Question

The ratio of the wavelengths of the first and second Balmer lines of the hydrogen spectrum is: 
 

• A. 4 : 3
• B. 36 : 5
• C. 16 : 3
• D. 27 : 20

Hint:

Remember that the series limit for the Balmer series is n = 2. This formula applies to all lines where \( n>2 \) in the Balmer series.

Answer 1

The Correct Option is D

The wavelengths of the Balmer series are given by the Balmer formula: \[ \frac{1}{\lambda} = R \left(\frac{1}{2^2} - \frac{1}{n^2}\right) \] where \( R \) is the Rydberg constant, \( \lambda \) is the wavelength, and \( n \) is the principal quantum number, with \( n>2 \). 
For the first Balmer line (H\( \alpha \), \( n = 3 \)): \[ \frac{1}{\lambda_1} = R \left(\frac{1}{2^2} - \frac{1}{3^2}\right) \] \[ \lambda_1 = \frac{1}{R \left(\frac{1}{4} - \frac{1}{9}\right)} = \frac{1}{R \times \frac{5}{36}} \] For the second Balmer line (H\( \beta \), \( n = 4 \)): \[ \frac{1}{\lambda_2} = R \left(\frac{1}{2^2} - \frac{1}{4^2}\right) \] \[ \lambda_2 = \frac{1}{R \left(\frac{1}{4} - \frac{1}{16}\right)} = \frac{1}{R \times \frac{3}{16}} \] The ratio of these wavelengths: \[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{1}{R \times \frac{5}{36}}}{\frac{1}{R \times \frac{3}{16}}} = \frac{\frac{16}{3}}{\frac{36}{5}} = \frac{16 \times 5}{3 \times 36} = \frac{80}{108} = \frac{20}{27} \]