Atoms Ques 9
9. In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is
[2015 RS, 2013]
(a) $\frac{9}{4}$
(b) $\frac{27}{5}$
(c) $\frac{5}{27}$
(d) $\frac{4}{9}$
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Answer:
Correct Answer: 9.(c)
Solution:
- (c) For Lyman series $(2 \to 1)$
$\frac{1}{\lambda_L}=R[1-\frac{1}{2^{2}}]=\frac{3 R}{4}$
For Balmer series $(3 \to 2)$
$ \begin{aligned} & \frac{1}{\lambda_B}=R[\frac{1}{4}-\frac{1}{9}]=\frac{5 R}{36} \\ & \Rightarrow \frac{\lambda_L}{\lambda_B}=\frac{\frac{4}{3 R}}{\frac{36}{5 R}}=\frac{4}{36}(\frac{5}{3})=\frac{5}{27} \end{aligned} $
The wavelength of spectral lines increases with the increases of order of the series.
$\lambda _{\text{PFund }}>\lambda _{\text{Brackett }}>\lambda _{\text{Paschen }}>\lambda _{\text{Balmer }}>\lambda _{\text{Lymen }}$
BETA
