JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

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
D
2.68

Solution

Einstein's photoelectric equation in the stopping-potential form: $e V_{s} = \frac{h c}{λ} - ϕ$ , so $ϕ = \frac{h c}{λ} - e V_{s}$ .

$\frac{h c}{λ} = \frac{( 6.62 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{331 \times 1 0 ^{- 9}} = 6.00 \times 1 0^{- 19}$ J.

$e V_{s} = (1.6 \times 1 0^{- 19}) (0.2) = 0.32 \times 1 0^{- 19}$ J.

$ϕ = (6.00 - 0.32) \times 1 0^{- 19} = 5.68 \times 1 0^{- 19}$ J, so $α = 5.68$ .

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