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 , respectiv

Question

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 :NEET 2026 (Re-NEET)

TestHub · Accepted Answer

According to Einstein's photoelectric equation, the magnitude of the stopping potential $V$ is given by:

$e V = \frac{h c}{λ} - \frac{h c}{λ _{t h}}$

For photoelectric emission to take place, the incident wavelength $λ$ must be less than the threshold wavelength $λ_{t h}$ .

For cell 1, $λ > λ_{1}$ . No photoelectric emission occurs, so $V_{1} = 0$ .

For cell 2, $λ < λ_{2}$ . Photoelectric emission occurs, and $V_{2} = \frac{h c}{e} (\frac{1}{λ} - \frac{1}{λ _{2}}) > 0$ .

For cell 3, $λ ≪ λ_{3}$ . Photoelectric emission occurs, and $V_{3} = \frac{h c}{e} (\frac{1}{λ} - \frac{1}{λ _{3}}) > 0$ .

Since $λ_{3} ≫ λ$ and $λ_{2} > λ$ , it follows that $λ_{3} > λ_{2}$ , which gives $\frac{1}{λ _{3}} < \frac{1}{λ _{2}}$ .

Therefore, $\frac{1}{λ} - \frac{1}{λ _{3}} > \frac{1}{λ} - \frac{1}{λ _{2}}$ , which implies $V_{3} > V_{2}$ .

Thus, $V_{1} = 0$ and $V_{2} < V_{3}$ .

Answer: $V_{1} = 0, V_{2} < V_{3}$

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