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. 2
B. 3
C. 4
D. 5

Answer

A

Explanation

(a) Radius of `H^-` atoms in its ground state,
`r_1 = 5.3 xx 10^-11 meters`
Radius of Bohr atom in its nth orbit,
`r_n = 5.3 xx 10^-11 (n^2)/(z) meters`
For `H-"atom"`, `z=1` and `rn = 2.2 xx 10^-11 meters`
`n = ((21.2 xx 10^-11)/(5.3 xx 10^-10))^(1//2) = 2`.