Atoms And Nuclei Question 55
Question: The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is [MP PET 1999]
Options:
A) $ \frac{4}{3} $
B) $ \frac{525}{376} $
C) 25
D) $ \frac{900}{11} $
Show Answer
Answer:
Correct Answer: D
Solution:
Shortest wavelength comes from
$ \Rightarrow \frac{N}{10.38}={{( \frac{1}{2} )}^{5}}\Rightarrow N=10.38\times {{( \frac{1}{2} )}^{5}} $ to $ n_{2}=1 $
and longest wavelength comes from $ n_{1}=6 $ to $ n_{2}=5 $ in the given case.
Hence $ \frac{1}{{\lambda_{\min }}}=R,( \frac{1}{1^{2}}-\frac{1}{{{\infty }^{2}}} )=R $
$ \frac{1}{{\lambda_{\max }}}=R,( \frac{1}{5^{2}}-\frac{1}{6^{2}} )=R,( \frac{36-25}{25\times 36} )=\frac{11}{900}R $
$ \therefore \frac{{\lambda_{\max }}}{{\lambda_{\min }}}=\frac{900}{11} $
BETA
