Let $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_{1}$ and $\lambda_{2}$, respectively are incident on a metallic surface. If $\lambda_{1}=3 \lambda_{2}$ then:

(B) $\mathrm{K}_{1}<\frac{\mathrm{K}_{2}}{3}$

(A) $\mathrm{K}_{1}>\frac{\mathrm{K}_{2}}{3}$

(C) $\mathrm{K}_{1}=\frac{\mathrm{K}_{2}}{3}$

(D) $\mathrm{K}_{2}=\frac{\mathrm{K}_{1}}{3}$

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Modern Physics

Let

K_{1}

and

K_{2}

be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength

λ_{1}

and

λ_{2}

, respectively are incident on a metallic surface. If

λ_{1} = 3 λ_{2}

then:

(A) $K_{1} > \frac{K _{2}}{3}$
(B) $K_{1} < \frac{K _{2}}{3}$ ✓ Correct
(C) $K_{1} = \frac{K _{2}}{3}$
(D) $K_{2} = \frac{K _{1}}{3}$

Answer: (B)

Solution

From Einstein's photoelectric equation:

K_{1} = \frac{h c}{λ _{1}} - ϕ

K_{2} = \frac{h c}{λ _{2}} - ϕ

Given

λ_{1} = 3 λ_{2}

, we have:

K_{1} = \frac{h c}{3 λ _{2}} - ϕ

Multiplying the first equation by 3:

3 K_{1} = \frac{h c}{λ _{2}} - 3 ϕ = K_{2} + ϕ - 3 ϕ = K_{2} - 2 ϕ

Since

ϕ > 0

(work function is positive):

3 K_{1} = K_{2} - 2 ϕ < K_{2}

K_{1} < \frac{K _{2}}{3}

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option

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