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. The value of $Z$ is

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

Answer: (A)

Shortest wavelength of Brackett series corresponds to the transition of electron between $n_{1}=4$ and $n_{2}=\in fty$ and the shortest wavelength of Balmer series corresponds to the transition of electron between $n_{1}=2$ and $n_{2}=\in fty$ , So,
$\left(z^{2}\right)\left(\frac{13.6}{16}\right)=\left(\frac{13.6}{4}\right)$
$\therefore \, \, Z^{2}=4$
or $Z=2$