Atoms And Nuclei Question 74
Question: The wavelength of Lyman series is [BHU 1997]
Options:
A) $ \frac{4}{3\times 10967}cm $
B) $ \frac{3}{4\times 10967}cm $
C) $ \frac{4\times 10967}{3}cm $
D) $ \frac{3}{4}\times 10967\ cm $
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Answer:
Correct Answer: A
Solution:
$ \frac{1}{\lambda }=R_{H}[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} ]. $
For Lyman series n1=1 and n2=2, 3, 4,
When n2=2, we get $ \lambda =\frac{4}{3R_{H}}=\frac{4}{3\times 10967}cm $
BETA
