Question

The shortest wavelength of the Bracken 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 $Z=2$