Atoms And Nuclei Question 135

Question: Taking Rydberg’s constant $ R_{H}=1.097\times 10^{7}m $ first and second wavelength of Balmer series in hydrogen spectrum is [Pb. PMT 2004]

Options:

A) 2000 Å, 3000 Å

B) 1575 Å, 2960 Å

C) 6529 Å, 4280 Å

D) 6552 Å, 4863 Å

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Answer:

Correct Answer: D

Solution:

$ \frac{1}{\lambda }=R[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} ]. $ For first wavelength, $ n_{1}=2 $ , $ n_{2}=3 $
Þ $ {\lambda_{1}}=6563,{\AA} $ . For second wavelength, $ n_{1}=2 $ , $ n_{2}=4 $
Þ $ {\lambda_{2}}=4861,{\AA} $



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