JEE Advanced 2022 Paper 2, Physics Section 3 Q2: Introduction and Photoelectric Effect

JEE Advanced2022Paper 2Physics Section 3

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, $1.20$ eV
B
$1.72 \times 1 0^{- 7}$ m, $5.60$ eV
C
$3.78 \times 1 0^{- 7}$ m, $5.60$ eV
D
$3.78 \times 1 0^{- 7}$ m, $1.20$ eV

Solution

The Einstein photoelectric equation gives the stopping potential $e V_{s} = \frac{h c}{λ} - ϕ$ . Intensity does not affect the stopping potential (only wavelength does).

First source: $\frac{h c}{λ} - ϕ = 6.0$ eV. Second source (wavelength $4 λ$ ): $\frac{h c}{4 λ} - ϕ = 0.6$ eV.

Subtracting: $\frac{h c}{λ} - \frac{h c}{4 λ} = 5.4$ eV, i.e. $\frac{3 h c}{4 λ} = 5.4$ eV.

Solving for $λ$ using $h c / e = 1.24 \times 1 0^{- 6}$ J m C $^{- 1}$ : $λ = \frac{3 \times 1.24 \times 1 0 ^{- 6}}{4 \times 5.4} \approx 1.72 \times 1 0^{- 7}$ m.

Work function: $ϕ = \frac{h c}{λ} - 6.0$ eV. With $h c / λ \approx 7.21$ eV, $ϕ \approx 1.21$ eV.

Hence answer is $1.72 \times 1 0^{- 7}$ m and $1.20$ eV: option (A).

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JEE Advanced 2022 Paper 2, Physics Section 3 Q2: Introduction and Photoelectric Effect

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