1.

If the shortest wavelength of H atom in Layman series is x, then the longest wavelength in Balmer series of He^(+) is

Answer»

`(36x)/5`
`(5x)/9`
`(9x)/5`
`x/5`

SOLUTION :For the SHORTEST wavelength of Lyman series in H-atom.
`1/(lamda)=R(1/(1^(2))-1/(prop^(2)))=R` or `lamda=1/R=x`………..(i)
For largest wavelength of Balmer series.
`1/(lamda.)=Rxx2^(2)xx(1/4-1/9)=4Rxx5/36`
or `1/(lamda.)=(5R)/9`…………(ii)
or `lamda.=9/(5R)=(9x)/5""( :. 1/R=x)`


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