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The shortest wavelength of H-atom in Lym...

The shortest wavelength of H-atom in Lyman series is x, then longest wavelength in Balmer series of `He^(+)` is

A

`(9x)/(5)`

B

`(36x)/(5)`

C

`(x)/(4)`

D

`(5x)/(9)`

Text Solution

Verified by Experts

The correct Answer is:
A

`bar(nu) = (1)/(lambda) = R'Z^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
For shortest wavelength (maximum energy) in Lyman series of hydrogen Z = 1, `n_(2) = 1, n_(2) rarr oo` and
`lambda = x`
`(1)/(x) = R'`
For longest wavelength (minimum energy) in Balmer series of `He^(+), Z = 2` and `n_(1) = 2, n_(2) = 3`.
`(1)/(lambda) = R'2^(2)[(1)/(2^(2))-(1)/(3^(2))]`
`(1)/(lambda) = (4)/(x)[(1)/(4)-(1)/(9)]`
`(1)/(lambda) = (4)/(x)(5)/(36)`
`lambda = (9x)/(5)`
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