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. |
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Answer» Frequency of revolution, `f=(v)/(2pir)` The moving charge is equilvalent to a current loop, given by `" " I=fxxe or I=(ev)/(2pir)` If A be the area of the orbit, then the magnetic moment of the orbiting electrons is `" " M=IA=((ev)/(2pir))(pir^(2))=(evr)/(2)` Putting the value, we get `" " M=((1.6xx10^(-19))(2.2xx10^(6))(5.3xx10^(-11)xx10^(-2)))/(2)` `" " M==9.3xx10^(-26)A-m^(2)` |
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