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## Q. 12.4

Recall from Chapter 7 that magnetic fields affect atoms because of the magnetic moments of electrons, resulting in the Zeeman effect and the Stern-Gerlach effect. Consider a proton with magnetic moment $\vec{\mu}$ in an applied magnetic field $\vec{B}$ of magnitude 2.0 T. Find (a) the energy difference between the two proton magnetic moment orientations and (b) the frequency and wavelength of the electromagnetic radiation (photons) that “flip” the proton spins.

Strategy The proton has nuclear spin 1/2, and there are two possible orientations for the magnetic moment, up and down. We use Equation (7.35) from Section 7.5 in which we discussed this previously for the electron. We have $\mu_{p}=2.79 \mu_{ N }$ for the proton. We find the frequency of the photons from $\Delta E=h f$.

$V_{B}=-\vec{\mu}_{s} \cdot \vec{B}=+\frac{e}{m} \vec{S} \cdot \vec{B}$ (7.35)

## Verified Solution

(a) From Equation (7.35) we have $V_{B}=\pm \mu_{S} B$, and the energy difference ΔE between the up and down proton states is

\begin{aligned}\Delta V_{B} &=2 \mu_{S} B=2 \mu_{p} B=2\left(2.79 \mu_{ N }\right) B \\&=2(2.79)\left(3.15 \times 10^{-8} eV / T \right)(2.0 T )=3.5 \times 10^{-7} eV\end{aligned}

(b) The photon frequency associated with this proton spin flip is

$f=\frac{\Delta E}{h}=\frac{3.5 \times 10^{-7} eV }{4.14 \times 10^{-15} eV \cdot s }=85 MHz$

This is in the RF (radio frequency) range. The wavelength is

$\lambda=\frac{c}{f}=\frac{3.00 \times 10^{8} m / s }{85 \times 10^{6} Hz }=3.5 m$