Atoms And Nuclei Question 321
Question: If the wavelength of the first line of the Balmer series in the hydrogen spectrum is K, then the wavelength of the first line of the Lyman series is
Options:
A) $ ( 27/5 )\lambda $
B) $ ( 5/27 )\lambda $
C) $ ( 32/27 )\lambda $
D) $ ( 27/32 )\lambda $
Show Answer
Answer:
Correct Answer: B
Solution:
- For first line of Balmer series $ \frac{1}{\lambda }=R( \frac{1}{4}-\frac{1}{9} )\Rightarrow R=\frac{36}{5\lambda } $
$ \therefore $ Wavelength of the first line, $ {\lambda_{L}} $ of the Lyman series is given by $ \frac{1}{{\lambda_{L}}}=R( 1-\frac{1}{4} )=\frac{36}{5\lambda }\times \frac{3}{4}=\frac{27}{5\lambda } $
$ \Rightarrow {\lambda_{L}}=\frac{5\lambda }{27} $
