Physics - Modern Physics Question with Solution | TestHub

The wavelength of a spectral line is given by Rydberg's formula:

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For the shortest wavelength in the Balmer series of hydrogen ($Z=1$), $n_1=2$ and $n_2=\infty$:

$\frac{1}{{\lambda }_1}=R(1{)}^2\left(\frac{1}{2^2}-0\right)=\frac{R}{4}$

For the shortest wavelength in the Brackett series of a hydrogen-like atom of atomic number $Z$, $n_1=4$ and $n_2=\infty$:

$\frac{1}{{\lambda }_2}=R{Z}^2\left(\frac{1}{4^2}-0\right)=\frac{R{Z}^2}{16}$

Given ${\lambda }_1={\lambda }_2$, we have:

$\frac{R}{4}=\frac{R{Z}^2}{16}$

$Z^2=4$

$Z=2$

Answer: $2$