Question 463: Atoms & Nuclei

Question

If ' $\lambda_1$ ' and ' $\lambda_2$ ' are the wavelengths of the first line of the Lyman and Paschen series respectively, then $\lambda_2: \lambda_1$ is

Accepted Answer

In the hydrogen atom spectrum, the Lyman and Paschen series are defined by the electron transition levels. The Lyman series involves transitions where the final energy level is $n_1 = 1$, while the Paschen series involves transitions to $n_1 = 3$.
The wavelength of light emitted during a transition between two energy levels in a hydrogen atom can be determined using the Rydberg formula:
$$ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $$
where $ R_H $ is the Rydberg constant, $ n_1 $ is the lower energy level, and $ n_2 $ is the higher energy level.
For the first line of the series (the first transition):
Lyman series: $ n_1 = 1 $, $ n_2 = 2 $
Paschen series: $ n_1 = 3 $, $ n_2 = 4 $
Calculating the wavelength for the Lyman series:
$$ \frac{1}{\lambda_1} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \cdot \frac{3}{4} $$
Thus,
$$ \lambda_1 = \frac{4}{3 R_H} $$
Calculating the wavelength for the Paschen series:
$$ \frac{1}{\lambda_2} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{9} - \frac{1}{16} \right) $$
Simplifying the above expression:
$$ \frac{1}{\lambda_2} = R_H \left( \frac{16 - 9}{144} \right) = R_H \cdot \frac{7}{144} $$
Thus,
$$ \lambda_2 = \frac{144}{7 R_H} $$
Finding the ratio $\frac{\lambda_2}{\lambda_1}$:
$$ \frac{\lambda_2}{\lambda_1} = \frac{\frac{144}{7 R_H}}{\frac{4}{3 R_H}} = \frac{144 \cdot 3}{7 \cdot 4} = \frac{432}{28} = \frac{108}{7} $$
Therefore, the ratio $\lambda_2 : \lambda_1$ is
$$ 108:7 $$
Thus, the correct answer is
Option D: $108: 7$
.

Answer

$3: 1$

Answer

$30: 1$

Answer

$50: 7$

Answer

$108: 7$

Atoms & Nuclei

If ' \lambda_1 ' and ' \lambda_2 ' are the wavelengths of the first line of the Lyman and Paschen series respectively, then \lambda_2: \lambda_1 is

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