Q42P - Fundamentals Of Physics

Question

A  $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{T}$magnetic field is applied to a paramagnetic gas whose atoms have an intrinsic magnetic dipole moment of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathbf{J}\mathbf{/}\mathbf{T}$. At what temperature will the mean kinetic energy of translation of the atoms equal the energy required to reverse such a dipole end for end in this magnetic field?

 

Step-by-Step Solution

Answer

$0.48 \mathrm{K}$

Step 1: Listing the given quantities

Magnitude of magnetic field,$\mathrm{B}=0.50 \mathrm{T}$

Magnetic dipole moment, $\mathrm{\mu }=1.0\times {10}^{-23}\mathrm{J}/\mathrm{T}$

Step 2: Understanding the concepts of magnetic dipole moment

We have the formula which relates temperature with mean translational kinetic energy. And also, we know the energy difference between up and down orientation of magnetic dipole moment. We can find the required temperature by using these two relations.

Formula:$\mathrm{T}=\frac{4\mathrm{\mu B}}{3{\mathrm{k}}_{\mathrm{B}}}$

 

Step 3: Calculations of the temperature at which the mean kinetic energy of translation of the atoms equals the energy required to reverse such a dipole end for end in this magnetic field

We want to find the temperature at which the mean translational kinetic energy is equal to the energy of dipole reversal. Therefore,

$\begin{array}{c}⟨\mathrm{K}.\mathrm{E}.⟩=\frac{3}{2}\cdot {\mathrm{k}}_{\mathrm{B}}\mathrm{T}\\ =|\mathrm{\mu }.\mathrm{B}-(-\mathrm{\mu }.\mathrm{B})|\\ =2\mathrm{\mu B}\end{array}$

From the above equation, we get 

$\begin{array}{c}\mathrm{T}=\frac{4\mathrm{\mu B}}{3{\mathrm{k}}_{\mathrm{B}}}\\ =\frac{4\times 1.0\times {10}^{-23}\times 0.50}{3\times 1.38\times {10}^{-23}}\\ =\frac{4\times {10}^{-23}\times {10}^{23}}{6\times 1.38}\\ =0.48 \mathrm{K}\end{array}$

The temperature at which the mean kinetic energy of translation of the atoms equals the energy required to reverse such a dipole end for end in this magnetic field is $0.48 \mathrm{K}$.