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`. The principal quantum number of the final state of the atom is.

Options

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

Answer

D

Explanation

In Bohr atom , `r_n prop n^2`
In this case the transition is form n = 1 to n=n
`therefore (r_n)/(r_1) = (n/1)^2 = n^2 therefore (r_n)/(r_1) = (21.1xx10^(-11))/(5.3xx10^(-11)) = 4`
`therefore n^2 = 4 " or " n = 2`