InterviewSolution
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InterviewSolution
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The electron in hydrogen atom moves with a speed of `2.2xx10^(6)m//s` in an orbit of radius `5.3xx10^(-11)cm`. Find the magnetic moment of the orbiting electron.A. `8.3 xx 10^(-23)A-m^(2)`B. `9.3 xx 10^(-24)A-m^(2)`C. `7.2 xx 10^(-24)A-m^(2)`D. `6 xx 10^(-24)A-m^(2)` |
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Answer» Correct Answer - B Frequency of revolution , ` f=(v)/( 2pir )` The moving charge is equivalent to a current loop , given by `i=f xx e or i=(ev)/( 2pir )` If A be the area of the orbit , then the magnetic moment of the orbiting electron is , `M=iA=((ev)/( 2pi r))(pir^(2))=(evr)/(2)` Putting the values , we have `M=((1.6 xx 10^(-19))(2.2 xx 10^(6))(5.3 xx 10^(-11)))/(2)` `=9.3 xx 10^(-24)A-m^(2)` |
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