The shortest wavelength in the Balmer series of hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen like atom of atomic number \(z\). The value of \(\mathrm{z}\) is

Using Rydberg's formula,
\(\frac{1}{\lambda}=R_H Z^2\left[\frac{1}{\mathrm{n}^2}-\frac{1}{\mathrm{~m}^2}\right]\)
where \(R_H\) is the Rydberg's constant
For calculating the shortest wavelength in the Balmer series of hydrogen atom,
\(\begin{aligned}
& \mathrm{n}=2, \mathrm{~m}=\infty \text { and } \mathrm{Z}=1 \\
& \therefore \quad \lambda_1=\frac{4}{\mathrm{R}_{\mathrm{H}}}
\end{aligned}\)
For hydrogen like atom, the shortest wavelength is given by,
In Brackett series
\(\begin{array}{ll}
& \mathrm{n}=4, \mathrm{~m}=\infty \\
\therefore \quad & \frac{1}{\lambda_2}=\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2\left(\frac{1}{16}\right) \\
\therefore \quad & \lambda_2=\frac{16}{\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2} \\
& \text { Given: } \lambda_1=\lambda_2, \\
\therefore \quad & \frac{4}{\mathrm{R}_{\mathrm{H}}}=\frac{16}{\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2} \Rightarrow \mathrm{Z}^2=4 \\
\therefore \quad & \mathrm{Z}=2
\end{array}\)