Question

The wavelength of the ultraviolet region of the hydrogen spectrum is 122 nm. The wavelength of the second sperctral line in the Balmer series of singly ionized helium atom is (a) `1215 Å` (b) `1640 Å` (C ) `2430 Å` (d) `4687 Å`

Answer 1

Correct Answer - A
First line of
Balmer series. Second line of
Balmer series. For hydrogen type atoms,
`(1)/(lambda) = RZ^2 ((1)/(n_f^2) - (1)/(n_i^2))`
In the transition form `n_1rarr n_f,`
`:. lambda prop (1)/(Z^2 ((1)/(n_f^2) - (1)/(n_i^2))
`:. lambda_2/lambda_1 = (Z_1^2((1)/(n_f^2)-(1)/(n_i^2))_1)/(Z_2^2((1)/(n_f^2)-(1)/(n_i^2))_2`
`lambda_2 = (lambda_1Z_1^2((1)/(n_f^2)-(1)/(n_i^2))_1)/(Z_2^2((1)/(n_f^2)-(1)/(n_i^2))_2`
Substituting the values we have
`=((6561 Å)(1)^2 ((1)/(2^2)-(1)/(3^2)))/((2)^2((1)/(2^2)- (1)/(4^2))) =1215 Å`
`:.` Correct option is (a).