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The wavelength of high energy transition...

The wavelength of high energy transition of H atoms is `91.2 nm` Calculate the corresponding wavelength of He atom.

Text Solution

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For H atom: `1/(lamda_(H))=R_(H)[1/(n_(1)^(2))-1/(n_(2)^(2))]`…….i
For `He^(+)` ion: `1/(lamdaHe^(+))=R_(H)xxZ^(2)[1/(n_(1)^(2))-1/(n_(2)^(2))]`……….ii
By eqs I and ii `lamda_(He^(+))=lamda_(H)xx1/(Z^(2))`
`:.lamda_(H)=91.2nm, Z` for `He^(+)=2`
`:.lamda_(He^(+))=91.2xx1/(2^(2))=22.8nm`
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