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. `8.27 xx 10^(-26) Am^(2)`B. `9.27 xx 10^(-27) Am^(2)`C. `9.3 xx 10^(-26)Am^(2)`D. `8.8 xx 10^(-27)Am^(2)` |
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Answer» Correct Answer - C Frequency of revolution , ` f=(v)/( 2pir )` The moving charge is equivalent to a current loop , given by `l=f xx e or l=(ev)/( 2pir )` If A be the area of the orbit , then the magnetic moment of the orbiting electron is , `M=lA=((ev)/( 2pi r))(pir^(2))=(evr)/(2)` Putting the values , we get `M=((1.6 xx 10^(-19))(2.2 xx 10^(6))(5.3 xx 10^(-11) xx 10^(-2)))/(2)` `=9.3 xx 10^(-26)A-m^(2)` |
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