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The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11} \mathrm{~m}$. What are the radii of the $\mathrm{n}=2$ and $\mathrm{n}=3$ orbits?
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The radius of $\mathrm{n}^{\text {th }}$ orbit is given by,
$$
\begin{aligned}
&r_n=4 \pi \varepsilon_0 \frac{n^2 h^2}{4 \pi^2 m e^2} \Rightarrow r_n \propto n^2 \\
&\Rightarrow \frac{r_1}{r_2}=\frac{n_1^2}{n_2^2} \text { (for } 1^{\text {st }} \text { and } 2^{\text {nd }} \text { orbit) }
\end{aligned}
$$
$\begin{aligned} \Rightarrow r_2=r_1 \times \frac{n_2^2}{n_1^2}=r_1 \times \frac{2^2}{1^2}=4 r_1=4 \times 5.3 \times 10^{-11} \mathrm{~m} \\ \Rightarrow r_2 &=21.2 \times 10^{-11}=2.12 \times 10^{-10} \mathrm{~m} \\ \text { And } r_3 &=r_1 \times \frac{n_3^2}{n_1^2}=r_1 \times \frac{3^2}{1^2}=9 r_1 \\ \Rightarrow r_3 &=9 \times 5.3 \times 10^{-11} \mathrm{~m}=4.77 \times 10^{-10} \mathrm{~m} . \end{aligned}$
$$
\begin{aligned}
&r_n=4 \pi \varepsilon_0 \frac{n^2 h^2}{4 \pi^2 m e^2} \Rightarrow r_n \propto n^2 \\
&\Rightarrow \frac{r_1}{r_2}=\frac{n_1^2}{n_2^2} \text { (for } 1^{\text {st }} \text { and } 2^{\text {nd }} \text { orbit) }
\end{aligned}
$$
$\begin{aligned} \Rightarrow r_2=r_1 \times \frac{n_2^2}{n_1^2}=r_1 \times \frac{2^2}{1^2}=4 r_1=4 \times 5.3 \times 10^{-11} \mathrm{~m} \\ \Rightarrow r_2 &=21.2 \times 10^{-11}=2.12 \times 10^{-10} \mathrm{~m} \\ \text { And } r_3 &=r_1 \times \frac{n_3^2}{n_1^2}=r_1 \times \frac{3^2}{1^2}=9 r_1 \\ \Rightarrow r_3 &=9 \times 5.3 \times 10^{-11} \mathrm{~m}=4.77 \times 10^{-10} \mathrm{~m} . \end{aligned}$
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