A hydrogen atom in the ground state emits a photon of energy 12.1 eV. Its orbita

Question

A hydrogen atom in the ground state emits a photon of energy 12.1 eV. Its orbital angular momentum changes by

Δ

L. Then

Δ

L is equal to

Options:

A . 1.05×10−34 Js

B . 2.11×10−34 Js

C . 3.16×10−34 Js

D . 4.22×10−34 Js

Answer: Option B
:
B

Δ E = 12.1 = E_{1} - E_{n}

E_{n} = 13.6 e V - 12.1

E_{n} = 1.5

\Rightarrow n = 3

L_{1} = \frac{2}{2 π}

L_{3} = \frac{3 h}{2 π}

Δ L = L_{3} - L_{1} = \frac{3 h}{2 π} - \frac{h}{2 π} = \frac{2 h}{2 π} = \frac{h}{π} = \frac{6.6}{3.14} \times 10^{- 34} s^{- 1}

Δ L = 2.11 \times 0^{- 34} J s

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