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The radius of hydrogen a tomin its ground state is $5.3 \times 10^{-11} \mathrm{~m}$. After collision with an electron
it is found to have a radius of $21.2 \times 10^{-11} \mathrm{~m}$.
What is the prindpal quantum number $n$ of the final state of atom?
Options:
it is found to have a radius of $21.2 \times 10^{-11} \mathrm{~m}$.
What is the prindpal quantum number $n$ of the final state of atom?
Solution:
1216 Upvotes
Verified Answer
The correct answer is:
$n=2$
$r \propto n^{2}$
ie,
$$
\frac{r_{f}}{r_{i}}=\left(\frac{n_{f}}{n_{i}}\right)^{2}
$$
$$
\begin{array}{l}
\Rightarrow \frac{21.2 \times 10^{-11}}{5.3 \times 10^{-11}}=\left[\frac{n}{1}\right]^{2} \\
\Rightarrow \quad n^{2}=4 \\
\Rightarrow \quad n=2
\end{array}
$$
ie,
$$
\frac{r_{f}}{r_{i}}=\left(\frac{n_{f}}{n_{i}}\right)^{2}
$$
$$
\begin{array}{l}
\Rightarrow \frac{21.2 \times 10^{-11}}{5.3 \times 10^{-11}}=\left[\frac{n}{1}\right]^{2} \\
\Rightarrow \quad n^{2}=4 \\
\Rightarrow \quad n=2
\end{array}
$$
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