Question

Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is :

• A. $ \frac{9}{31}$
• B. $ \frac{5}{27}$
• C. $ \frac{3}{23}$
• D. $ \frac{7}{29}$

Answer 1

The Correct Option is B

To solve the problem of finding the ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum, we will follow these steps:

Understand the concept of spectral series in the hydrogen atom: 

• The Lyman series corresponds to electronic transitions from higher energy levels to the first energy level (\(n_1 = 1\)).
• The Balmer series corresponds to transitions to the second energy level (\(n_1 = 2\)).

Utilize the formula for the wavelength of spectral lines in the hydrogen spectrum, which is given by the Rydberg formula:

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

where \(\lambda\) is the wavelength, \(R\) is the Rydberg constant, \(n_1\) and \(n_2\) are the principal quantum numbers with \(n_2 > n_1\).

Find the longest wavelength (smallest energy transition) for each series:

For the Lyman series, the smallest energy transition (longest wavelength) occurs when \(n_2 = 2\):

\[\frac{1}{\lambda_{\text{Lyman}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left(1 - \frac{1}{4}\right) = \frac{3}{4}R\]\[\lambda_{\text{Lyman}} = \frac{4}{3R}\]

For the Balmer series, the smallest energy transition occurs when \(n_2 = 3\):

\[\frac{1}{\lambda_{\text{Balmer}}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left(\frac{1}{4} - \frac{1}{9}\right) = R \left(\frac{5}{36}\right)\]\[\lambda_{\text{Balmer}} = \frac{36}{5R}\]

Calculate the ratio of the longest wavelengths of the Lyman and Balmer series:

\[\text{Ratio} = \frac{\lambda_{\text{Lyman}}}{\lambda_{\text{Balmer}}} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3} \times \frac{5}{36} = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} = \frac{5}{27}\]

Therefore, the ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is \(\frac{5}{27}\).