KCET · Physics · Atomic Physics
Frequency of revolution of an electron revolving in nth orbit of H-atom is proportional to
- A \( n \) independent of \( n \)
- B \( \frac{1}{n^{2}} \)
- C \( \frac{1}{n^{3}} \)
- D \( n \)
Answer & Solution
Correct Answer
(C) \( \frac{1}{n^{3}} \)
Step-by-step Solution
Detailed explanation
\( (\mathrm{C}) \)
Time period of revolution, \( T=\frac{2 \pi n}{v} \)
Frequency of revolution, \( f=\frac{v}{2 \Pi n} \)
\( \mathrm{v} \propto \frac{1}{n} \)
\( r \propto n^{2} \) and hence \( f \propto \frac{1}{n^{3}} \)
Time period of revolution, \( T=\frac{2 \pi n}{v} \)
Frequency of revolution, \( f=\frac{v}{2 \Pi n} \)
\( \mathrm{v} \propto \frac{1}{n} \)
\( r \propto n^{2} \) and hence \( f \propto \frac{1}{n^{3}} \)
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