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?

Respuesta :

Answer:

[tex]\Delta U= 4.8204\times10^{-24} J[/tex]

Explanation:

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

[tex]U_1-U_2= (-\mu Bcos(\theta_2))-(-\mu Bcos(\theta_1))[/tex]

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.

[tex]U_1-U_2= (\mu B)-(-\mu B)[/tex]

[tex]U_1-U_2= 2\mu B[/tex]

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

put this value we get

[tex]U_1-U_2= 2\times 9.27\times10^{-24}(0.26)[/tex]

[tex]\Delta U= 4.8204\times10^{-24} J[/tex]

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