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. 3
B. 4
C. 5
D. 2

Answer

D

Explanation

According to the question,
`" " lambda_(("shortest"))"Brackett"=(lambda_("shortest"))"Balmer"`"
`rArr" "(1)/((lambda_("shortest"))"Brackett")=(1)/((lambda_("shortest"))"Balmer")`
`rArrZ^(2)R((1)/(4^(2)))=R((1)/(4))rArrZ^(2)=4 rArrZ=2`