Question

What is the energy difference between parallel and antiparallel alignment of the z component of an electron's spin magnetic dipole moment with an external magnetic field of magnitude 0.26 T, directed parallel to the z axis?

Answer 1

Answer:

$\Delta U= 4.8204\times10^{-24} J$

Explanation:

The difference between parallel and anti parallel alignment of the z component of an electron's spin magnetic dipole moment  is given by

$U_1-U_2= (-\mu Bcos(\theta_2))-(-\mu Bcos(\theta_1))$

where  μ= dipole moment B= strength of magnetic field and θ= angle between direction of magnetic field and dipole

here θ_2 and θ_1 are 180 and 0° respectively.

$U_1-U_2= (\mu B)-(-\mu B)$

$U_1-U_2= 2\mu B$

here μ is bohr magnetron or  magnetic moment of an electron = 9.27×10^{-24}

put this value we get

$U_1-U_2= 2\times 9.27\times10^{-24}(0.26)$

$\Delta U= 4.8204\times10^{-24} J$