Question

The shortest wavelength of the Brackett series of 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. 3
• B. 4
• C. 5
• D. 2

Answer 1

Answer: (D)

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