Question

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

• A. 4 : 1
• B. 1 : 2
• C. 1 : 4
• D. 2 : 1

Answer 1

The Correct Option is A

The wavelength \( \lambda \) of a hydrogen spectral line is determined by:

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

where \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

For the shortest wavelength in the Balmer series:

\[ n_1 = 2, \quad n_2 = \infty \]

This results in:

\[ \frac{1}{\lambda_B} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = RZ^2 \left( \frac{1}{4} \right). \]

For the shortest wavelength in the Lyman series:

\[ n_1 = 1, \quad n_2 = \infty \]

This yields:

\[ \frac{1}{\lambda_L} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = RZ^2 (1). \]

The ratio of these wavelengths is calculated as follows:

\[ \frac{\lambda_B}{\lambda_L} = \frac{\frac{1}{RZ^2 \frac{1}{4}}}{\frac{1}{RZ^2 (1)}} = 4 : 1. \]

Therefore, the ratio is:

\[ \lambda_B : \lambda_L = 4 : 1. \]

Final Answer: 4:1 (Option 1)