Precessional frequency in H-NMR

Question

Suppose we have a hydrogen nucleus. Now, let's apply an external magnetic field $B$. As the magnetic field is applied, the hydrogen nucleus undergoes precessional motion about its own axis with an angular frequency of, say, $p$. Now suppose, we supply an electromagnetic radiation of frequency '$\nu$' to the nucleus.

Then,according to Spectroscopy (third edition) by Pavia, Lampman and Kriz, resonance occurs when frequency of applied electromagnetic radiation, $\nu$ becomes equal to precessional frequency, that is: $\nu=p$.

But in my textbook, it is written that $p=2\pi \nu$. This is possible if precessional frequency equals the angular frequency ($\omega$) of magnetic field component of external electromagnetic radiation (as $\omega =2\pi \nu$) and not to the frequency($\nu$)of EM radiation.

So,here is my doubt:At resonance, is it correct to say that precessional frequency($p$) is equal to angular frequency of magnetic field component of applied electromagnetic radiation ($\omega$)?

That is: $p=2\pi \nu$

Is this correct?

Answer 1

Short answer: We say, that the excitation frequency (frequency of the EM radiation) $\nu_e$ is at resonance if it is exactly the precession frequency also known as the Larmor frequency $\nu_l$. (So no extra factor or transformation of the larmor frequency) This is also written in this Wikipedia article.

Brief explanation: We first need to accept, that nuclei with non-zero spin with an angular momentum of $\boldsymbol{J}$ induce a (nuclear) magnetic dipole, where $\boldsymbol{\mu} = \gamma \boldsymbol{J}$. Then, using this fact, we can treat the nucleus (proton in this case) inside magnetic field classically. We see that (for static, homogenous magnetic field) this is Larmor precession. Assuming a static, homogenous magnetic field with field strength $B_0$ in the $z$-direction, we see that the magnetic dipole precesses around the field with Larmor frequency $\nu_l = \gamma B_0/(2\pi)$.

Now also adding an oscillating magnetic field (as the EM-Wave) $\boldsymbol{B_1(t)} = (2B_1\cos{\omega_d t}, 0, 0)$: We can derive (with a minor approximation requring $B_1 \ll B_0$) that the magnetic moment precesses around $\boldsymbol{B_{eff}} = \boldsymbol{e_z}(B_0-\omega_d/\gamma) + \boldsymbol{e_x}B_1$ in the rotating frame of reference rotating with the angular frequency $\omega_d$ around the $z$-axis.

Image of the rotating frame of reference for the system in the general case:

Now here comes the explanation of what we mean when we say resonance in NMR:

If we set the angular frequency of the EM-Wave to the angular Larmor frequency so $\omega_d = 2\pi\nu_e = 2\pi\nu_l$, we get that $\boldsymbol{B_{eff}} = \boldsymbol{e_x}B_1$ (in the rotating frame of reference). When the EM-Wave / pulse is over (no $\boldsymbol{B_1}$ field anymore), the precession around $\boldsymbol{e_x}B_1$ (in rotating frame) stops and the magnetic dipole continues the "normal" Larmor precession around $\boldsymbol{B_0}$. By controlling the pulse duration, one thus can control at which angle the dipole (and thus the angular momentum) is stopped. This property is used to manipulate the spin direction.

Image of the rotating frame of reference for the system in the resonance case:

Disclaimer: This explanation relies heavily on the classical approach. Other resonance arguments can be made using the quantum approach by considering Fermis golden rule and Rabi Oscillation.