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)/(lambda_(He))=RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`(1)/(m)= R xx4[(1)/(9) -(1)/(16)]=(7R)/(36)" " …(i)`
`(1)/(lambda_(Be^(3+)))= R xx 16 [ (1)/(9) -(1)/(oo)]=(16R)/(9)" "…(ii)`
Dividing eq. (i) by (ii)
`(lambda_(Be^(3+)))/(m)=(7xx9)/(16xx36), lambda_(Be^(3+))=(7)/(64)m`