The radius of hydrogen atom in its ground state is `5.3 xx 10^(-11)m`. After collision with an electron it is found to have a radius of `21.2 xx 10^(-11)m`. What is the principle quantum number of `n` of the final state of the atom ?

Options

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

Answer

B

Explanation

`r propn^(2) ` i.e., `(r_(f))/(r_(i))=((n_(f))/(n_(i)))^(2)`
`implies(21.2xx10^(-11))/(5.3xx10^(-11))=[(n)/(1)]^(2)`
`implies n^(2)=4implies n=2`