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Obtain an experssion for the magnetic dipole moment of a revolving electron. |
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Answer» Solution :Magnetic dipole moment of revolving electron : Suppose an electron udergoes circular motion around the nucleus. The circulating electron in a loop is like current in a circular loop (since flow of charge is current ). The magnetic dipole moment due to current carrying circular loop is `vec(mu_(L)) = I vec(A)"" `.........(1) In Magnitude , `mu_(L) = IA` IF T is the time period ofan 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 ciruclar 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) = - (EV R)/(2)""` ...(4) where A = `pi R^(2)` is the area of the circular loop. By definition, angular momentum of the electron about O is `vec(L) = vec(R) xxvec(P)` In magnitude , L = PR = mvR `""`...(5) Using equation (4) and equation (5), we get `(mu_(L))/(L) = -(2)/(mv R) = - (e)/(2m) rArr vec(mu_(L)) = - (e)/(2m) vec(L)` The negative sign indicates that the magnetic moment and angular momentum are in opposite direction. In magnitude. `(mu_(L))/(L) = (e)/(2m) = (1.60 xx 10^(-19))/(2 xx 9.11 xx 10^(-31)) = 0.0878xx 10^(12)` `(mu_(L))/(L) = 8.78 xx 10^(10) C kg^(-1)` = constant 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. According to Neil.s Bohr quatization rule, the angular momentum of an electron moving in a stationary orbit is quantized, which means, l = nh = n `(h)/(2pi)` where, h is the planck.s constant `(h = 6.63 xx 10^(-34) J s)` and number n takes natural numbers(i.e, n = 1, 2,3....). hence, `mu_(L) = (e)/(2m) L = n (eh)/(4pi m) A m^(2)` `mu_(L) = n((1.60 xx 10^(-19))h)/(4pi m) Am^(2) = n ((1.60 xx 10^(-19)(6.63 x 10^(-34)))/(4xx 3.14xx (9.11 xx 10^(-31))))` `mu_(L) = n xx 9.27 xx 10^(-24) A m^(2)` the minimum magnetic moment can be obtained by substituting n = 1, `mu_(L) = 9.27 xx 10^(-24) A m^(2) = 9.27 xx 10^(-24) J T^(-1) = (mu_(L))_(min) = mu_(B)` Where, `mu_(B) = (eh)/(4pi m) = 9.27 xx 10^(-24) A m^(2)` is called Bohr magneton. This is a convenient unit with which ONE can measure atomic magnetic MOMENTS. 
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