JAMB 2023 Physics · Question 8

When light of a certain frequency is incident on a metal surface, no photoelectrons are emitted. If the…

Question

When light of a certain frequency is incident on a metal surface, no photoelectrons are emitted. If the frequency of the light is increased, what happens to the stopping potential?

A.The stopping potential does not change.
B.The stopping potential decreases.
C.The stopping potential can either increase or decrease, depending on the intensity of the light.
D.The stopping potential increases.

Answer: D. The stopping potential increases.

Explanation written with AI and checked against the answer key

1. Analyze the Question

The question asks about the relationship between the frequency of incident light and the stopping potential in the photoelectric effect. Specifically, it asks what happens to the stopping potential when the frequency of the incident light is increased.

2. Explain the Concept

The photoelectric effect is governed by Einstein’s photoelectric equation:

hf = \Phi + K_{max}

Where:

$h$ is Planck’s constant.
$f$ is the frequency of the incident light.
$\Phi$ (Work function) is the minimum energy required to remove an electron from the metal surface.
$K_{max}$ is the maximum kinetic energy of the emitted photoelectrons.

The stopping potential ( $V_s$ ) is the negative potential applied to the anode to stop the most energetic photoelectrons. It is directly related to the maximum kinetic energy by the equation:

K_{max} = eV_s

(where $e$ is the charge of an electron).

Combining these, we get:

eV_s = hf - \Phi

V_s = \frac{hf}{e} - \frac{\Phi}{e}

From this equation, you can see that the stopping potential ( $V_s$ ) is directly proportional to the frequency ( $f$ ) of the incident light. As the frequency increases, the energy of the photons increases, which gives the emitted electrons more kinetic energy, requiring a higher (more negative) potential to stop them.

3. Show the Solution

Step 1: Identify the relationship: $V_s = \frac{hf - \Phi}{e}$ .
Step 2: Observe the effect of increasing $f$ : Since $h$ , $e$ , and $\Phi$ are constants for a given metal, increasing $f$ makes the term $(hf - \Phi)$ larger.
Step 3: Conclude the result: Because $(hf - \Phi)$ increases, $V_s$ must also increase to effectively stop the faster-moving electrons.

4. Identify the Answer

The correct option is D. The stopping potential increases as the frequency of the incident light increases.

5. Exam Tip

Always remember the "Frequency-Energy Connection": In the photoelectric effect, frequency determines the energy (and thus the stopping potential) of the electrons, while intensity determines the number of electrons emitted. If you see "frequency" in a question, think "energy/potential."

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