The wavelength of the first spectral line in the Balmer series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom is

Options

A. `1215Å`
B. `1640Å`
C. `2430Å`
D. `4687Å`

Answer

A

Explanation

The wavelength `(lambda)` of a spectral line in the Balmer series is given by
`1/lambda=RZ^(2)[1/(2^(2))-1/(n^(2))]`
where n=3,4,5,6.....
In case of hydrogen atom, for first spectral line in the Balmer series, Z=1, n=3 , `lambda=lambda_(H)` (say)
`:. 1/(lambda_(H))=Rxx(1)^(2)[1/(2^(2))-1/(3^(2))]=Rxx5/36.......(i)`
In case of singly -ionized helium atom, for second spectral line in the Balmer series, Z=2 , n=4,
`lambda=lambda_(He^(+))`
`1/(lambda_(He^(+)))=Rxx(1)^(2)[1/(2^(2))-1/(4^(2))]= 4Rxx3/16.......(ii)`
Dividing (i) by (ii), we get
`(lambda_(He^(+)))/(lambda_(H))=(5//36)/(4xx3//16)=5/27`
`lambda_(H^(+))= lambda_(H) xx5/27=(6561Å)xx5/27=1215Å`