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

`\Delta U= 4.8204*10^(-24) J`

Step-by-step 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* 9.27*10^(-24)(0.26)`

`\Delta U= 4.8204*10^(-24) J`