The electron in a hydrogen atom makes a transition from `n=n_(1)` to `n=n_(2)` state. The time period of the electron in the initial state `(n_(1))` is eigh times that in the final state `(n_(2))`. The possible values of `n_(1)` and `n_(2)` are

Options

A. `n_(1)=4, n_(2)=2`
B. `n_(1)=8, n_(2)=2`
C. `n_(1)=8, n_(2)=1`
D. `n_(1)=6, n_(2)=2`

Answer

A

Explanation

In the `n^("th")` orbit, let `r_(n)=` radius and `v_(n)=` speed of electron.
Time period, `T_(n)=(2pir_(n))/(v_(n)) prop (r_(n))/(v_(n))`
Now, `r_(n) prop n^(2) and v_(n)=(1)/(n)`
`therefore" "(r_(n))/(v_(n)) prop n^(3) or T_(n) prop n^(3)`
Here, `8=((n_(1))/(n_(2)))^(3) or (n_(1))/(n_(2))=2 or n_(1)=2n_(2)`