KCET · Physics · Atomic Physics
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
- A \(n=2\)
- B \(n=3\)
- C \(n=4\)
- D \(n=1\)
Answer & Solution
Correct Answer
(A) \(n=2\)
Step-by-step Solution
Detailed explanation
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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