Conservation of energy and loss of mass in nuclear radiation

Question

I'm learning about mechanical energy from Kleppner and Kolenkow (second edition), and in Chapter 5 (section 5.10), they introduce the notion of conservation of energy and connect it to the Einstein relation $E = mc^2$. Kleppner and Kolenkow give the following example:

In the 1930s, experimenters were able to measure nuclear masses with enough accuracy to show that the energy released in a nuclear reaction agrees with the known mass difference $\Delta m$ according to $\Delta E = \Delta mc^2$. For example, an atom of radium-226 spontaneously emits an $\alpha$-ray (a nucleus of helium-4) having a kinetic energy of $4.78$ MeV, leaving a residual nucleus of radon-222: $$ ^{226}\mathrm{Ra} \to ^{222}\mathrm{Rn} + ^4\mathrm{He} $$ The difference between the initial mass $^{226}\mathrm{Ra}$ and the final masses $^{222}\mathrm{Rn}$ plus $^4\mathrm{He}$ is $8.80 \times 10^{-30}$ kg and the mass energy accounts closely for the kinetic energy of the $\alpha$-ray plus the small kinetic energy of the recoiling $^{222}\mathrm{Rn}$ nucleus.

I've only started learning about this stuff recently, so my intuition is relatively limited, but this seems really weird to me. The obvious question to me is, where does the mass go? I can accept that the mass gets converted to mechanical energy, but what part of the atoms have lost mass? The number of protons and neutrons are the same, and assuming the charge of the atoms is $0$ on both sides, the number of electrons should be the same as well, so do the protons/neutrons or other particles have less mass after the $\alpha$-particle is emitted?

Answer 1

This is a very important question, because it can be really easy to get a false intuition and understanding on this topic.

I humbly submit these other answers https://physics.stackexchange.com/a/747510/313823

https://physics.stackexchange.com/a/667132/313823

for you to read, and even more so this video https://youtu.be/g39nwNm0Xfw

In short, to answer your question: "mass" is "a measure of the amount of energy confined in a particular space." That's it. Mass is not a separate thing that is converted to energy. It is a measure of the energy in an object. $E=mc^2$. You could set $c=1$ by choice of units, and $E=m$.

The nuclear reaction in your question releases energy (i.e. the products are in a lower potential energy state than the reactants), so the products weigh less the reactants. Mixing methane and oxygen and burning it also releases energy. Therefore, the products weigh less than the reactants. If you heat up a cup of tea, that tea weighs slightly more than it did when it was cold. If it cools down again, it weighs less again. Mass is a measure of confined energy.

(I say "weighs more" here as a shorthand for "is more massive." Although they are related in some ways, gravity has nothing to do with $E=mc^2$).

Answer 2

This is a physical phenomenon called the "Mass Defect" or the "Mass Deficit". Quantitatively, the mass defect $\Delta m$ is the difference between the mass of an object and the sum of the masses of its constituent particles.

The mass defect can be explained with the knowledge of relativistic physics that the energy of the stationary particle can be read from the mass: the binding energy of the nucleons reduces the total mass that would result from the sum of the masses of the individual nuclear building blocks. Thus, the binding energy $E_{B}$ of the nucleons released during the construction of an atomic nucleus is equal to the mass defect $\Delta m$ multiplied by the square of the speed of light $c$ according to the equation $E_{B} = \Delta m \cdot c^{2}$.

So what you call "the mass" here is actually just the total energy of the system in mass equivalent (if nothing moves). (this is what @Jon Custer noted)

You can observe this not only in nuclear reactions, but also in chemical reactions (even if it is extremely small there), but also if you simply compare the sum of the masses of the prototes and neutrons in the nucleus with the mass of the nucleus.

You can also calculate the mass defect for nucleus without using Einstein's mass-energy-equivalence formula: $$ \begin{align*} \Delta m_{\text{nucleus}} &= P \cdot m_{p} + N \cdot m_{n} - m_{k}\\ \end{align*} $$ where $P$ is the number of protons and N the number of neutrons in the nucleus, $m_{p}$ is the mass of a proton (without movement), $m_{n}$ is the mass of a neutron (without movement) and $m_{k}$ is the mass of the nucleus (without movement). In reality, however, one does not often find simply large atomic nuclei without even a single electron in the orbitals. In reality, however, one does not often find simply large atomic nuclei without even a single electron in the orbitals. Therefore there is also a similar formula for nuclear and charge-technically neutral atoms given by: $$ \begin{align*} \Delta m_{A} &= P \cdot m\left( \rm{_{1}^{1}H} \right) + N \cdot m_{n} - m_{k}\\ \end{align*} $$ where $m\left( \rm{_{1}^{1}H} \right)$ is the mass of is the mass of neutral Hydrogen-$1$.