The longest wavelength of `He^(+)` when its electron return to third Bohr orbit is ‘m' then shortest wavelength of `Be^(3+)` in the corresponding transition series in terms of 'm' is

Options

A. `(7)/(64)`m
B. `(5)/(36)`m
C. `(64)/(7)` m
D. `(53)/(8) m`

Answer

A

Explanation

`(1)/(lamda_(He^(+)))=RZ^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`(1)/(m) = R xx 4[(1)/(9)-(1)/(16)]=(7R)/(36)" ...(i)"`
`(1)/(lamda_(Be^(3+)))=Rxx16[(1)/(9)-(1)/(oo)]=(16 R)/(9)" ....(ii)"`
Dividing eq. (i) by (ii)
`(lamda_(Be^(3+)))/(m) =(7xx9)/(16xx36)," "lamda_(Be^(3+))=(7)/(64) m`