JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Modern Physics

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Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is

α \times 1 0^{- 19}

J. The value of

α

is ______.
(

h = 6.62 \times 1 0^{- 34}

J·s,

e = 1.6 \times 1 0^{- 19}

C and

c = 3 \times 1 0^{8}

m/s)

(A)3.68
(B)4.68
(C)5.68✓ Correct
(D)2.68

Answer: (C)

Solution

The photoelectric equation gives the work function as

ϕ = \frac{h c}{λ} - e V_{s}

. The photon energy is

\frac{h c}{λ} = \frac{6.62 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{331 \times 1 0 ^{- 9}} = 6.0 \times 1 0^{- 19}

J, and the stopping-potential energy is

e V_{s} = 1.6 \times 1 0^{- 19} \times 0.2 = 0.32 \times 1 0^{- 19}

J. Thus

ϕ = 6.0 \times 1 0^{- 19} - 0.32 \times 1 0^{- 19} = 5.68 \times 1 0^{- 19}

J, so

α = 5.68

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Modern Physics

Solution

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