Question

An electron in an atom revolves around the nucleus in an orbit of radius $0.25A$. If frequency of revolution of electron is $6.8\times {10}^{9}MHz$ then magnetic moment of electron is ____

• A. $2\times {10}^{-24}A{m}^2$
• B. $2.1\times {10}^{-24}A{m}^2$
• C. $3\times {10}^{-24}A{m}^2$
• D. $3.5\times {10}^{-24}A{m}^2$

Answer 1

Answer: (B)

$2.1\times {10}^{-24}A{m}^2$

Given: r=0.25, f=$6.8\times {10}^9\times {10}^6$

Solution: Formula used :M=NIA (N=1)

$M=\frac{QA}{t}=QfA=1.6\times {10}^{-19}\times 6.8\times {10}^9\times {10}^6\times 3.14\times 0.25\times 0.25\times {10}^{-20}$

M=$2.1\times {10}^{-24}$

Hence the correct option is B