Atoms And Nuclei Question 325
Question: If the series limit wavelength of Lyman series for the hydrogen atom is $ 912\overset{o}{\mathop{A}}, $ , then the series limit wavelength for Balmer series of hydrogen atoms is
Options:
A) $ 912\overset{o}{\mathop{A}}, $
B) $ 912\times 2\overset{o}{\mathop{A}}, $
C) $ 912\times 4\overset{o}{\mathop{A}}, $
D) $ \frac{912}{2}\overset{o}{\mathop{A}}, $
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Answer:
Correct Answer: C
Solution:
- $ \frac{1}{\lambda }=R[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} ] $
For limiting wavelength of Lyman series $ n_{1}=1,,n_{2}=\infty \frac{1}{{\lambda_{L}}=R} $
For limiting wavelength of Balmer series $ n_{1}=1,n_{2}=\infty $
$ \frac{1}{{\lambda_{B}}}=R( \frac{1}{4} )\Rightarrow {\lambda_{B}}=\frac{4}{R} $
$ \therefore {\lambda_{B}}=4{\lambda_{1}}=4\times 912\overset{o}{\mathop{A}},. $
BETA
