Question

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

Answer 1

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

For Hydrogen atom (Lymen series) $\frac{1}{({\lambda }_{min}{)}_H}=R[\frac{1}{1^2}-\frac{1}{\infty }]=R$

$\Rightarrow ({\lambda }_{min}{)}_H=\frac{1}{R}.......(i)$

For hydrogen like atom (Balmer series) ${\left(\frac{1}{{\lambda }_{min}}\right)}_{atom}=R{Z}^2(\frac{1}{2^2}-\frac{1}{\infty })$

$\Rightarrow ({\lambda }_{min}{)}_{atom}=\frac{4}{R{Z}^2}.....(ii)$

From equation (i) and (ii) $\frac{1}{R}=\frac{4}{R{Z}^2}\Rightarrow Z=2.$