Question

derive the resonance condition for nuclear magnetic resonance.calculate the require magnetic field strength at which proton spin comes into resonance with 500 MHz radiation

Answer 1

Answer on Question #70531, Physics / Electromagnetism

Question. Derive the resonance condition for nuclear magnetic resonance. Calculate the require magnetic field strength at which proton spin comes into resonance with $500\,\mathrm{MHz}$ radiation.

Given.

$$\nu_{L} = 500\,\mathrm{MHz}.$$

Find.

$$B_{0} - ?.$$

Solution.

In a few words (in more detail see D. Freude Spectroscopy), the energy of a magnetic moment $\vec{p}_m$ in a magnetic field $\vec{B}_0$ is given by:

$$E = - \vec{p}_{m} \cdot \vec{B}_{0}.$$

If the $z$ axis is chosen along $\vec{B}_0$ then

$$E = - p_{mz} B_{0}$$

or

$$E = - \gamma m \frac{h}{2\pi} B_{0},$$

where $\gamma$ is the gyromagnetic ratio, $m$ is the magnetic quantum number of the atomic nucleus, $h$ is the Planck constant. For uneven mass numbers $m = \pm \frac{1}{2}$ and we get two levels with an energy difference of

$$\Delta E = \gamma \frac{h}{2\pi} B_{0}.$$

The energy absorbed by the proton is then $E = h\nu_{L}$, where $\nu_{L}$ is the resonance radio frequency. Hence, a magnetic resonance absorption will only occur when $\Delta E = h\nu_{L}$, which is when

$$\nu_{L} = \frac{\gamma}{2\pi} B_{0}.$$

This most important equation of NMR relates the magnetic fields to the resonant frequency. For the proton $\frac{\gamma}{2\pi} = 42.58\ MHz/T$.

Finally

$$B_{0} = \frac{\nu_{L}}{\gamma \cdot \frac{\gamma}{2\pi}} = \frac{500}{42.58} = 11.74\ T.$$

Answer: $B_{0} = 11.74\ T$.

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