Atoms - Result Question 15

16. 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: 16. (c)

Solution:

  1. (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.

$\lambda _{\text{PFund }}>\lambda _{\text{Brackett }}>\lambda _{\text{Paschen }}>\lambda _{\text{Balmer }}>\lambda _{\text{Lymen }}$



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