# $E=mc^2$ Shouldn't high energy decay GAIN mass not lose it? [closed]

I feel I've hit a bit of a paradox. I've learned that alpha decay releases a lot of energy, and that energy comes from the mass of the daughter products via $$E=mc^2$$. But, if the decay has more energy, by that same equation doesn't that mean there should be MORE mass, not less?

I see online that things with more energy have more mass. Ok. But then I see that "mass decreases and converts to energy". That seems contradictory!

• I am not sure I am following here. If you convert mass into energy then obviously you will be losing mass and gaining energy. Are you reading $E=mc^2$ as "If you convert $m$ mass you will be left with $E$ energy"? I think you need to add more detail to your question. Jan 28, 2021 at 16:44
• "I see online that things with more energy have more mass." Perhaps you can give a reference to this statement as that would allow us to better address your apparent paradox. Jan 28, 2021 at 21:51

Consider a nucleus of $$^{238}$$U. It is radioactive, which means it can spontaneously transform into other elements, accompanied by the emission of energy, either EM radiation (gamma) or the kinetic energy of the final set of particles. The question would be "where does this energy come from?

$$^{238}$$U has a couple of radioactive modes: alpha decay (mostly), and spontaneous fission (rarely).

In alpha decay, we have the reaction $$^{238}U \to \alpha + ^{234}Th.$$ In this reaction, the alpha particle has kinetic energy of about 4.187 MeV, and the thorium has much less, $$0.08$$ MeV. If we calculate the difference in the nuclear masses before and after the reaction, we see that it's the same as the total kinetic energy/c$$^2$$. So the total energy of the system remains constant: $$m_{U-238}c^2 = m_{\alpha}c^2 + m_{Th-234}c^2 + KE$$

The mass of the product is lower than the original mass, so the kinetic energy of the products is higher than the original; mass-energy changed to kinetic energy.

One could do the same type calculation for the spontaneous fission channel, but the reaction products are not so well defined, but you should get an excess energy around 200 MeV.

The combined rest masses of the fragments is smaller than the rest mass of the original particle. As total energy is conserved, and the original particle's energy is only its rest mass, the energy difference must be some other kind of energy, usually kinetic energy of the fragments and radiation like photons with Energy $$E=pc$$.

The thing is that Helium-4 particle has high binding energy. Due to that : $$M_{^4\text{He}} < 2m_p + 2m_n$$ where $$m_p, m_n$$ is proton and neutron masses respectively. Thus this energy excess :

$$E_{out} = (2m_p + 2m_n - M_{^4\text{He}})~c^2$$

is usually released as an alpha particle kinetic energy in a nucleus radioactive disintegration.

• What kind of reaction are you talking about? 2 individual protons and 2 individual neutrons don't react to form an alpha. In solar fusion alphas are formed from two helium-3 nuclei. Energy of nuclear disintegration is the difference in the parent mass and the product masses. Your example doesn't happen in radioactivity. Jan 29, 2021 at 1:15
• If you check alpha decay, you'll find there stated, that "Part of the reason is the high binding energy of the alpha particle, which means that its mass is less than the sum of the masses of two protons and two neutrons". They give such alpha decay disintegration energy : $$E=(m_{\text{i}}-m_{\text{f}}-m_{\text{p}})c^{2}$$, which usually is released as alpha particle kinetic energy. Jan 29, 2021 at 7:08
• I don't meant reaction of two protons and neutrons, just the sum of their masses Jan 29, 2021 at 7:36

Energy is always conserved.
What happens during the decay process is that energy that was "locked" into "mass form" gets converted into other forms of energy e.g. kinetic energy, thermal energy, ecc.

The reactants and products have the same amount of total energy, i.e. some mass-energy in the reactants is converted to kinetic energy in the products (this conversion is typically what's termed "release of energy" in nuclear reactions)

Your confusion probably arises from a misinterpretation of $$E=mc^2$$. Here, we take $$m$$ as relativistic mass, $$m=\frac{m_0}{\sqrt{1-(v/c)^2}}$$, where $$m_0$$ is rest mass. Defining $$\gamma\equiv\frac{1}{\sqrt{1-(v/c)^2}}$$, we can rewrite: $$E=\gamma m_0c^2$$ As you can see, an object has some baseline energy when it is at rest, but gains more energy as it starts moving faster.

• Isn't "relativistic mass" an outdated concept? Jan 28, 2021 at 16:47
• Yeah, but it keeps showing up in $E=mc^2$ nonetheless... Jan 28, 2021 at 16:49
• Mathematically, yes, but physically you run into issues with relativistic mass. Jan 28, 2021 at 16:50
• How would one explain why $E=mc^2$ gives total energy without some $m\to\gamma m_0$ kind of relation though? Jan 28, 2021 at 16:51
• Even though Einstein initially used the expressions "longitudinal" and "transverse" mass in two papers (see previous section), in his first paper on ${\displaystyle E=mc^{2}}$(1905) he treated m as what would now be called the rest mass.[2] Einstein never derived an equation for "relativistic mass", and in later years he expressed his dislike of the idea:[27] oln Barnett, 19 June 1948 (quote from L.B. Okun (1989), p. 42[5])
– Gert
Jan 28, 2021 at 16:54