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 $