The frequency of light emitted for the transition `n = 4` to `n =2` of `He^+` is equal to the transition in `H` atom corresponding to which of the following ?

Options

A. `n = 3 " to " n = 1`
B. `n = 2 " to " n = 1`
C. `n = 3 " to " n = 2`
D. `n = 4 " to " n = 3`

Answer

B

Explanation

`bar(v) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) Z^(2)`
`v = (c)/(lamda) = cbar(v) or bar(v) = (v)/(c)`
`:. (v)/(c) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) Z^(2)`
or `v = Rc ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) Z^(2) = R' ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) Z^(2)`
For `He^(+)`, for transition `n = 4 " to " n = 2`
`v = R' ((1)/(2^(2)) - (1)/(4^(2))) xx 2^(2) = R' ((1)/(1^(2)) - (1)/(2^(2)))`
For H, Z = 1, therefore for same value of v, transition will be form `n = 2 " to " n = 1`