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According to Bohr's postulate, the angular momentum of an electron , mvr=(nh)/(2pi) Using this formula show that magnetic moment of the atom is M=nmu_B Here what is mu_B and and what is its value ? |
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Answer» Solution :According to Bohr.s postulate , the ANGULAR MOMENTUM of an electron, `mvr=(nh)/(2pi)` where m - MASS of the electron , v - velocity of the electron , h - Planck.s constant and n - 1 , 2 etc, the no of orbits . Since `v=romega,` we have `r^2=(nh)/(2pi) "":.M=1/2e.(nh)/(2pim)=n.(EH)/(4pim)` ie, `M=nmu_B` where `(eh)/(4pim)=mu_B` , called Bohr Magneton. Bohr magneton is the UNIT of atomic magnetic dipole moment. |
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