I was reading about Deuterium and came to know that its binding energy is $$2.22$$ MeV. I am curious whether this energy is in the natural unit system where $$\hbar = c =1$$.

For instance, the energy of a SHO is given by $$E = \hbar\omega \, (n+1/2)$$ where the symbols have their usual meaning. In the natural unit system, this becomes $$E = \omega \, (n+1/2)$$. So how do I know whether the energy of the Deuterium is in the natural unit system or not if not explicitly stated?

The “MeV” is the unit for that “2.22 MeV” quantity: millions of electron-volts, a unit of energy.

The result of a calculation is the same no matter what units you express it in, but the number you use to express it depends on the units. How many meters tall are you? How many inches? How many centimeters? How many light years? Those are all different ways of expressing the same quantity.

For more on natural units, a good place to start is the Wikipedia article. Constants like c and hbar always appear, they just have different numbers in different systems.

• Let's say, I am calculating the binding energy theoretically. If I do separate calculations using conventional units (i.e.; leaving $\hbar$ and $c$ intact) and natural units, do I get the same result? The answer is yes if the energy expression is independent of $\hbar$ and $c$. Isn't it? – rainman Jan 27 '19 at 0:21
• Added a bit on that. – Bob Jacobsen Jan 27 '19 at 0:28

The unit $$\mathrm{MeV}$$ is an energy unit, just like joule or erg or kilowatt-hour, but has a different "calibration" for defining $$1\ \mathrm{MeV}$$. First, the $$\mathrm{M}$$ is the metric prefix "mega" meaning $$\times 10^6$$.

Now for the $$\mathrm{eV}$$ part: $$1~\mathrm{eV}$$ is the energy change of a charge of $$1~\mathrm{e}$$ when it is accelerated across a potential difference of $$1~\mathrm{V}$$. For example, if an alpha particle (q=$$2~\mathrm{e}$$) is accelerated from a potential of $$4000~\mathrm{V}$$ to $$2000~\mathrm{V}$$, it will gain kinetic energy of $$4000~\mathrm{eV}$$. (Positive charges gain energy when they are accelerated across dropping potentials.)

The unit equivalence in SI units is $$1~\mathrm{eV} = 1.602\times 10^{-19}\ \mathrm{J}.$$ When calculating binding energies, one may encounter a variety of units, but most often the atomic mass unit, $$\mathrm{u}$$ (or "old-school" $$\mathrm{amu}$$), is used. The equivalence of $$1~\mathrm{u}$$ in electron volt style units is $$1\ \mathrm{u} = 931.468 \frac{\mathrm{MeV}}{c^2}.$$

The $$c^2$$ is there to remind us that $$\mathrm{u}$$ is a mass unit.

Binding energies can be calculated using tables of masses in $$\mathrm{u}$$ and multiplying the mass result by $$c^2$$, leaving units of $$\mathrm{u}c^2$$. Multiply by the previous conversion factor gives binding energies in $$\mathrm{MeV}$$ (the $$c^2$$ factors divide out to a "pure" 1.)

how do I know whether the energy of the Deuterium is in the natural unit system or not if not explicitly stated?

You can't, really.

The use of natural units allows you to blur the lines between units, but it doesn't fully remove them the way the Planck units do. Thus, you still have one remaining unit, which can be electronvolts, meters, seconds, or any other unit with nontrivial weight in natural units.

You'd be able to tell that I was using natural units if I said something like

The energy states of the system are $$E_0=1\:\rm eV$$ and $$E_1=3\:\rm eV$$, so the period of oscillation between the two is $$\displaystyle T=\frac{2\pi}{E_1-E_0}$$

instead of adding a correct factor of $$\hbar$$, i.e. giving the period as $$T=\frac{2\pi \hbar}{E_1-E_0}$$.

Similarly, with things like $$E_n=\hbar \omega(n+1/2)$$ changing to $$E_n=\omega(n+1/2)$$, the use of natural units allows you to quote both $$E_n$$ and $$\omega$$ in both electronvolts and rad/s $$-$$ whatever is most convenient.