Atoms And Nuclei Question 73

Question: The first line of Balmer series has wavelength 6563 Å. What will be the wavelength of the first member of Lyman series [RPMT 1996]

Options:

A) 1215.4 Å

B) 2500 Å

C) 7500 Å

D) 600 Å

Show Answer

Answer:

Correct Answer: A

Solution:

$ \frac{1}{{\lambda_{Balmer}}}=R[ \frac{1}{2^{2}}-\frac{1}{3^{2}} ]=\frac{5R}{36} $ , $ \frac{1}{{\lambda_{Lyman}}}=R[ \frac{1}{1^{2}}-\frac{1}{2^{2}} ]=\frac{3R}{4} $
$ \therefore {\lambda_{Lyman}}={\lambda_{Balmer}}\times \frac{5}{27}=1215.4\ {\AA} $



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