Which transition in the hydrogen atomic spectrum will have the same wavelength as the balmer transition, n = 4 to n = 2 of `He^(+)` spectrum ?

Options

A. a) n = 4 to n = 3
B. b) n = 3 to n = 2
C. c) n = 4 to n = 2
D. d) n = 2 to n = 1

Answer

D

Explanation

According to Rydberg.s formula
`(1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))Z^(2)`
For `He^(+)` ion, Z = 2, `n_(1) = 2 and n_(2) = 4`
`therefore" "(1)/(lambda) = R ((1)/(2^(2)) - (1)/(4^(2))) xx 2^(2)`
`= R xx (3)/(16) xx 4`
`(1)/(lambda) = (3R)/(4)`
or `lambda = (4)/(3R)`
For getting the same value of wavelength in hydrogen atomic spectrum, the values of `n_(1) and n_(2)` must be 1 and 2 respectively.
`therefore" "(1)/(lambda) = R((1)/(1^(2)) - (1)/(2^(2)))xx 1^(2)`
`(1)/(lambda) = (3R)/(4)`
or `lambda = (4)/(3R)`