The kinetic energy of emitted electron is E when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy, the incident light must have wavelength :

(B) $\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$

(A) $\frac{\mathrm{hc}}{\mathrm{E} \lambda-\mathrm{hc}}$

(C) $\frac{\mathrm{h} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$

(D) $\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda-\mathrm{hc}}$

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Modern Physics

The kinetic energy of emitted electron is E when the light incident on the metal has wavelength

λ

. To double the kinetic energy, the incident light must have wavelength :

(A) $\frac{hc}{E λ - hc}$
(B) $\frac{hc λ}{E λ + hc}$ ✓ Correct
(C) $\frac{h λ}{E λ + hc}$
(D) $\frac{hc λ}{E λ - hc}$

Answer: (B)

Solution

From the photoelectric equation:

E = \frac{h c}{λ} - ϕ

and

2 E = \frac{h c}{λ ^{'}} - ϕ

. Subtracting the first from the second:

E = h c (\frac{1}{λ ^{'}} - \frac{1}{λ})

. Solving for

λ^{'}

λ^{'} = \frac{h c λ}{E λ + h c}

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option

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