MHT CET · Physics · Atomic Physics
Ratio of centripetal acceleration for an electron revolving in \(3^{\text {rd }}\) and \(5^{\text {th }}\) Bohr orbit of hydrogen atom is
- A \(425: 18\)
- B \(625: 81\)
- C \(125: 27\)
- D \(221: 36\)
Answer & Solution
Correct Answer
(B) \(625: 81\)
Step-by-step Solution
Detailed explanation
Centripetal acceleration \(\mathrm{a}=\frac{\mathrm{v}^2}{\mathrm{r}}\)
For hydrogen atoms, \(\mathrm{v} \propto \frac{1}{\mathrm{n}}\) and \(\mathrm{r} \propto \mathrm{n}^2\)
\(
\begin{aligned}
& \therefore \mathrm{a} \propto \frac{1}{\mathrm{n}^4} \\
& \therefore \frac{\mathrm{a}_3}{\mathrm{a}_5}=\left(\frac{5}{3}\right)^4=\frac{625}{81}
\end{aligned}
\)
For hydrogen atoms, \(\mathrm{v} \propto \frac{1}{\mathrm{n}}\) and \(\mathrm{r} \propto \mathrm{n}^2\)
\(
\begin{aligned}
& \therefore \mathrm{a} \propto \frac{1}{\mathrm{n}^4} \\
& \therefore \frac{\mathrm{a}_3}{\mathrm{a}_5}=\left(\frac{5}{3}\right)^4=\frac{625}{81}
\end{aligned}
\)
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