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Question

Physics Question on Atomic Spectra

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

Answer

4

Explanation

Solution

For last Balmer series
1λb=R[12212]\frac{1}{\lambda_{b}}=R\left[\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right]
λb=4R\lambda_{b}=\frac{4}{R}
For last Lyman series
1λl=R[11212]\frac{1}{\lambda_{l}}=R\left[\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right]
λI=1R\lambda_{I}=\frac{1}{R}
λbλl=4R1R\frac{\lambda_{b}}{\lambda_{l}}=\frac{\frac{4}{R}}{\frac{1}{R}}
λbλl=4\frac{\lambda_{b}}{\lambda_{l}}=4