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} $


जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक