Question

If the frequency of incident light is doubled, what happens to the stopping potential? ____.

• A. Doubles
• B. Becomes zero
• C. Remains same
• D. Quadruples

Hint:

The stopping potential depends linearly on frequency, but not proportionally. Doubling the frequency always results in a stopping potential that is more than double the original.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Einstein's photoelectric equation, the energy of an incident photon is used to overcome the work function of the metal and the remainder is converted into the maximum kinetic energy of the photoelectron.

Step 2: Key Formula or Approach:
The photoelectric equation is: \[ h\nu = \phi_0 + eV_0 \] Where: - $h\nu$ = Energy of incident light (frequency $\nu$) - $\phi_0$ = Work function (constant for a metal) - $eV_0$ = Maximum kinetic energy ($V_0$ is the stopping potential)

Step 3: Detailed Explanation:
1. Let the initial state be $eV_{01} = h\nu - \phi_0$. 2. If frequency is doubled ($2\nu$), the new stopping potential $V_{02}$ is: \[ eV_{02} = h(2\nu) - \phi_0 \] \[ eV_{02} = 2h\nu - \phi_0 \] 3. Comparing the two, $eV_{02} = 2(eV_{01} + \phi_0) - \phi_0 = 2eV_{01} + \phi_0$. 4. Since $\phi_0$ is positive, $V_{02}$ is actually greater than $2V_{01}$. In multiple-choice contexts where "increases" or "doubles" are the only logical growth options, (1) is the standard selection, though technically it more than doubles.

Step 4: Final Answer:
The stopping potential increases (doubles/more than doubles).