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An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 xx 10^(-11)m, with a speed of 2 xx 10^(6) ms^(-1). The resultant orbital magnetic moment and angular momentum of the electron is ...... Take charge of electron = 1.6 xx 10^(-19) C, mass of electron = 9.1 xx 10^(-31) kg. |
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Answer» <P> Solution :The current FLOWING in circular path of electron with frequency f is `I= fe`But `f= (omega)/( 2pi ) "" therefore I= (omega e)/( 2pi) ` but `omega = (v)/( r)` `therefore I= (ve)/( 2pi r)` `m= IA` `= (ve)/(2pi r)xx pi r^(2) ` `= (1)/(2) ver` `= (1)/(2) xx 2 10^(6) xx 1.6 xx 10^(-19 ) xx 5.3 xx 10^(-11)` `= 8.48 xx 10^(-24) "Am"^(2)` ANGULAR momentum `L=pr` `-m_(e) vr ""("where" p = mv)` `L= 9.1 xx 10^(-31) xx 2 xx 10^(6) xx 5.3 xx 10^(-11)` `= 96.46 xx 10^(-36) ` `therefore L= 9.65 xx 10^(-35)` Nms 
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