1.

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

Answer»

1215Å
1640Å
2430Å
4687Å

Solution :`1/(LAMBDA H_(2))=RZ_(H)^(2) [1/4-1/9] =R(1)^(2)[(5)/(36)]`
`(1)/(lambda_(He))=RZ_(He)^(2) [1/4-1/(16)]=R(4) [3/(16)]`
`(lambda_(He))/(lambda_(H_(2)))=1/4 [(16)/(3) xx(5)/(36)]=(5)/(27)`
`lambda_(He)=(5)/(27) xx 6561 =1215Å`


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