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The revolution of the electron in the first Bohr orbit of a hydrogen atom constitutes a current loop of area 8.8 xx 10^(-21)m^(2). If the frequency of revolution is 6.6 xx 10^(15) Hz, calculate the equivalent magnetic moment and the orbital angular momentum. [y_(0) = 8.975 xx 10^(10)C//kg] |
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Answer» Solution :Data: `A=8.8 xx 10^(-21) m^(2), f=6.6 xx 10^(15)Hz, y_(0)= 8.975 xx 10^(10) C//Kg, e=1.6 xx 10^(-19) C` The equivalent magnetic moment, `M=IA = efA = (1.6 xx 10^(19))(6.6 xx 10^(15))(8.8 xx 10^(-21))= 1.6 xx 48 xx 1.21= 9.293xx 10^(24)A.m^(2)` GYROMAGNETIC ratio, `y_(0)=(M_(0))/(L_(0))` `THEREFORE` The orbital angular momentum, `I= (M_(0))/y_(0) = (9.293 xx 10^(-24))/(8.795 xx 10^(10)) = 1.056 xx 10^(-34) kg.m^(2)//s` |
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