The radius of hydrogen atom in the ground state is 0.53 mathringA . After collision with an electron, it is found to have a radius of 2.12 mathringA , the principle quantum number ' n ' of the final state of the atom is

Question

The radius of hydrogen atom in the ground state is 0.53 mathringA . After collision with an electron, it is found to have a radius of 2.12 mathringA , the principle quantum number ' n ' of the final state of the atom is (A) n =3 (B) n =4 (C) n=1 (D) n =2

Tardigrade Admin · Accepted Answer

r propto n 2 ( r 1/ r 2)=(( n 1/ n 2))2 0.25=(1/ n 22) n 22=(1/0.25)=(100/25)=4 n 2=2

Q. The radius of hydrogen atom in the ground state is $0.53 \overset{˚}{A}$ . After collision with an electron, it is found to have a radius of $2.12 \overset{˚}{A}$ , the principle quantum number ' $n$ ' of the final state of the atom is

$n = 3$

$n = 4$

$n = 1$

$n = 2$

$r \propto n^{2}$
$\frac{r _{1}}{r _{2}} = (\frac{n _{1}}{n _{2}})^{2}$
$0.25 = \frac{1}{n _{2}^{2}}$
$n_{2}^{2} = \frac{1}{0.25} = \frac{100}{25} = 4$
$n_{2} = 2$

Q. The radius of hydrogen atom in the ground state is 0.53A˚. After collision with an electron, it is found to have a radius of 2.12A˚, the principle quantum number ' n ' of the final state of the atom is

n=3

n=4

n=1

n=2