Question

The longest wavelength of ` He^(+)` in paschen series is "m", then shortest wavelenght of `Be^(+3)` in Pacchen series is( in terms of m):
A. `(7)/(64)m`
B. `(5)/(36)m`
C. m
D. `(53)/(8)m`

Answer 1

Correct Answer - A
`(1)/(lamda_(He^(+)))=RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`(1)/(m)=Rxx4[(1)/(9)-(1)/(16)]=(7R)/(36)`. . . .(i)
`(1)/(lamda_(Be^(3+)))=Rxx16[(1)/(9)-(1)/(infty)]=(16R)/(9)` .. . .(ii)
Dividing eq. (i) by (ii)
`(lamdaBe^(3+)))/(m)=(7xx9)/(16xx36)`
`lamda_(Be^(3+))=(7)/(64)m`