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What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition `n=4` to `n=2` of `He^(+)` spectrum ?A. n=4 to n=3B. n=3 to n=2C. n=4 to n=2D. n=2 to n=1

Answer» Correct Answer - D
For `He^+` ion, `1/lambda=Z^2R[1/n_1^2-1/n_2^2]`
`(2)^2 R[1/2^2 -1/4^2] ="3R"/4`
For hydrogen atom , `1/lambda=R[1/n_1^2-1/n_2^2]`
`"3R"/4=R[1/n_1^2-1/n_2^2]` or `1/n_1^2-1/n_2^2=3/4`
`n_1=1` and `n_2=2`


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