COMEDK · Physics · 28. Atomic Physics
An electron has a mass of \(9.1 \times 10^{-31} \mathrm{~kg}\). It revolves round the nucleus in a circular orbit of radius \(0.529 \times 10^{-10} \mathrm{~m}\) at a speed of \(2.2 \times 10^6 \mathrm{~ms}^{-1}\). The magnitude of its angular momentum is
- A \(2.06 \times 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}\)
- B \(1.06 \times 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}\)
- C \(2.06 \times 10^{-24} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}\)
- D \(1.06 \times 10^{-24} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(B) \(1.06 \times 10^{-34} \mathrm{Kg} \mathrm{m}^2 \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
The angular momentum \(L\) of a particle moving in a circular orbit is given by the formula \(L = mvr\), where \(m\) is the mass of the particle, \(v\) is its speed, and \(r\) is the radius of the orbit. Given values are: \(m = 9.1 \times 10^{-31} \text{ kg}\)…
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