The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of…
The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number $Z$ of hydrogen like ion is
4
1
2
3
Solution
Lyman series for $\mathrm{H}$-ion
$\frac{h c}{\lambda}=R h c\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$
And for H-like ion
$\begin{aligned}
\frac{h c}{\lambda} & =Z^2 R h c\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\
\therefore \quad\left(\frac{1}{1^2}-\frac{1}{2^2}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\
\left(1-\frac{1}{4}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\
Z & =2
\end{aligned}$