Question

If the series limit of Lyman series for Hydrogen atom is equal to the series limit of Balmer series for a hydrogen like atom, then the atomic number of this hydrogen like atom is -

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

By using,

$\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

For Hydrogen atom :

$\frac{1}{{\lambda }_1}=R[\frac{1}{1^2}-\frac{1}{\infty }]=R$

$\Rightarrow {\lambda }_1=\frac{1}{R}...\left(1\right)$

For hydrogen like atom :

$\frac{1}{{\lambda }_2}=R{Z}^2[\frac{1}{2^2}-\frac{1}{\infty }]$

$\Rightarrow {\lambda }_2=\frac{4}{R{Z}^2}...\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$,

$\frac{1}{R}=\frac{4}{R{Z}^2}[\text{Given}{\lambda }_1={\lambda }_2]$

​​​​$\Rightarrow Z=2$

Hence, option $\left(B\right)$ is the correct answer.