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

Answer

1

Explanation

`hv=DeltaE=13.6 z^(2)(1/n_(1)^(2)-1/n_(2)^(2))rArr v_(He^(+))=v_(H)xxz^(2)(1/((n_(1)/2)^(2))-1/((n_(2)/2)^(2)))=v_(H)(1/((2/2)^(2))-1/((4/2)^(2)))`
For H-atom `rArr n_(1)=1, n_(2)=2`