An electron in a `H_2` atom makes a transition from `n_(1) to n_(2)`. The time period of electron in the initial state is eight times that in the final state. Then ratio of `n_(1)" to "n_(2)`:

Options

A. `1:2`
B. `2:1`
C. `4:1`
D. `8:1`

Answer

B

Explanation

`T^(2) alpha r^(3), (T_(1)^(2))/(T_(2)^(2))=(r_(1)^(3))/(r_(2)^(3)), r_(1)/r_(2)=8^(2/3)=4`
Now `r alpha n^(2)`
`r_(1)/r_(2)=(n_(1))/(n_(2))^(2)`
`n_(1)/n_(2)=sqrt((r_(1))/r_(2))=sqrt(4/1)=2/1`