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

Answer

A

Explanation

For Balmer series of hydrogen spectrum
`1/lamda = R(1/2^2 - 1/n^2)` for the first line n=3
`therefore 1/lamda_1 = R [1/4 - 1/9] = (5R)/(36) therefore lamda_1 = (36)/(5R) " " ..(1)`
For the second line of the balmer series, of the singly ionised helium atom ,
`1/lamda_2 = RZ^2 [ 1/2^2 - 1/n^2]` where Z =2, and n' = 4
`therefore 1/lamda_2 = R(4) [1/4-1/16] = 4Rxx3/16= (3R)/(4)`
`therefore lamda_2 = (4)/(3R)`
`therefore lamda_2/lamda_1 = (4)/(3R) xx(5R)/(36) = 5/27`
`therefore lamda_2 = 5/27 xx 6561 = 5xx243=1215 Å`