The radius of hydrogen atom in its ground state is `5.3xx10^(-11)`m. After collision with an electron it is found to have a radius `21.2xx10^(-11)` m, then principle quantum number n of the final state of the atom ......

Options

A. n=4
B. n=2
C. n=16
D. n=3

Answer

B

Explanation

`rpropn^(2)`
`(r_(2))/(r_(1))=((n_(2))/(n_(1)))^(2)`
`r_(2)=` radius in final state
`r_(1)=` radius in ground state
`:.(21.2xx10^(-11))/(5.3xx10^(-11))=((n_(2))/(1))^(2)`
`:.4=n_(2)^(2)`
`:.n_(2)=2`