JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Modern Physics

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A hydrogen atom in is ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of : (Given, Plank's constant

= 6.6 \times 1 0^{- 34} Js

)

(A) $2.10 \times 1 0^{- 34} Js$
(B) $1.05 \times 1 0^{- 34} Js$ ✓ Correct
(C) $3.15 \times 1 0^{- 34} Js$
(D) $4.2 \times 1 0^{- 34} Js$

Answer: (B)

Solution

The energy absorbed is

13.6 (\frac{1}{1 ^{2}} - \frac{1}{n ^{2}}) = 10.2

eV, giving

n = 2

. The angular momentum in the

n

th orbit is

L = \frac{nh}{2 π}

. Initial:

L_{i} = \frac{h}{2 π}

, Final:

L_{f} = \frac{2 h}{2 π}

. Change:

Δ L = \frac{h}{2 π} = \frac{6.6 \times 1 0 ^{- 34}}{2 π} \approx 1.05 \times 1 0^{- 34}

Js.

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option

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