The shortest wavelength in the Balmer series of hydrogen atom is equal to the shortest wavelength in the…
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
$2$
$3$
$4$
$6$
Solution
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
$\begin{array}{ll}
& \mathrm{n}=4, \mathrm{~m}=\infty \\
\therefore \quad & \frac{1}{\lambda_2}=\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2\left(\frac{1}{16}\right) \\
\therefore \quad & \lambda_2=\frac{16}{\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2} \\
& \text { Given: } \lambda_1=\lambda_2, \\
\therefore \quad & \frac{4}{\mathrm{R}_{\mathrm{H}}}=\frac{16}{\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2} \Rightarrow \mathrm{Z}^2=4 \\
\therefore \quad & \mathrm{Z}=2
\end{array}$