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Question: Energy of an electron in an excited hydrogen atom is – 3.4 eV. Its angular momentum will be: \[h = ...

Energy of an electron in an excited hydrogen atom is – 3.4 eV. Its angular momentum will be:

h=6.626×10−34J−sh = 6.626 \times 10^{- 34}J - s

A

1.11×1034J 6musec⁡1.11 \times 10^{34}J\mspace{6mu}\sec

B

1.51×10−31J 6musec⁡1.51 \times 10^{- 31}J\mspace{6mu}\sec

C

2.11×10−34J 6musec⁡2.11 \times 10^{- 34}J\mspace{6mu}\sec

D

3.72×10−34J 6musec⁡3.72 \times 10^{- 34}J\mspace{6mu}\sec

Answer

2.11×10−34J 6musec⁡2.11 \times 10^{- 34}J\mspace{6mu}\sec

Explanation

Solution

The electron is in the second orbit

Hence L=nh2π=2h2π=6.6×10−34π=2.11×10−34J−sec⁡L = \frac{nh}{2\pi} = \frac{2h}{2\pi} = \frac{6.6 \times 10^{- 34}}{\pi} = 2.11 \times 10^{- 34}J - \sec