JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Modern Physics

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In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V , what is the wavelength of the em-wave?
(Given hc

= 1242 eVnm

where h is the Planck's constant and c is the speed of light in vaccum.)

(A)400 nm
(B)600 nm
(C)200 nm
(D)300 nm✓ Correct

Answer: (D)

Solution

Einstein's photoelectric equation gives the incident photon energy as

E = ϕ + e V_{s} = 2.14 + 2 = 4.14 eV

Using

E = \frac{h c}{λ}

λ = \frac{h c}{E} = \frac{1242 eV nm}{4.14 eV} = 300 nm

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option

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