One of the lines in the emission spectrum of `Li^(2+)` has the same wavelength as that of the `2nd` line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is:

Options

A. `n = 4 rarr n = 2`
B. `n = 8 rarr n = 2`
C. `n = 8 rarr n = 4`
D. `n = 12 rarr n= 6`

Answer

D

Explanation

For `2nd` line of Balmer series in hydrogen spectrum,
`(1)/(lambda) = R(1) ((1)/(2^(2)) - (1)/(4^(2))) = (3)/(16)R`
For `Li^(2+) : (Li^(2+))`
Which is satisfied by only `(d)`.