The radius of the 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. The principal quantum number of final state of the atom is:

Options

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

Answer

A

Explanation

`r_2/r_1=((n_2)/(n_1))^2 , (21.2 xx10^(-11))/(5.3 xx10^(-11))=(n_2/1)^(2) , " " 4 = n_2^2 , n_2 = 2`