MHT CET · Physics · Atomic Physics
Ratio of centripetal acceleration for an electron revolving in \(3^{\text {rd }}\) orbit to \(5^{\text {th }}\) Bohr orbit of hydrogen atom is
- A \(\frac{424}{21}\)
- B \(\frac{625}{81}\)
- C \(\frac{125}{4}\)
- D \(\frac{775}{61}\)
Answer & Solution
Correct Answer
(B) \(\frac{625}{81}\)
Step-by-step Solution
Detailed explanation
Centripetal acceleration \(\mathrm{a}=\frac{\mathrm{v}^{2}}{\mathrm{r}}\)
\(v \propto \frac{1}{n}\) and \(r \alpha n^{2}\)
\(\therefore a=\frac{1}{n^{4}}\)
\(\therefore \frac{a_{5}}{a_{3}}=\frac{(3)^{4}}{(5)^{4}}=\frac{81}{125}\)
\(v \propto \frac{1}{n}\) and \(r \alpha n^{2}\)
\(\therefore a=\frac{1}{n^{4}}\)
\(\therefore \frac{a_{5}}{a_{3}}=\frac{(3)^{4}}{(5)^{4}}=\frac{81}{125}\)
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