91Ó°ÊÓ

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

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

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

The magnetization per particle is given as:

$\frac{M}{N}=\mathrm{μ}\tanh \frac{\mathrm{μ}B}{kT}$

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

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

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.

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

The wavelength of such a photon is given as:

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

The magnetization per particle is calculated as:

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

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

$E=1.008×{10}^{-26}\mathrm{J}$

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

$\mathrm{λ}=19.7\mathrm{m}$