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A lithium nucleus has four independent spin orientations, conventionally labeled by the quantum number m = -3/2, -1/2, 1/2, 3/2. In a magnetic field B, the energies of these four states are E = mμB, where μ the constant is 1.03 x 10-7 eV/T. In the Purcell-Pound experiment described in Section 3.3, the maximum magnetic field strength was 0.63 T and the temperature was 300 K. Calculate the probability of a lithium nucleus being in each of its four spin states under these conditions. Then show that, if the field is suddenly reversed, the probabilities of the four states obey the Boltzmann distribution for T =-300 K.

Short Answer

Expert verified

Therefore, the probability of a lithium nucleus is:

P1=0.2500009413P2=0.2500003138P3=0.2499996862P4=0.2499990587

Step by step solution

01

Given information

A lithium nucleus has four independent spin orientations, conventionally labeled by the quantum number m = -3/2, -1/2, 1/2, 3/2. In a magnetic field B, the energies of these four states are E = mμB, where μ the constant is 1.03 x 10-7 eV/T. In the Purcell-Pound experiment described in Section 3.3, the maximum magnetic field strength was 0.63 T and the temperature was 300 K.

02

Explanation

The partition function is: because none of the levels are degenerate.

Z=se-E(s)/kT

Where,

E(s)=mμBm=-32,-12,12,32

So,

Z=e3μB/2kT+eμB/2kT+e-μB/2kT+e-3μB/2kT

Let x=μB/kT

Z=e3x/2+ex/2+e-x/2+e-3x/2(1)

Therefore the value of x is:

x=1.03×10-7eV/T(0.63T)8.617×10-5eV/K(300K)=2.51×10-6

Substitute x into (1):

Z=e32.51×10-6/2+e2.51×10-6/2+e-2.51×10-6/2+e-32.51×10-6/2Z=4

03

Calculations

The probability of first state is:

P=1Ze3x/2

Substitute x and z:

P1=14e32.51×10-6/2P1=0.2500009413

The probability of second state is:

P=1Zex/2

Substitute x and z:

P2=14e2.51×10-6/2P2=0.2500003138

The probability of third state is:

P=1Ze-x/2

Substitute x and z:

P3=0.2499996862

The probability of fourth state is:

P=1Ze-3x/2

Substitute x and z:

P4=14e-32.51×10-6/2P4=0.2499990587

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