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 Question:
The problem tests our understanding of Einstein's Photoelectric Equation. We must determine how the stopping potential ($V_s$) behaves when the incident light frequency ($f$) is strictly multiplied by 2.

Step 2: Key Formula or Approach:
Einstein's photoelectric equation relates the maximum kinetic energy of emitted photoelectrons to the stopping potential:
$$K_{max} = eV_s = hf - \Phi$$
Where $h$ is Planck's constant, $f$ is incident frequency, and $\Phi$ is the metal's work function.
Therefore, the initial stopping potential is: $V_s = \frac{hf}{e} - \frac{\Phi}{e}$.

Step 3: Detailed Explanation:
Let the initial frequency be $f$, yielding stopping potential $V_1$:
$$V_1 = \frac{hf}{e} - \frac{\Phi}{e}$$
Now, the frequency is doubled to $2f$. Let the new stopping potential be $V_2$:
$$V_2 = \frac{h(2f)}{e} - \frac{\Phi}{e} = \frac{2hf}{e} - \frac{\Phi}{e}$$
We want to compare $V_2$ to exactly double the original potential ($2V_1$).
Calculate $2V_1$:
$$2V_1 = 2 \left( \frac{hf}{e} - \frac{\Phi}{e} \right) = \frac{2hf}{e} - \frac{2\Phi}{e}$$
Compare the expression for $V_2$ and $2V_1$:
$$V_2 = \left( \frac{2hf}{e} - \frac{2\Phi}{e} \right) + \frac{\Phi}{e}$$
$$V_2 = 2V_1 + \frac{\Phi}{e}$$
Since the work function $\Phi$ and elementary charge $e$ are both inherently positive quantities, the term $\frac{\Phi}{e}$ is strictly positive.
Therefore, $V_2 > 2V_1$. The new stopping potential is greater than twice the original.

Step 4: Final Answer:
The stopping potential becomes more than double, matching option (c).