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Deduce and expression for Magnetic dipole moment of revolving electron around the nucleus in a circular orbit. |
Answer» Solution :Suppose an electron undergoes circular motion AROUND the nuleus as shown in Figure. The circulating electron in a loop is like current ina circular loop ( since flow of charge is known as current). The magnetic DIPOLE moment due to current carrying circular loop is `vecmu_(L) =I vecA`......(1) In magnitude , ` mu_(L) = I A` If T is the TIME period of an electron , the current due to circular motion of the electron is ` I = (-e)/T ` .....(2) where - e is the charge of an electron . If R is the radius of the circular orbit and v is the velocity of the electron in the circular orbit , then ` T = (2 pi R)/v ` ......(3) Using equation (2) and equation (3) in equation (1), we get `mu_(L)=- e/((2pi R)/v) pi R^(2) = ( evR)/2 ` .....(4) where ` A = pi R^(2)` is the area of the circular loop. By definition, angular momentum of the electron about O is ` vecL = vecr xx vecp` In magnitude, ` L = Rp = MV R` ......(5) Using equation (4) and equation (5) , we get , `mu_(L)/L = - ((evR)/2)/(mvR) = e/(2 m) rArrvecmu_(L) = - e/(2 m ) vecL ` ......(6) The negative sign indicates that the magnetic moment and angular momentum are in OPPOSITE direction. The ratio `mu_(L)/L ` is a constant and also known as gyro - magnetic ratio `(e/(2m)) ` . It must be noted that the gyro- magnetic ratio is a constant of proportionality which connects angular momentum of the electron and the magnetic moment of the electron. |
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