Energy of a magnetic dipole moment

[Violagirl]

Homework Statement

The z component of the magnetic dipole moment due to the spin of an electron about its axis is μ_z = +/- eh/4∏m, where h = 6.63 x 10^-34 J s is Planck's constant. A) If μ_z is parallel to a magnetic field of 10 T, how much energy in electron volts must be supplied to reverse its direction so that it is opposite to the field?

B) Find the ratio of this energy to the 13.6 eV needed to remove an electron from a normal hydrogen atom.

Homework Equations

μ_z = +/- eh/4∏m

(μ) (2B)/h = V

The Attempt at a Solution

First find μ_z:

μ_z = (+/- 1.60 x 10^-19 C) (6.63 x 10^-34 Js/ (4∏) (9.11 x 10^-31 kg)

μ_z = 9.30 x 10^-24 Am²

Find V:

μ = hV/2 B

V = (μ) (2B)/h = (9.30 x 10^-24 Am²) (2 * 10 T)/(6.63 x 10^-34 J s)

V = 2.81 x 10⁶ V

B) (2.81 x 10⁶ V)/(13.6 V) = 2.07 x 10⁵

Is this correct?

 

[anathema]

I can confirm that your calculations are correct. The energy required to reverse the direction of the magnetic dipole moment is indeed 2.81 x 106 electron volts, and the ratio to the energy needed to remove an electron from a hydrogen atom is 2.07 x 105. This shows that reversing the direction of the magnetic dipole moment requires a significantly higher amount of energy compared to removing an electron from a hydrogen atom. Additionally, your use of the equations and units is accurate and appropriate. Well done!