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

Step 1: Understanding the Concept:
According to Einstein's photoelectric equation, the energy of the incident photon is used to overcome the work function and provide kinetic energy to the emitted electron. Stopping potential is directly proportional to the maximum kinetic energy.
Step 2: Key Formula or Approach:
\[ hf = \Phi + eV_{0} \]
Where \(hf\) is photon energy, \(\Phi\) is work function, and \(V_{0}\) is stopping potential.
: Detailed Explanation:
Initial state: \(eV_{0} = hf - \Phi\) \dots (i)
Final state (frequency doubled): \(hf' = 2hf\).
\[ eV_{0}' = h(2f) - \Phi = 2hf - \Phi \]
From equation (i), we have \(hf = eV_{0} + \Phi\). Substitute this into the new equation:
\[ eV_{0}' = 2(eV_{0} + \Phi) - \Phi \]
\[ eV_{0}' = 2eV_{0} + 2\Phi - \Phi \]
\[ eV_{0}' = 2eV_{0} + \Phi \]
Dividing by \(e\):
\[ V_{0}' = 2V_{0} + \frac{\Phi}{e} \]
Since the work function \(\Phi\) is a positive constant, \(V_{0}'>2V_{0}\).
Step 3: Final Answer:
The stopping potential becomes more than double.