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)

2

Answer 2

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 $Z=2$