JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Modern Physics

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A metal exposed to light of wavelength 800 nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The work function of the metal is (Take hc

= 1230 eV - nm

(A)1.537 eV
(B)2.46 eV
(C)0.615 eV✓ Correct
(D)1.23 eV

Answer: (C)

Solution

Using the photoelectric equation

K = \frac{h c}{λ} - ϕ

. For 800 nm:

K_{1} = \frac{1230}{800} - ϕ

. For 500 nm:

K_{2} = \frac{1230}{500} - ϕ

. Given

K_{2} = 2 K_{1}

, substituting:

\frac{1230}{500} - ϕ = 2 (\frac{1230}{800} - ϕ)

. Simplifying:

\frac{1230}{500} - ϕ = \frac{2 \times 1230}{800} - 2 ϕ

, so

ϕ = \frac{2 \times 1230}{800} - \frac{1230}{500} = \frac{1230 \times 2}{800} - \frac{1230}{500} = 1230 (\frac{1}{400} - \frac{1}{500}) = 1230 \times \frac{100}{200000} = 0.615 eV

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Modern Physics

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