[Physics] Tritium decay is spontaneous even if the binding energy of tritium is higher than the binding energy of 3He. Why

binding-energynuclear-physicsradiation

Given this nuclear reaction:

$^3_1\mathrm H\to {}^3_2\mathrm{He}+e^-+\bar{\nu}$

and knowing the binding energies:

$BE(^3_1\mathrm H)=8.48 \,\mathrm{MeV}$

$BE(^3_2\mathrm{He})=7.72 \,\mathrm{MeV}$

If I calculate the mass defect (obviously considering the binding energies in the mass calculation) I obtain a positive value:

$M(^3_1\mathrm H)c^2=2809.08 \,\mathrm{MeV} > M( ^3_2\mathrm{He})c^2+M(e^-)c^2=2808.991 \,\mathrm{MeV}$

as expected for a spontaneous decay.
Considering the binding energies I have written above I expect the $^3\mathrm{H}$ to be more stable than $^3_2\mathrm{He}$.

My question is: why does this decay occur?

Best Answer

Binding energy simply isn't the right metric (because it is calculated from different starting points on account of the differing masses of the constituent nucleons).

Proper energy (AKA mass) of the states is the right metric.

Wolfram Alpha gives the masses as

$$M_{\mathrm{T}} = 2809.432 \,\mathrm{MeV}$$ $$M_{^3\mathrm{He}} = 2809.413 \,\mathrm{MeV}$$

In other words, there is about 19 keV to be had in this decay.

Related Solutions

Nuclear Physics – Understanding Elemental Anomalies in Nuclear Binding Energy

To understand binding energy and mass defects in nuclei, it helps to understand where the mass of the proton comes from.

The news about the recent Higgs discovery emphasizes that the Higgs mechanism gives mass to elementary particles. This is true for electrons and for quarks which are elementary particles (as far as we now know), but it is not true for protons or neutrons or for nuclei. For example, a proton has a mass of approximately $938 \frac{\mathrm{MeV}}{c^2}$, of which the rest mass of its three valence quarks only contributes about $11\frac{\mathrm{MeV}}{c^2}$; much of the remainder can be attributed to the gluons' quantum chromodynamics binding energy. (The gluons themselves have zero rest mass.) So most of the "energy" from the rest mass energy of the universe is actually binding energy of the quarks inside nucleons.

When nucleons bind together to create nuclei it is the "leakage" of this quark/gluon binding energy between the nucleons that determines the overall binding energy of the nucleus. As you state, the electrical repulsion between the protons will tend to decrease this binding energy.

So, I don't think that it is possible to come up with a simple geometrical model to explain the binding energy of nuclei the way you are attempting with your $\left(1\right)$ through $\left(15\right)$ rules. For example, your rules do not account for the varying ratios of neutrons to protons in atomic nuclei. It is possible to have the same total number of nucleons as $\sideset{^{56}}{}{\text{Fe}}$ and the binding energies will be quite different the further you move away from $\sideset{^{56}}{}{\text{Fe}}$ and the more unstable the isotope will be.

To really understand the binding energy of nuclei it would be necessary to fully solve the many body quantum mechanical nucleus problem. This cannot be done exactly but it can be approached through many approximate and numerical calculations. In the 1930's, Bohr did come up with the Liquid Drop model that can give approximations to the binding energy of nuclei, but it does fail to account for the binding energies at the magic numbers where quantum mechanical filled shells make a significant difference. However, the simple model you are talking about will be incapable of making meaningful predictions.

EDIT: The original poster clarified that the sign of the binding energy seems to be confusing. Hopefully this picture will help:
$\hspace{75px}$.

This graph shows how the potential energy of the neutron and proton that makes up a deuterium nucleus varies as the distance between the neutron and proton changes. The zero value on the vertical axis represents the potential energy when the neutron and proton are far from each other. So when the neutron and proton are bound in a deuteron, the average potential energy will be negative which is why the binding energy per nucleon is a negative number - that is we can get fusion energy by taking the separate neutron and proton and combining them into a deuteron. Note that the binding energy per nucleon of deuterium is $-1.1 \, \mathrm{MeV}$ and how that fits comfortably in the dip of this potential energy curve.

The statement that $\sideset{^{56}}{}{\text{Fe}}$ has the highest binding energy per nucleon means that lighter nuclei fusing towards $\text{Fe}$ will generate energy and heavier elements fissioning towards $\text{Fe}$ will generate energy because the $\text{Fe}$ ground state has the most negative binding energy per nucleon. Hope that makes it clear(er).

By the way, this image is from a very helpful article which should also be helpful for understanding this issue.

Nuclear Physics – Why Are Alpha Particles Such a Prominent Form of Radiation?

When people say that the decay rate depends critically on the $Q$ value, they're talking about alpha decays compared to other alpha decays. When you compare alpha decay to emission of other small clusters, the dependence on the atomic number $Z_c$ of the emitted cluster is much more prominent. The reason is as follows.

In the Gamow model of beta decay, we assume that the decay rate is the product of three factors: (1) a hand-wavy probability of preformation of the cluster; (2) the frequency with which a cluster assaults the Coulomb barrier; and (3) the probability of transmission through the barrier. (Re #1, don't take Ohanian too seriously when he says this factor is 0.1 to 1. Actually the literal existence of any cluster bouncing around inside an atomic nucleus would violate the exclusion principle. The whole thing is just a model.)

The critical factor is the tunneling probability $P$, which can be estimated using the WKB approximation, which looks like $\exp(-\int\ldots)$, where the integral is over the classically forbidden region. The integral depends on the Q value, because a higher Q value both shrinks the classically forbidden region and reduces the value of the integrand within that region. However, the height of the Coulomb barrier is proportional to the product $Z_cZ_d$ of the atomic numbers of the cluster and the daughter nucleus. If you want all the gory details, you can google the Geiger-Nuttall equation. But the result turns out to be of the form $\ln P=a-b$, where the $Z$-dependence is dominated by the term $a=(1/\hbar)\sqrt{32Z_cZ_d m_c R ke^2}$. For the alpha decay of uranium, we have $a\approx 74$. In Yval's example, decay by emission of 9Be basically doubles the value of $a$, which reduces the decay rate by a factor of $e^{-74}\approx10^{-32}$.

In the question, Yuval estimated that emission of Be should only be down by a factor of $e^{-2}$ relative to emission of alphas. This was an algebra mistake. We have an expression of the form $e^{-Zu}$, where $u$ is a constant. Changing this expression from $e^{-2u}$ to $e^{-4u}$ doesn't just reduce its value by a factor of $e^{-2}$, it reduces it by $e^{-2u}$, which is a huge factor.

Actually, as pointed out by JoeHobbit, the real mystery isn't why we don't emit larger clusters, it's why we don't emit lighter objects like protons or deuterons. A proton doesn't have to worry about preformation, and its tunneling probability would be much higher. Possibly this is due to the lower $Q$ value for proton emission. This comes into the Geiger-Nuttall equation because $b\propto Z_cZ_d/\sqrt{Q}$. In fact proton emission does occur, but it's only competitive for extremely proton-rich nuclei. There is also neutron emission, which doesn't involve any Coulomb barrier at all; as you'd expect, its half-life is very short (on the order of the assault frequency) when it's energetically allowed.

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Related Solutions

Nuclear Physics – Understanding Elemental Anomalies in Nuclear Binding Energy

Nuclear Physics – Why Are Alpha Particles Such a Prominent Form of Radiation?

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