TS EAMCET · Physics · Atomic Physics
Consider an electron revolving in a circular orbit of hydrogen atom, whose quantum number is \(n=2\). The velocity of the electron in that orbit is
- A \(1.1 \times 10^6 \mathrm{~m} / \mathrm{s}\)
- B \(2.2 \times 10^7 \mathrm{~m} / \mathrm{s}\)
- C \(4.4 \times 10^6 \mathrm{~m} / \mathrm{s}\)
- D \(2.2 \times 10^5 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(A) \(1.1 \times 10^6 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Velocity of electron of \(\mathrm{H}\) - atom in \(n^{\text {th }}\) orbit is \(v_n=\frac{v_1}{n}=\frac{22 \times 10^6}{2}=1.1 \times 10^6 \mathrm{~ms}^{-1}\)
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