Question

The shortest wavelength of the Brackett series of a hydrogen-like atom (atomic number $Z$) is the same as the shortest wavelength of the Balmer series of hydrogen atom. Find the value of $Z$.

Answer 1

Shortest wavelength for the Balmer-series for hydrogen is given by,

$1/{\lambda }_1=R(1/2^2-1/{\infty }^2)=R/4$

Shortest wavelength for the Brackett-series for atom with atomic number $Z$ is given by,

$1/{\lambda }_2=R{Z}^2(1/4^2-1/{\infty }^2)=R{Z}^2/16$

As both the wavelengths are equal,

$R/4=R{Z}^2/16\Rightarrow Z^2=4\Rightarrow Z=2$