Question

In the experiment of Purcell and Pound, the maximum magnetic field strength was $0.63\mathrm{T}$and the initial temperature was $300\mathrm{K}$. Pretending that the lithium nuclei have only two possible spin states (in fact they have four), calculate the magnetization per particle, $M/N$, for this system. Take the constant $\mu$to be $5\times {10}^{-8}\mathrm{eV}/\mathrm{T}$. To detect such a tiny magnetization, the experimenters used resonant absorption and emission of radio waves. Calculate the energy that a radio wave photon should have, in order to flip a single nucleus from one magnetic state to the other. What is the wavelength of such a photon?

Short answer

The required values are:

$$\begin{gathered} \frac{M}{N}=9.68\times {10}^{-33}{\mathrm{JT}}^{-1} \\ E=1.008\times {10}^{-26}\mathrm{J} \\ \mathrm{\lambda }=19.7\mathrm{m} \end{gathered}$$

Step-by-step solution

Given Information

Magnetic field strength $=B=0.63\mathrm{T}$

Temperature $=T=300\mathrm{K}$

Constant related to the magnetic moment$=\mu =5\times {10}^{-8}{\mathrm{eVT}}^{-1}=8\times {10}^{-27}{\mathrm{JT}}^{-1}$

Calculation

The magnetization per particle is given as:

$\frac{M}{N}=\mu \tanh \frac{\mu B}{kT}$

By substituting the values in the above equation, we get,

$$\begin{gathered} \frac{M}{N}=8\times {10}^{-27}\times \tanh \frac{8\times {10}^{-27}\times 0.63}{1.38\times {10}^{-23}\times 300} \\ \frac{M}{N}=(8\times {10}^{-27})(1.21\times {10}^{-6}) \\ \frac{M}{N}=9.68\times {10}^{-33}{\mathrm{JT}}^{-1} \end{gathered}$$

The magnitude of magnetism per molecule in this system is extremely low. As a result, the energy of a radio wave photon used to detect magnetization in the experimenters must be equivalent to the energy required to flip a single nucleus.

$$\begin{gathered} E=2\mu B \\ E=2\times 8\times {10}^{-27}\times 0.63 \\ E=1.008\times {10}^{-26}\mathrm{J} \end{gathered}$$

The wavelength of such a photon is given as:

$$\begin{gathered} E=\frac{hc}{\lambda } \\ \lambda =\frac{hc}{E} \\ \lambda =\frac{\left(6.626\times {10}^{-34}\right)\times \left(3\times {10}^8\right)}{1.008\times {10}^{-26}} \\ \lambda =19.7m \end{gathered}$$

Final answer

The magnetization per particle is calculated as:

$\frac{M}{N}=9.68\times {10}^{-33}{\mathrm{JT}}^{-1}$

To flip a nucleus from one magnetic field to another, the energy of the radio waves of the experimenter is calculated to be:

$E=1.008\times {10}^{-26}\mathrm{J}$

For the photon to be in the radio wave region its wavelength is calculated to be:

$\lambda =19.7\mathrm{m}$