The shortest wavelength of the Brackett series of a hydrogen-like atom (atomic number of `Z` ) is the same as the shortest wavelength of the Balmer series of hydrogen atom. The value of `z` is

Options

A. 2
B. 3
C. 4
D. 6

Answer

A

Explanation

Shortest wavelength of Brackett series corresponds to the transition of electron between `n_(1) = 4and n_(1) = oo`. So,
`(Z^(2)) ((13.6)/(16)) = ((13.6)/(4))`
`:. z^(2) = 4`
or `z = 2`