MHT CET · Physics · Atomic Physics
In a hydrogen atom, the electron is making \(6.6 \times 10^{15} \mathrm{revs}^{-1}\) around the nucleus in an orbit of radius \(0.528 \mathrm{~A}\). The magnetic moment \(\left(\mathrm{A}-\mathrm{m}^{2}\right)\) will be
- A \(1 \times 10^{-15}\)
- B \(1 \times 10^{-10}\)
- C \(1 \times 10^{-23}\)
- D \(1 \times 10^{-27}\)
Answer & Solution
Correct Answer
(C) \(1 \times 10^{-23}\)
Step-by-step Solution
Detailed explanation
Current, \(I=6.6 \times 10^{15} \times 1.6 \times 10^{-19}\)
\(=10.5 \times 10^{-4} \mathrm{~A} .\)
Area \(A=\pi R^{2}=3.142 \times(0.528)^{2} \times 10^{-20} \mathrm{~m}^{2}\)
So, Magnetic moment \(M=I A=10.5 \times 10^{-4} \times 3.142\) \(\times(0.528)^{2} \times 10^{-20} \)
\(= 10 \times 10^{-24}=10^{-23} \text {units}\)
\(=10.5 \times 10^{-4} \mathrm{~A} .\)
Area \(A=\pi R^{2}=3.142 \times(0.528)^{2} \times 10^{-20} \mathrm{~m}^{2}\)
So, Magnetic moment \(M=I A=10.5 \times 10^{-4} \times 3.142\) \(\times(0.528)^{2} \times 10^{-20} \)
\(= 10 \times 10^{-24}=10^{-23} \text {units}\)
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