CBSE Class 12-science Answered
dual
Asked by chandant | 01 Oct, 2010, 05:00: PM
Expert Answer
Dear Student
For any hydrogen like element:
where
is the wavelength of the light emitted in vacuum;
is the Rydberg constant for this element;
is the atomic number, i.e. the number of protons in the atomic nucleus of this element;
and 
are integers such that 
.
For hydrogen, Z = 1
For Helium, Z = 2
Putting in the above expression:
λhelium = λhydrogen / 4 (for same transition) = 4860/4 = 1215 angstrom
We hope that satisfies your query.
Regards
Team
Topperlearning
Answered by | 01 Oct, 2010, 10:33: PM
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