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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

`(36lamda_1)/5`

B

`(5lamda_1)/9`

C

`(9lamda_1)/5`

D

`(27lamda_1)/5`

Text Solution

Verified by Experts

The correct Answer is:
C
For hydrogen atom
For Lyman series `n_1 =1 and n_2 = oo`
`1/(lamda_H)=R_H[1/1-1/oo]` So, `lamda=1/R_H`
For `He^(+)` ion
Balmer series `n_1 = 2 and n_2 = 3`
`1/(lamda_(He^(+)))=R_HxxZ^2[1/4-1/9]`
`1/(lamda_(He^(+)))=R_Hxx4xx5/36`
`1/(lamda_(He^(+)))=5/9 R_H=(5/9)1/lamda`
`(lamda_(He^(+)))=9/5 lamda`
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