Question:

The first emission of hydrogen atomic spectrum in Lyman series appears at a wavelength of

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The first emission in the Lyman series corresponds to the transition from $ n = 2 $ to $ n = 1 $ in the hydrogen atom. Use the Rydberg formula to calculate the wavelength for the transition.

Updated On: May 3, 2025

$ \frac{1}{4} \, \text{cm}^{-1} $
$ \frac{4}{3R} \, \text{cm} $
$ \frac{2}{3} \, \text{cm}^{-1} $
$ \frac{1}{3} \, \text{cm}^{-1} $

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The Correct Option is B

Approach Solution - 1

The hydrogen atomic spectrum in the Lyman series corresponds to electron transitions from higher energy levels ($ n > 1 $) to the first energy level ($ n = 1 $). The wavelength of the emission can be calculated using the Rydberg formula for hydrogen:

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

where:

$ \lambda $ is the wavelength of the emitted light.
$ R_H $ is the Rydberg constant ($ \approx 1.097 \times 10^7 \, \text{m}^{-1} $).
$ n_1 $ is the lower energy level.
$ n_2 $ is the higher energy level.

For the first emission in the Lyman series, $ n_1 = 1 $ and $ n_2 = 2 $. Therefore, the formula becomes:

$ \frac{1}{\lambda} = R_H \left(\frac{1}{1^2} - \frac{1}{2^2}\right) $

$ \frac{1}{\lambda} = R_H \left(1 - \frac{1}{4}\right) $

$ \frac{1}{\lambda} = R_H \times \frac{3}{4} $

So, $ \lambda = \frac{4}{3R_H} $

The wavelength expressed in terms of the Rydberg constant is $ \frac{4}{3R} \, \text{cm} $, which matches the correct answer: $ \frac{4}{3R} \, \text{cm} $

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Approach Solution -2

The Lyman series in the hydrogen atomic spectrum involves electronic transitions from higher energy levels to the first energy level ($n_1 = 1$). The wavelength for these transitions is given by the Rydberg formula:

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

where:

$R$ is the Rydberg constant ($1.097 \times 10^{5} \, \text{cm}^{-1}$)
$n_1 = 1$ for the Lyman series
$n_2 > n_1$, corresponds to different lines of the series

The first line of the series occurs when $n_2 = 2$, leading to:

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

The wavelength $\lambda$ then becomes:

$$\lambda = \frac{4}{3R}$$

The first emission in the Lyman series thus appears at a wavelength of $\frac{4}{3R} \, \text{cm}$, matching the correct option.