JEE Advanced 2022 — Paper 2 — PHY-III solution

JEE Advanced2022Paper 2PHY-III

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).

← Back to full paper

Practice more PHY-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →