Question

If the frequency of incident light in a photoelectric experiment is doubled, then stopping potential will ______.

• A. be doubled.
• B. be halved.
• C. become more than double.
• D. become less than double.

Hint:

This non-linear behavior occurs because the metal always demands a "fixed tax" (the work function $\Phi$) before releasing an electron. When you double the input energy, you only pay that tax once, leaving a disproportionately larger chunk of energy for the electron's kinetic speed!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Einstein’s Photoelectric Equation: $h\nu = \phi_0 + eV_s$, where $V_s$ is the stopping potential and $\phi_0$ is the work function.

Step 2: Formula Application:
Initially: $eV_1 = h\nu - \phi_0$. When frequency is doubled ($2\nu$): $eV_2 = h(2\nu) - \phi_0 = 2h\nu - \phi_0$.

Step 3: Explanation:
We can rewrite $eV_2$ as: $eV_2 = 2(h\nu - \phi_0) + \phi_0 = 2eV_1 + \phi_0$. Since the work function $\phi_0$ is positive, $V_2$ must be greater than $2V_1$.

Step 4: Final Answer:
The stopping potential becomes more than double.