

Step 3:Compare with \(2V_0\) \[ 2V_0 = \frac{2h\nu - 2\phi}{e} \]

Step 4:Compare \(V_0'\) and \(2V_0\) \[ V_0' - 2V_0 = \frac{2h\nu - \phi - (2h\nu - 2\phi)}{e} = \frac{\phi}{e}>0 \] \[ \Rightarrow V_0'>2V_0 \] Conclusion \[ {V_0'>2V_0} \Rightarrow \text{more than double} \]