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CBSE Class 12 Physics 2026 All Sets Paper

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Question : 1 of 16
Marks: +1, -0
The energy of an electron in an orbit in hydrogen atom is -3.4 eV. Its angular momentum in the orbit will be:
Solution:  
Given that En = −3.4 eV:
−3.4=−  13.6n2-3.4 = -\;\frac{13.6}{n^2}
n2=  13.63.4=4n^2 = \;\frac{13.6}{3.4} = 4
n=2n = 2
For the second orbit (n = 2), the angular momentum becomes:
L=  2h2π=  hπL = \;\frac{2h}{2\pi} = \;\frac{h}{\pi}
The angular momentum of the electron in this orbit is   hπ\;\frac{h}{\pi}
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