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The longest wavelength of He^(+) in pas...

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`

Text Solution

Verified by Experts

The correct Answer is:
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`
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