Atoms Ques 5
5. The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is :-
(a) 1
(b) 4
(c) 0.5
(d) 2
[2017]
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Answer:
Correct Answer: 5.(b)
Solution:
- (b) For last line of Balmer series:
$ n_1=2 \text{ and } n_2=\infty $
$\frac{1}{\lambda_B}=RZ^{2}[\frac{1}{n_1^{2}}-\frac{1}{n_2{ }^{2}}]=R_1{ }^{2}[\frac{1}{2^{2}}-\frac{1}{\infty^{2}}]$
$\lambda_B=\frac{4}{R}$
For last line of Lyman series: $n_1=1$ and $n_2=\infty$ $\frac{1}{\lambda_L}=RZ^{2}[\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}]=RZ^{2}[\frac{1}{1^{2}}-\frac{1}{\infty^{2}}]$
$\lambda_L=\frac{1}{R}$
Dividing equation (i) by (ii)
$\frac{\lambda_B}{\lambda_L}=\frac{\frac{4}{R}}{\frac{1}{R}}$
Ratio of wavelengths is $\frac{\lambda_B}{\lambda_L}=4$
BETA
