JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Modern Physics

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When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is ____ m.
(h = 6.63 × 10⁻³⁴ J·s, e = 1.6 × 10⁻¹⁹ C, c = 3 × 10⁸ m/s)

(A) $2.9 \times 1 0^{- 8}$
(B) $2.2 \times 1 0^{- 8}$
(C) $3.1 \times 1 0^{- 7}$
(D) $2.5 \times 1 0^{- 7}$ ✓ Correct

Answer: (D)

Solution

Einstein's photoelectric relation gives

e V_{0} = \frac{h c}{λ} - ϕ

. For the first light

e (3.2) = \frac{h c}{λ} - ϕ

, and for light of wavelength

2 λ

e (0.7) = \frac{h c}{2 λ} - ϕ

. Subtracting removes the work function:

e (3.2 - 0.7) = \frac{h c}{λ} - \frac{h c}{2 λ} = \frac{h c}{2 λ}

, so

\frac{h c}{λ} = 5 e

and

λ = \frac{h c}{5 e} = \frac{6.63 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{5 \times 1.6 \times 1 0 ^{- 19}} \approx 2.5 \times 1 0^{- 7}

m. The wavelength of the first light is about

2.5 \times 1 0^{- 7}

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Modern Physics

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