A) 3
B) 4
C) 5
D) 2
Correct Answer: D
Solution :
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 \[Z=2\]
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