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CBSE Class 12 Physics 2020 Delhi Set 3 Paper

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Question : 5
Total: 7
Calculate the de Broglie wavelength associated with the electron in the 2‌nd ‌ excited state of hydrogen atom. The ground state energy of the hydrogen atom is 13.6‌eV.
Solution:  
The energy of the n‌th ‌ state of Hydrogen atom
=En=‌
−13.6
n2
For ground state, n=1
When atom is in second excited state, n=3
∴‌‌E=‌
−13.6
32
−1.51‌eV
Now, the de Broglie wavelength associated with an electron is
λ=‌
h
p
h= Planck's constant
p= Momentum of electron
m= Mass of electron
‌p‌=√2mE
∴‌λ‌=‌
h
p
‌ Or ‌‌‌λ‌=‌
h
√2mE
‌ Or, ‌‌‌λ‌=‌
6.6×10−34
√2×9.1×10−31×1.51×1.6×10−19

∴‌‌λ‌=‌
6.6×10−34×1025
6.631

‌=0.995×109m
‌=9.95A∘
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