CBSE Class 12 Physics 2017 Outside Delhi Set 3 Paper

Question : 4

Total: 8

The ground state energy of hydrogen atom is 13.6 eV. If an electron makes a transition from an energy level −1.51‌eV to −3.4‌eV, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.

Solution:

(a) Calculation of energy difference
(b) Formula
(c) Calculation of wavelength
(d) Name of the series of spectral lines
Energy difference =3.4‌eV−1.51‌eV=1.89 eV=3.024×10−19J
Energy =‌

=3.024×10−19J
Wavelength =6.57×10−7m
Series is Balmer series.

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