Question

The radius of hydrogen atom in its ground state is $5.3\times {10}^{-11}m$. After collision with an electron it is found to have a radius of $21.2\times {10}^{-11}m$. What is the principal quantum number $n$ of the final state of atom?

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

Answer 1

The correct option is B $n=2$

$r\propto n^2$
ie, $\frac{r_f}{r_i}={\left(\frac{n_f}{n_i}\right)}^2$
$\Rightarrow \frac{21.2\times {10}^{-11}}{5.3\times {10}^{-11}}={\left[\frac{n}{1}\right]}^2$
$\Rightarrow n^2=4$
$\Rightarrow n=2$