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 prop n^(2)` i.e `(r_(f))/(r_(i)) = ((n_(f))/(n_(i)))^(2)`
`rArr (21.2 xx 10^(-11))/(5.3 xx 10^(-11)) = ((n)/(1))^(2) rArr n^(2) = 4 rArr n = 2`