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 )/(lambda) =c bar(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_(1) = 4` to `n_(2) = 2`
`v = R"[(1)/(2^(2))-(1)/(4^(2))] xx 2^(2) = R' [(1)/(1^(2))-(1)/(2^(2))]`
For H, Z = 1, for the same valu of v, transition will be from n = 2 to n = 1.