Atoms - Result Question 11

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12. The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is :-

======= ####12. The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is :-

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/atoms/atoms—result-question-11.md (a) 1

(b) 4

(c) 0.5

(d) 2

[2017]

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Answer:

Correct Answer: 12. (b)

Solution:

  1. (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 $\frac{\lambda_B}{\lambda_L}=4$