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}} $
