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Atoms and Nuclei

Section: Physics

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Question : 7 of 33

Marks: +1, -0

The ratio of the wavelengths of the first Lyman line and the second Balmer line of hydrogen atom is

[AP EAPCET 22 May 2025 S2]

$3 : 4$
$1 : 4$
$2 : 3$
$1 : 3$

Solution:

The formula for wavelength is:

\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

First Line of Lyman Series:

For Lyman series,

n_1 = 1

. The first line means

n_2 = 2

.

\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) = \frac{3R}{4}

So,

\lambda_L = \frac{4}{3R}

Second Line of Balmer Series:

For Balmer series,

n_1 = 2

. The second line means

n_2 = 4

.

\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) = \frac{3R}{16}

So,

\lambda_B = \frac{16}{3R}

Find the Ratio:

Now, find

\frac{\lambda_L}{\lambda_B}

:

\frac{\lambda_L}{\lambda_B} = \frac{\frac{4}{3R}}{\frac{16}{3R}} = \frac{4}{16} = \frac{1}{4}

This means

\lambda_L : \lambda_B = 1 : 4

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