A ray of light with wavelength lambda is incident on three different photo-electric cells namely 1, 2 and 3. The threshold wavelength of these photo-electric cells are lambda_1, lambda_2, and lambda_3, respectively and the magnitude of stopping potentials of these cells are V_1, V_2 and V_3, respectively. The relation between lambda and threshold wavelengths are lambda_1 lambda and lambda_3 >> lambda. The correct option is :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q18 – NEET-UG 2026

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Home Physics Dual Nature of MatterQuestion #87

A ray of light with wavelength

λ

is incident on three different photo-electric cells namely 1, 2 and 3. The threshold wavelength of these photo-electric cells are

λ_{1}

λ_{2}

, and

λ_{3}

, respectively and the magnitude of stopping potentials of these cells are

V_{1}

V_{2}

and

V_{3}

, respectively. The relation between

λ

and threshold wavelengths are

λ_{1} < λ

λ_{2} > λ

and

λ_{3} >> λ

. The correct option is :

Clue 1 of 3: Photoelectric Effect Principle

Photoelectric emission occurs only if the incident photon energy is greater than or equal to the work function of the material, which means the incident wavelength must be less than or equal to the threshold wavelength.

🥷

Ninja Strategy•Threshold Wavelength Analysis

First, determine which cells will exhibit photoelectric emission based on the incident and threshold wavelengths. Then, compare the stopping potentials by considering how the threshold wavelength affects the kinetic energy of emitted electrons.

🧠 Full Solution Path

Step 1: Determine Stopping Potential for Cell 1✦ Active

For cell 1, the relation is $λ_{1} < λ$ . This means the threshold wavelength $λ_{1}$ is less than the incident wavelength $λ$ . According to the photoelectric effect, emission occurs only if the incident wavelength is less than or equal to the threshold wavelength ( $λ \leq λ_{0}$ ). Since $λ > λ_{1}$ , the incident light does not have enough energy to cause photoelectric emission from cell 1. Therefore, the stopping potential $V_{1} = 0$ .

Step 2: Determine Stopping Potentials for Cell 2 and Cell 3○ Expand

For cell 2, $λ_{2} > λ$ . This means $λ < λ_{2}$ , so photoelectric emission occurs. The stopping potential is $V_{2} = \frac{h c}{e} (\frac{1}{λ} - \frac{1}{λ_{2}})$ . For cell 3, $λ_{3} >> λ$ . This also means $λ < λ_{3}$ , so photoelectric emission occurs. The stopping potential is $V_{3} = \frac{h c}{e} (\frac{1}{λ} - \frac{1}{λ_{3}})$ .

💡 Teacher's Secret Hint

Remember that $h c / e$ is a positive constant. The stopping potential's magnitude depends on the term in the parenthesis.

Step 3: Compare

V_{2}

and

V_{3}

○ Expand

We are given $λ_{2} > λ$ and $λ_{3} >> λ$ . This implies that $λ_{3}$ is significantly larger than $λ_{2}$ . Therefore, $\frac{1}{λ_{3}}$ will be significantly smaller than $\frac{1}{λ_{2}}$ . Comparing the terms for $V_{2}$ and $V_{3}$ : $(\frac{1}{λ} - \frac{1}{λ_{3}})$ vs $(\frac{1}{λ} - \frac{1}{λ_{2}})$ Since $\frac{1}{λ_{3}} < \frac{1}{λ_{2}}$ , it follows that $(\frac{1}{λ} - \frac{1}{λ_{3}}) > (\frac{1}{λ} - \frac{1}{λ_{2}})$ . Thus, $V_{3} > V_{2}$ . Combining with $V_{1} = 0$ , the correct relation is $V_{1} = 0, V_{2} < V_{3}$ .

💡 Teacher's Secret Hint

A larger threshold wavelength means a smaller work function, leading to higher kinetic energy of emitted electrons and thus a higher stopping potential for the same incident light.

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Physics Question 18 – NEET-UG 2026

Clue 1 of 3: Photoelectric Effect Principle

🧠 Full Solution Path