The radii of two Bhor's orbits of the hydrogen atom are in the ratio `4:9`. The energy difference between
then may be:

Options

A. `1.89 eV`
B. `3.4 eV`
C. `12.09 eV`
D. `0.472 eV`

Answer

A

Explanation

`r_(n_(1))/(r_(n_(2)))=(4)/(9)rArr (n_(1)^(2))/(n_(2)^(2))rArrn_(1)=(2)/(3)n_(2)`
`n_(2)=3,n_(2)=6,n_(2)=9`
`n_(2)=2,n_(2)=4,n_(2)=6`" "and so" "on"
`DeltaE=13.6[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
case I `n_(1)=2, n_(2)=3`
`DeltaE=1.89 eV` Ans
Case II `n_(1)=4, n_(2)=6`
`DeltaE=0.472 eV` Ans