KCET · Physics · Atomic Physics
If an electron is revolving in its Bohr orbit having Bohr radius of \(0.529 Å\), then the radius of third orbit is
- A \(4496 Å\)
- B \(4.761 Å\)
- C \(5125 \mathrm{~nm}\)
- D \(4234 \mathrm{~nm}\)
Answer & Solution
Correct Answer
(B) \(4.761 Å\)
Step-by-step Solution
Detailed explanation
Given, Bohr radius, \(r_1=0.529 Å\)
We know that, radius of revolving electron in \(n\)th orbit is given as
\[
r \propto n^2
\]
\[
\begin{array}{ll}
\Rightarrow & r_1=\left(\begin{array}{l}
n_1 \\
n_2
\end{array}\right)^2 \\
\Rightarrow & \frac{r_1}{r_2}=\left(\frac{1}{3}\right)^2=\frac{1}{9} \quad\left[\because n_1=1 \text { and } n_2=3\right] \\
\Rightarrow & r_2=9 r_1=9 \times 0.529=4.761 Å
\end{array}
\]
We know that, radius of revolving electron in \(n\)th orbit is given as
\[
r \propto n^2
\]
\[
\begin{array}{ll}
\Rightarrow & r_1=\left(\begin{array}{l}
n_1 \\
n_2
\end{array}\right)^2 \\
\Rightarrow & \frac{r_1}{r_2}=\left(\frac{1}{3}\right)^2=\frac{1}{9} \quad\left[\because n_1=1 \text { and } n_2=3\right] \\
\Rightarrow & r_2=9 r_1=9 \times 0.529=4.761 Å
\end{array}
\]
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