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} $
BETA
