KCET · Physics · Atomic Physics
The period of revolution of an electron in the ground state of hydrogen atom is \( T \). The period of
revolution of the electron in the first excited state is
- A 2T
- B 4T
- C 1T
- D \( 8 \mathrm{~T} \)
Answer & Solution
Correct Answer
(D) \( 8 \mathrm{~T} \)
Step-by-step Solution
Detailed explanation
Time period of revolution of electron is given as
\(T=\frac{2 \pi r}{v}\)
We know \(r \propto n^{2}\) and \(v \propto \frac{1}{n}\)
\(\Rightarrow T \propto n^{3}\)
Time period of revolution in ground state \(=\mathrm{T}\)
Time period of revolution in first excited state \(=T_{1} \propto 2^{3}=8\)
Therefore, \(\frac{T}{T}=8\)
\(\Rightarrow T_{1}=8 T\)
\(T=\frac{2 \pi r}{v}\)
We know \(r \propto n^{2}\) and \(v \propto \frac{1}{n}\)
\(\Rightarrow T \propto n^{3}\)
Time period of revolution in ground state \(=\mathrm{T}\)
Time period of revolution in first excited state \(=T_{1} \propto 2^{3}=8\)
Therefore, \(\frac{T}{T}=8\)
\(\Rightarrow T_{1}=8 T\)
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