Question

The radius of a hydrogen atom in its ground state is $5.3\times {10}^{–11}\text{m}$. After collision with an electron, it is found to have a radius $21.2\times {10}^{–11}\text{m}.$ Find the value of $x$ where $x=6\times n$ and (where $n$ is principal quantum number) (integer only)

Answer 1

Given,
$\text{Radius in ground state}=5.3\times {10}^{–11}\text{m}\text{Radius of nth state}=21.2\times {10}^{–11}\text{m}{n}_1=1\left(\text{ground state}\right)$

The Bohr radius of the $n^{th}$ state is,

$r=0.529\frac{n^2}{Z}\mathring{A}$

$r\propto n^2$

$\Rightarrow \frac{r_1}{r_2}=\frac{K{n}_1^2}{K{n}_2^2}$

$\frac{5.3\times {10}^{-11}}{21.2\times {10}^{-11}}=\frac{1}{n^2}$

$\Rightarrow n=2$

$\therefore x=6\times n=6\times 2=12$