Question

The potential energy of a magnetic moment in an external magnetic field is U = - ??B. U is smallest when B is parallel to ? and largest when it is antiparallel. Since the magnetic moment of the electron is directed opposite to its spin (because it has a negative charge), the electron’s energy is highest when the spin is parallel to B. The two possible energy states for the electron are the spin up state (m_s = 1212) and the spin down state (m_s = - 1212). Calculate the energy difference ?E (in eV) between these states if the electron is placed in a magnetic field of B = 1.1 T. Assume that n = 10 and l = 0 for the electron.

If these electrons are bombarded with photons that have an energy equal to this energy difference ?E, then the electrons can be excited from the spin down state to the spin up state. This is referred to as a “spin flip” transition and the technique is called Electron Spin Resonance (ESR). Find the wavelength of the photons (in meters) that are needed to excite such spin flip transitions.

What type of electromagnetic wave has this wavelength?

Answer 1

The potential energy of a magnetic moment in an external magnetic field B is

Where is the spin magnetic moment and is the angle between and B.

Since Cos has maximum value +1 at =0 and minimum value -1 at =respectively then from above expression maximum energy corresponds to = with value E_max= B.

And minimum energy corresponds to =0 with value E_min= - B.

So we can conclude that E is smallest when B is parallel (=0to and largest when it is antiparallel (=).

If the two possible energy states for the electron are the spin up state (m_s=1/2) and spin down state (m_s=-1/2).

Then the energy of each state can be written as E=m_sg_sB B. Where g_s is the spin g-factor which has value 2.0023 for electron and _B is Bohr magneton. For spin up state E is positive and for spin down state E is negative (depending on the sign of m_s).

So, the energy difference between the up and down state is E= (g_sB B)/2 - (-g_sB B)/2 =g_sB B

We know _B has value 9.274 *10^-24 J/T and B= 1.1 T. Replacing these values we get energy difference

E= 2.04*10^-23 J

If these electrons are bombarded with photons that have an energy equal to this energy difference then the electrons can be excited from the spin down state to the spin up state. Let photon has energy hthen we can write

E= h = hc/

where h is planck constant which has value 6.626*10^-34 J.s and c is the speed of light (3*10⁸ m/s) and be the wavelength of photon. Now

E= hc/ = 2.04*10^-23

i.e. = ( 6.626*10^-34 *3*10⁸) /  2.04*10^-23 = 9.74*10^-3 m

This is the required wavelength which is in the microwave region.