JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is $3.2$ V. If a second light having wavelength twice of first light is used, the stopping potential drops to $0.7$ V. The wavelength of first light is ______ m.

( $h = 6.63 \times 1 0^{- 34}$ J·s, $e = 1.6 \times 1 0^{- 19}$ C, $c = 3 \times 1 0^{8}$ m/s)

A
$2.9 \times 1 0^{- 8}$
B
$2.2 \times 1 0^{- 8}$
C
$3.1 \times 1 0^{- 7}$
D
$2.5 \times 1 0^{- 7}$

Solution

Einstein's photoelectric equation for the two wavelengths:

$e (3.2) = \frac{h c}{λ} - ϕ$ ... (1)

$e (0.7) = \frac{h c}{2 λ} - ϕ$ ... (2)

Subtract (2) from (1) to eliminate $ϕ$ :

$e (2.5) = \frac{h c}{λ} - \frac{h c}{2 λ} = \frac{h c}{2 λ}$ .

So $λ = \frac{h c}{2 e ( 2.5 )} = \frac{12400}{5} \overset{˚}{A} = 2480 \overset{˚}{A}$ .

$λ \approx 2.48 \times 1 0^{- 7} m$ .

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