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Question

A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by

A
1.05×1034 Js
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B
2.11×1034 Js
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C
3.16×1034 Js
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D
4.22×1034 Js
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Solution

The correct option is A 1.05×1034 Js
The energy of an electron present in ground state of a hydrogen atom will be,

E=13.6 eV

After absorbing 10.2 eV, the energy of the electron becomes

E=13.6+10.2=3.4 eV

E=3.4 eV corresponds to the energy of the first excited state (n=2)

Thus, an electron jumps from ground state (n1=1) to first excited state(n2=2)

According to Bohr's postulate, the angular momentum of the stationary states is given by,

L=nh2π

Increase in Angular momentum is given by

ΔL=n2h2πn1h2π

ΔL=2h2πh2π

ΔL=h2π

ΔL=6.6×10342×3.14

ΔL=1.05×1034 Js

Hence, option (A) is correct.
Note: It's a very common question in both JEE & NEET.

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