Atoms And Nuclei Question 89

Question: If $ {\lambda_{max}} $ is 6563 Å, then wave length of second line for Balmer series will be [RPMT 2000]

Options:

A) $ \lambda =\frac{16}{3R} $

B) $ \lambda =\frac{36}{5R} $

C) $ \lambda =\frac{4}{3R} $

D) None of the above

Show Answer

Answer:

Correct Answer: A

Solution:

For Balmer series $ \frac{1}{\lambda }=R( \frac{1}{2^{2}}-\frac{1}{n^{2}} ) $ where n = 3, 4, 5

For second line n = 4

So $ \frac{1}{\lambda }=R( \frac{1}{2^{2}}-\frac{1}{4^{2}} )=\frac{3}{16}R\Rightarrow \lambda =\frac{16}{3R} $



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