The wave number of a photon in Brakett series of hydrogen atom is `(9)/(400)` R. the electron has transited from the orbit having quantum number.

Options

A. 5
B. 6
C. 4
D. 7

Answer

A

Explanation

`(1)/(lamda) = R ((1)/(n_(1)^(2))- (1)/(n_(2)^(2)))`
For Brackett series, `n_(1)= 4`
`rArr (9)/(400) R= R ((1)/(4^(2))- (1)/(n^(2)))` [As `(1)/(lamda)`= Wave number]
`rArr (1)/(n^(2))= (1)/(16)- (9)/(400) = (16)/(400) rArr n= 5`