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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
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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
$$
\begin{aligned}
\frac{h c}{\lambda} & =Z^2 R h c\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\
\therefore \quad\left(\frac{1}{1^2}-\frac{1}{2^2}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\
\left(1-\frac{1}{4}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\
Z & =2
\end{aligned}
$$
$$
\frac{h c}{\lambda}=R h c\left(\frac{1}{1^2}-\frac{1}{2^2}\right)
$$
And for H-like ion
$$
\begin{aligned}
\frac{h c}{\lambda} & =Z^2 R h c\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\
\therefore \quad\left(\frac{1}{1^2}-\frac{1}{2^2}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\
\left(1-\frac{1}{4}\right) & =Z^2\left(\frac{1}{4}-\frac{1}{16}\right) \\
Z & =2
\end{aligned}
$$
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