CBSE Class 12 Physics 2020 Delhi Set 1 Paper

Question : 29

Total: 37

Calculate the de Broglie wavelength associated with the electron revolving in the first excited state of hydrogen atom. The ground state energy of hydrogen atom is −13.6‌eV.

Solution:

The energy of the n‌th ‌ state of Hydrogen atom
En=‌

−13.6

‌eV
For ground state, n=1
When atom is in first excited state, n=2
∴‌‌E=‌

−13.6

=−3.4‌eV
Now, the de Broglie wavelength associated with an electron is
λ=‌

h= Planck's constant
p= Momentum of electron
m= Mass of electron
‌p=√2mE
∴‌λ=‌

‌ Or ‌‌λ=‌

√2mE

‌ Or, ‌‌λ‌=‌

6.6×10−34

√2×9.1×10−31×3.4×1.6×10−19

∴‌‌λ‌=‌

6.6×10−34×1025

9.95

=0.66×10−9
‌=6.6A0

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