The first four spectral in the Lyman series of a H-atom are λ= 1218 Å, 1028 Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider deuterium, calculate the shift in the wavelength of these lines.
The first four spectral in the Lyman series of a H-atom are λ= 1218 Å, 1028 Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider deuterium, calculate the shift in the wavelength of these lines.
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1 Answer
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This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation- as we know total energy in stationary orbit is
En=- where sign have usual meaning.
According to bohr third postulate h
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Where is the reduced mass
Reduced mass for H=H=;me(1-me/M)
D= D; me(1-me/2M)
=me(1-me/2M)(1+me/2M)
If for hydrogen deuterium, the wavelength
= (1+ )-1= (1- )
so lines emitted are 1217.7A0,1027.7A0,974.04A0
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Kindly go through the solution
Change in surface energy = work done
|DE0| = –10.2
]
= 3 m/s
n = 4
Number of transitions =
Kinetic energy: Potential energy = 1 : –2
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