AP EAMCET · PHYSICS · Atomic Physics
The radius of orbit of an electron and the speed of electron in the ground state of hydrogen atom are \(5.5 \times 10^{-11} \mathrm{~m}\) and \(4 \times 10^6 \mathrm{~ms}^{-1}\) respectively. Then, the orbital period of this electron in the first excited state will be
......... .
- A \(6.90810^{-16} \mathrm{~s}\)
- B \(9.608 \times 10^{-16} \mathrm{~s}\)
- C \(7.806 \times 10^{-16} \mathrm{~s}\)
- D \(8.9068 \times 10^{-16} \mathrm{~s}\)
Answer & Solution
Correct Answer
(A) \(6.90810^{-16} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
\(r_1=5.5 \times 10^{-11} \mathrm{~m}\) \(v_1=4 \times 10^6 \mathrm{~m} / \mathrm{s}\) As \(r_n \propto n^2\) So, \(r_2=5.5 \times 10^{-11} \times 2^2=22.0 \times 10^{-11} \mathrm{~m} / \mathrm{s}\) As \(v_n \propto \frac{1}{n}\) So,…
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