Question

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

• A. $n =3$
• B. $n =4$
• C. $n=1$
• D. $n =2$

Answer 1

Answer: (D)

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