Atoms And Nuclei Question 131

Question: The shortest wavelength in the Lyman series of hydrogen spectrum is 912 Å corresponding to a photon energy of 13.6 eV. The shortest wavelength in the Balmer series is about [MP PMT 2004]

Options:

A) 3648 Å

B) 8208 Å

C) 1228 Å

D) 6566 Å

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Answer:

Correct Answer: A

Solution:

In Lyman series $ {{({\lambda _{\min }})} _{L}}=\frac{1}{R} $ and $ {{({\lambda _{\min }})} _{B}}=\frac{4}{R} $
Þ $ {{({\lambda _{\min }})} _{B}}=4\times {{({\lambda _{\min }})} _{L}}=4\times 912=3648,{\AA} $



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