**NMR(Nuclear Magnetic Resonance) spectroscopy** is one well known method for determining the molecular structure of compounds. It is used in chemistry, biology and in physics. When a sample inside a tube is placed in a homogeneous magnetic field, the nuclei of the atoms constituting the sample are aligned in parallel or anti-parallel direction to the external magnetic field because the nuclei are like tiny magnets. The application of the external magnetic field causes separation of energy level of the parallel and anti-parallel nuclei. Here it is shown how to mathematically derive the energy levels in such nuclear magnetic resonance(NMR) process.

An atom consist of dense positively charged nucleus surrounded by cloud of negatively charged electrons. The positively charged nucleus spins around the nuclear axis. The spinning of this nuclear charge generates a magnetic dipole with magnetic moment \(\mu\). Hence nucleus behaves like a tiny magnetic needle. An oversimplified view of nucleus as magnetic needle is depicted below.

Also the spinning of this dipole generates angular momemtum \(P\). The relation between magnetic moment \(\mu\) and angular momemtum \(P\) is as follows,

\(\mu = \gamma P\) ------->(1)

where the constant \(\gamma\) is called gyromagnetic ratio and is characteristics of particular nucleus.

Below is a table showing gyromagnetic ratio of some nuclei.

Like an electron, the spinning nucleus also has a spin quantum number. An electron has spin quantum number +1/2 and -1/2. The number indicate that electron can have two possible spin orientation. These spin states are popularly indicated using the up and down arrow. Likewise, the spinning nucleus also has nuclear spin quantum number \(I\). The actual nuclear spin quantum number depends upon the mass number and atomic number. That is the number of proton and neutron inside the nucleus. If the number of proton and neutron are equal then the nuclear spin quantum number is zero and the nucleus will be inactive in NMR. If the number of proton and neutron are unequal then the nucleus will react with external applied magnetic field and the nucleus will be active in NMR.

From quantum physics, the angular momentum\(P\) is quantized and can have only discrete values. It is given by the following equation.

\(P = \frac{h}{2 \pi}m\) ----->(2)

where, \(h\) is the Planck constant and \(m\) is the magnetic quantum number.

The magnetic quantum numbers, \(m\) are related to the nuclear spin quantum numbers as follows.

\(m = 2I+1\) ----->(3)

For example, proton has spin quantum number \(I=1/2\) hence the magnetic quantum numbers from (3) are,

\(m = 2(1/2)+1 = 2\)

Hence the two different magnetic quantum number for the proton are:

\(m_1 = +\frac{1}{2}\) and \(m_2 = -\frac{1}{2}\).

Consider another example. Consider an element with spin quantum number \(I=2\). Then from (3) the magnetic quantum number is \(m=5\) and the 5 different magnetic quantum numbers are:

\(m_1 = +2\) , \(m_2 = +1\) , \(m_3 = 0\) , \(m_4 = -1\) and \(m_5 = -2\).

From(1) and (2) we can obtain the relationship between the magnetic quantum number \(m\) and the magnetic moment \(\mu\).

\(\mu = \frac{\gamma h}{2 \pi}m\) ---------->(4)

which shows that nuclear magnetic moment \(\mu\) depends directly on the magnetic quantum number \(m\) which is quantized.

Now when such nucleus with magnetic moment \(\mu\) is placed inside a static and homogeneous magnetic field \(H_0\) then the potential energy of the nucleus is,

\(E = \mu H_0\) -------->(5)

and using (4) in (5) we get,

\(E = \frac{\gamma h}{2 \pi}m H_0\) -------->(6)

From (6), we can see that if the magnetic field \(H_0\) is constant, the energy depends only on the magnetic quantum number since all other factors, \(h\), \(\gamma\) are constant. Hence the energy of nucleus is quantized and can have only certain values.

Consider the 1H proton and 13C carbon used mainly in NMR. These two nucleus have spin quantum number \(I=1/2\) and therefore the magnetic quantum numbers \(m_1 = +\frac{1}{2}\) and \(m_2 = -\frac{1}{2}\). Hence the two energy levels for these nucleus are:

\(E_1 = \frac{\gamma h H_0}{2 \pi} (+\frac{1}{2}) \) and \(E_2 = \frac{\gamma h H_0}{2 \pi} (-\frac{1}{2}) = \)

These energy levels are depicted diagrammatically below.

So in this way the energy levels in NMR is derived.

See next Resonance in NMR.

See also Types of Spectroscopy and their comparison.