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A ray of light of wavelength λ is incident on three different photoelectric cells, 1, 2 and 3. The threshold wavelengths of these cells are λ₁, λ₂ and λ₃ respectively, and the magnitudes of their stopping potentials are V₁, V₂ and V₃ respectively. The relations between λ and the threshold wavelengths are λ₁<λ, λ₂>λ and λ₃λ. The correct option is:

Asked in NEET Re-exam 2026 · What the stopping potential depends on

Answer: (2) V₁=0, V₂<V₃

Step-by-step solution

Emission needs the incident wavelength to be shorter than the cell's threshold wavelength, λ<λ₀.

Cell 1 has λ₁<λ, so the light is longer than its threshold and no electrons leave. Hence V₁=0.

Cells 2 and 3 have λ₂>λ and λ₃λ, so both of them emit.

For a cell that emits, eV₀=(hc)/λ-(hc)/(λ₀).

The incident term (hc)/λ is the same for both, so the cell with the larger λ₀ subtracts less and ends with the larger V₀.

Since λ₃λ₂, it follows that V₃>V₂, so V₁=0 and V₂<V₃.

Why the other options are wrong

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