JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Modern Physics

When a metal surface is illuminated by light of wavelength

λ

, the stopping potential is 8 V. When the same surface is illuminated by light of wavelength

3 λ

, stopping potential is 2 V. The threshold wavelength for this surface is:

(A) $5 λ$
(B) $3 λ$
(C) $9 λ$ ✓ Correct
(D) $4.5 λ$

Answer: (C)

Solution

Using Einstein's photoelectric equation

e V_{0} = \frac{h c}{λ} - \frac{h c}{λ _{0}}

8 e = \frac{h c}{λ} - \frac{h c}{λ _{0}} ...(i)

2 e = \frac{h c}{3 λ} - \frac{h c}{λ _{0}} ...(ii)

Subtracting (ii) from (i):

6 e = \frac{h c}{λ} - \frac{h c}{3 λ} = \frac{2}{3} \cdot \frac{h c}{λ} ⟹ \frac{h c}{λ} = 9 e

Substituting into (i):

\frac{h c}{λ _{0}} = 9 e - 8 e = e = \frac{1}{9} \cdot \frac{h c}{λ}

, so

λ_{0} = 9 λ

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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