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Show that the electron revolving around the nucleus in a radius 'r with orbital speed 'r' has magnetic moment evr/2. Hence, using Bohr's postulate of the quantization of angular momentum, obtain the expression for the magnetic moment of hydrogen atom in its ground state. |
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Answer» Solution :Magnetic momentum, `mu=-((e)/(2m))L` where, (-) INDICATES that direction of u is opposite to L. As per Bohr.s ATOMIC model `L=mvr` `:. Mu=-((e)/(2m))xx mvr` `mu=(evr)/(2)` Energy levels of hydrogen atom `E_(N)=(2pi^(2)mK^(2)e^(4))/(n^(2)h^(2))` For LOWEST energy level, n = 1 `E_(n)=(2pi^(2)mK^(2)e^(4))/(n^(2)h^(2))` `E_(2)=-(2pi^(2)mK^(2)e^(4))/(n^(2)h^(2))` `B_(2)=-(2pi^(2)mK^(2)e^(4))/(h^(2))` `E_(1)=-((4pi^(2)mKe^(2))/(h^(2)))(ke^(2))/(2)` While, `r=-(h^(2))/(4pi^(2)Kme^(2))` `E_(n)=-(Ke^(2))/(2r)=(-3.6)/(n^(2))eV` `:.mu=(evr)/(2)` `E_(1)=-(Kev.er)/(2vr^(2))=-13.6` `(Kmue)/(vr^(2))=-13.6` `mu=13.6((vr^(2))/(Ke))` |
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