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The radius of hydrogen atom in the ground state is $0.53 Å$. After collision with an electron, it is found to have a radius of $2.12 Å$, the principal quantum number $n$ of the final state of the atom is
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The correct answer is:
$n=2$
Given, radius of $\mathrm{H}$-atom in the ground state $\left(n_1=1\right)$
$$
r_1=0.53 Å
$$
Radius of excited state $\left(n_2\right)$
$$
r_2=212 Å
$$
We know that, $r \propto n^2$
$$
\begin{array}{ll}
\Rightarrow & \frac{r_1}{r_2}=\left(\frac{n_1}{n_2}\right)^2 \Rightarrow \frac{0.53}{212}=\left(\frac{1}{n_2}\right)^2 \quad\left[\because n_1=1\right] \\
\Rightarrow & \frac{1}{4}=\frac{1}{n_2^2} \Rightarrow\left(\frac{1}{2}\right)^2=\frac{1}{n_2^2} \Rightarrow n_2=2 \\
\text { i.e. } & n \simeq n=2
\end{array}
$$
$$
r_1=0.53 Å
$$
Radius of excited state $\left(n_2\right)$
$$
r_2=212 Å
$$
We know that, $r \propto n^2$
$$
\begin{array}{ll}
\Rightarrow & \frac{r_1}{r_2}=\left(\frac{n_1}{n_2}\right)^2 \Rightarrow \frac{0.53}{212}=\left(\frac{1}{n_2}\right)^2 \quad\left[\because n_1=1\right] \\
\Rightarrow & \frac{1}{4}=\frac{1}{n_2^2} \Rightarrow\left(\frac{1}{2}\right)^2=\frac{1}{n_2^2} \Rightarrow n_2=2 \\
\text { i.e. } & n \simeq n=2
\end{array}
$$
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