Atomic Structure Question 474

Question: Last line of Lyman series for H-atom has wavelength $ {\lambda_1}\overset{o}{\mathop{A}},, $

$ 2^{nd} $ line of Balmer series has wavelength $ {\lambda_2}\overset{o}{\mathop{A}}, $ then

Options:

A) $ \frac{16}{{\lambda_1}}=\frac{9}{{\lambda_2}} $

B) $ \frac{16}{{\lambda_2}}=\frac{9}{{\lambda_1}} $

C) $ \frac{4}{{\lambda_1}}=\frac{1}{{\lambda_2}} $

D) $ \frac{16}{{\lambda_1}}=\frac{3}{{\lambda_2}} $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ \frac{1}{{\lambda_1}}=R{{(1)}^{2}}( \frac{1}{1^{2}}-\frac{1}{{{\infty }^{2}}} ) $ and $ \frac{1}{{\lambda_2}}=R{{(1)}^{2}}( \frac{1}{2^{2}}-\frac{1}{4^{2}} ) $

$ \therefore $ $ {\lambda_1}=\frac{1}{R} $ and $ {\lambda_2}=\frac{16}{3R} $

$ \therefore $ $ \frac{16}{{\lambda_2}}=\frac{3}{{\lambda_1}} $



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