For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is. [2025]

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Solution

Q.
For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is. [2025]

1 5 : 36

2 5 : 27

3 3 : 4

4 27 : 5

Ans.
(2)

Largest wavelength of Lyman series,

$\frac{1}{λ_{1}} = R [\frac{1}{1} - \frac{1}{4}] = \frac{3 R}{4}$

$λ_{1} = \frac{4}{3 R}$ ... (i)

Largest wavelength of Balmer series

$\frac{1}{λ_{2}} = R [\frac{1}{4} - \frac{1}{9}] = \frac{5 R}{36}$

$λ_{2} = \frac{36}{5 R}$

Then, $\frac{λ_{1}}{λ_{2}} = \frac{5}{27}$

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