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InterviewSolution
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Obtain an expression for the magnetic dipole moment of a revolving electron in a hydrogen atom, and hence find the value of Bohr magneton. |
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Answer» Solution :Let e represent the charge on the particle electron. Let R be the radius of a circular path. Let the electron revolve in an anticlockwise direction. CONVENTIONAL current is taken as clockwise. So magnetic MOMENT can be visualised to be pointing inwards towards the centre and perpendicular to the plane of the orbit. By definition, electric current `""I="charge"/"time"` i.e., `""I=e/T` where, T is the time period of revoulution of electron The period of REVOLUTION of revolution of electron, `""T="circumference"/"speed"` `""T=(2pir)/v` `therefore ""I=(ev)/(2pir)` Magnetic moment associated with the orbiting electron, `""m=1A=mu_(l),` where, `""mu_(l)=` magnetic moment due to orbital motion. Hence, `""mu_(l)=((ev)/(2pir))(pir^(2))` i.e.,`""mu_(l)=(e/(2m_(e)))(m_(e)vr)""` where, angular momentum `l=m_(e)vr` `therefore ""mu_(l)=(e/(2m_(e)))l` The direction of `vec(mu_(l))` is opposite to the direction of angular momentum. The expression `mu_(l)=(evr)/2` for I orbit electron is called Bohr magneton. `mu_(l)=1.6 times 10^(-19) times 2.18 times 10^(6) times 0.53 times 10^(-10) div 2=9.24 times 10^(-24)Am^(2)`. 
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