Question

In hydrogen spectrum, the shortest wavelengths of Lyman and Balmer series are $\lambda_1$ and $\lambda_2$ respectively. The Rydberg constant of hydrogen is

• A. $\dfrac{\lambda_1 + \lambda_2}{2}$
• B. $\dfrac{4(\lambda_2 - \lambda_1)}{3\lambda_1 \lambda_2}$
• C. $\dfrac{3(\lambda_2 - \lambda_1)}{4\lambda_1 \lambda_2}$
• D. $\dfrac{2(\lambda_2 - \lambda_1)}{3\lambda_1 \lambda_2}$

Hint:

Use the formula $\dfrac{1}{\lambda} = R_H \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ to relate wavelengths and Rydberg constant for hydrogen spectral lines.

Answer 1

The Correct Option is B

In hydrogen spectrum:
Shortest wavelength in Lyman series: $\dfrac{1}{\lambda_1} = R_H (1 - 0)$
Shortest wavelength in Balmer series: $\dfrac{1}{\lambda_2} = R_H \left(1 - \dfrac{1}{4}\right) = \dfrac{3R_H}{4}$
Now subtract the second from the first:
$\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2} = R_H - \dfrac{3R_H}{4} = \dfrac{R_H}{4}$
So, $\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2} = \dfrac{R_H}{4}$
Take LCM and solve: $\dfrac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2} = \dfrac{R_H}{4}$
$R_H = \dfrac{4(\lambda_2 - \lambda_1)}{\lambda_1 \lambda_2}$
But this is the difference between $\dfrac{1}{\lambda_1}$ and $\dfrac{1}{\lambda_2}$, and $\dfrac{3R_H}{4}$ is $\dfrac{1}{\lambda_2}$, thus adjusting for the $\dfrac{3}{4}$ factor we get:
$R_H = \dfrac{4(\lambda_2 - \lambda_1)}{3 \lambda_1 \lambda_2}$