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

Answer: (A)

4 : 1

Answer 2

The wavelength of a spectral line in the hydrogen spectrum is given 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:

$n_1 = 2, \quad n_2 = \infty$

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

For the shortest wavelength in the Lyman series:

$n_1 = 1, \quad n_2 = \infty$

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

Taking the ratio of wavelengths:

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

Thus, the ratio is:

$\lambda_B : \lambda_L = 4 : 1.$

Final Answer: 4:1 (Option 1)