Question

The difference in the wavelength of the 1st line of Lyman series and 2nd line of Balmer series in a hydrogen atom is :

• A. $\frac{9}{2R}$
• B. $\frac{4}{R}$
• C. $\frac{88}{15R}$
• D. None of these

Answer 1

The correct option is B $\frac{4}{R}$

As we know, wavelength of photon emitted when electron jumps from $n_1$ to $n_2$ is

$\frac{1}{\lambda }=R_H\times Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

In hydrogen atom, for first line of lyman series $n_1=1$ and $n_2=2$.

$\frac{1}{\lambda }=R_H\times 1^2(\frac{1}{1^2}-\frac{1}{2^2})=R_H\times \left(\frac{3}{4}\right)$

Similarly, for 2nd line of balmer series $n_1=2$ and $n_2=4$.

$\frac{1}{\lambda }=R_H\times 1^2(\frac{1}{2^2}-\frac{1}{4^2}=R_H\times (\frac{3}{16})$

So, difference between wavelength is $\frac{12}{3}\times \frac{1}{R}=\frac{4}{R}$.