JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Modern Physics

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When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is

6.0 V

. This potential drops to

0.6 V

if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively?

[Take \frac{h c}{e} = 1.24 \times 1 0^{- 6} J m C^{- 1} .]

(A) $1.72 \times 1 0^{- 7} m, 1.20 e V$ ✓ Correct
(B) $1.72 \times 1 0^{- 7} m, 5.60 e V$
(C) $3.78 \times 1 0^{- 7} m, 5.60 e V$
(D) $3.78 \times 1 0^{- 7} m, 1.20 e V$

Answer: (A)

Solution

Using Einstein's photoelectric equation:

For first source:

\frac{h c}{λ} = ϕ + 6

eV

For second source:

\frac{h c}{4 λ} = ϕ + 0.6

eV

Subtracting:

\frac{3 h c}{4 λ} = 5.4

eV

Solving:

λ = \frac{3 h c}{4 \times 5.4 \times e} = \frac{3 \times 1.24 \times 1 0 ^{- 6}}{4 \times 5.4} = 1.72 \times 1 0^{- 7}

m

Substituting back:

ϕ = \frac{h c}{λ} - 6 = \frac{1.24 \times 1 0 ^{- 6}}{1.72 \times 1 0 ^{- 7} \times 1.6 \times 1 0 ^{- 19}} - 6 \approx 7.2 - 6 = 1.2

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Single correct option

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JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Modern Physics

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