Q145 / 150 · MHT CET 2009 · 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
    $1 \times 10^{-10}$
  • Correct answer
    C
    $1 \times 10^{-23}$
  • D
    $1 \times 10^{-27}$

Solution

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 $M=I A=10.5 \times 10^{-4} \times 3.142$ $\begin{aligned} \times(0.528)^{2} \times 10^{-20} & \\=& 10 \times 10^{-24}=10^{-23} \text { units } \end{aligned}$

Hence, the correct answer is option: 'C'.

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