Atoms And Nuclei Question 319
Question: Taking Rydberg’s constant $ R=1.097\times 10^{7}m $ , first and second wavelength of Balmer series in hydrogen spectrum is
Options:
A) $ 2000\overset{o}{\mathop{\text{A }}},,3000\overset{o}{\mathop{\text{A }}},~ $
B) $ \text{1575 }\overset{o}{\mathop{\text{A }}},\text{, }\text{ 2960 }\overset{o}{\mathop{\text{A }~}}, $
C) $ \text{6529 }\overset{o}{\mathop{A}},\text{, },\text{4280 }\overset{o}{\mathop{\text{A }~}}, $
D) $ 6552\overset{o}{\mathop{\text{A }}},,4863\overset{o}{\mathop{\text{A }}},~~ $
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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 $
$ \Rightarrow {\lambda_{1}}=6563\overset{o}{\mathop{A}},. $ For second wavelength, n, $ =2,n_{2}=4 $
$ \Rightarrow {\lambda_{2}}=4861\overset{o}{\mathop{A}}, $
BETA
