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 :

The frequency of light emitted for the transition n = 4 " to " n = 2 o

1.	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 :
Answer» `n = 3 " to " n = 1` `n = 2 " to " n = 1` `n = 3 " to " n = 2` `n = 4 " to " n = 3` Solution :`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`