Bemutatkozó Órára, szakkörre

# ENERGY MINIMUM – THE DRIVING FORCE OF NUCLEAR REACTIONS

The "energy valley" is excellent for determining the direction of nuclear transformations. Based on the model, we can say that the average energy of the nucleons (E/A) should decrease. It means that the studied isotope “drops” from the top of a higher column to a lower one in the energy valley while transforming. The change is analogous to a ball rolling down a hillside. As a result of this process, the particles that make up a nucleus become more stable and bound. The model is also suitable as a first approximation for determining the energy released during nuclear transformations. The columns in the 3D diagram show the average energy of each nucleon, but the energy of all the “components” must be considered during the transformation. In the case of beta decay, the mass number does not change, so in this case the value of the released energy is obtained by multiplying this "column difference" (in the order of parent-daughter isotope) by the number of particles (A). The result obtained must have a positive sign (this is the opposite of the negative energy change of the system). The problem is more complex for alpha decay, since 4 nucleons "jumps to the column" of $$^4He$$, while the remaining A-4 particle jumps to the "column" of the product. So the total energy change can be given by considering three columns and their associated mass numbers. In the case of nuclear fission, the situation is even more complicated, because there are many ways in which nuclear fission can take place. This case, usually we only consider average energy release. However, the calculation method based on the average energy of nucleons is limited, in some cases not accurate enough to describe the energetics of decays.

There are neutral atoms in our world, so the final state of every nuclear process results in neutral atoms. The fulfilment of a nuclear transmutation depends on whether the total mass of the final neutral atom(s and other particles) is less than the total mass of the initial neutral atom(s and other particles). So it is not enough to just look at the nuclei, we always have to take into consideration the mass of the electrons. This is especially important for beta decays. - Based on this line of reasoning, the energy released during nuclear processes can be determined with high accuracy. We can safely say that all nuclear transformations can be explained by the energy minimum principle. However, when calculating the "energy minimum" and the energy released, the mass of each component of the system and their full relativistic energy must be used.

In the derivations we are discussing now, we use that a neutral atom consists of a nucleus with Z atomic number and Z number of electrons, and therefore: $$m[_Z^AX] + Zm_e = m[X_{atom}]$$

The energetical description of $$\alpha$$-decay $$(_Z^AX \to _{Z-2}^{A-4}Y +_2^4\alpha)$$:
$$(m[_Z^AX] + Zm_e)c^2 = (m[_{Z-2}^{A-4}Y] + m[_2^4\alpha] + (Z - 2)m_e + 2m_e)c^2 + Q$$, where $$Q$$ is the energy released in the process.
During the decay, the nucleon composition (number of protons and neutrons) does not change. Thus, the system remains neutral with its $$Z$$ number of electrons. Based on this, the released energy: $$Q = (m[X_{atom}] - m[Y_{atom}] - m[He])c^2$$

The energetical description of $$\beta^-$$-decay $$(_Z^AX \to _{Z+1}^{A}Y + \beta^-)$$:
$$(m[_Z^AX] + Zm_e)c^2 = (m[_{Z+1}^{A}Y] + m_e + Zm_e)c^2 + Q$$
(the mass of the antineutrino generated in the process is negligibly small compared to the other members, so it is considered to be zero)
Although the electron formed during the decay leaves the atom, the positive ion, which formed due to the +1 increase in atomic number, traps an electron, thus neutralizing it. Based on this, the released energy: $$Q = (m[X_{atom}] - m[Y_{atom}])c^2$$

The energetical description of $$\beta^+$$-decay $$(_Z^AX \to _{Z-1}^{A}Y + \beta^+)$$:
$$(m[_Z^AX] + Zm_e)c^2 = (m[_{Z-1}^{A}Y] + m_e + m_e + (Z-1)m_e)c^2 + Q$$
(the mass of neutrino generated in the process is negligibly small compared to the other members, so it is considered to be zero)
The positron formed during the decay (its mass $$m_e$$ is the second term in the right parenthesis) and one of the electrons of the daughter isotope (its mass $$m_e$$ is the third term in the right parenthesis) annihilate, i.e. two gamma photons are created from them. The daughter isotope, resulting from the reduction of the atomic number by 1, will thus be neutral. The disappearance of the electron-positron pair also contributes to the energy released: $$Q = (m[X_{atom}] - m[Y_{atom}] - 2m_e)c^2$$ Hence, positive beta decay can only occur if the mass of the parent nucleus is more than two electrons greater than that of its daughter isotope.

Some negative beta decays seem energetically contradictory at first glance. An example is the $$^{14}C \to ^{14}N + \beta^-$$ process. Here, the average energy per nucleon for the daughter product is higher than for the nucleons of the parent nucleus. Although the difference is very small, during the transformation in the "energy valley", $$^{14}C$$ "jumps" upwards to become $$^{14}N$$ isotope. But keep in mind that the study of the average energy of nucleons is not enough to describe the transformation energetically! Typically, electron masses make a difference here. Since the weight of the daughter isotope is a little less than that of the parent isotope, the transformation is energetically advantageous, so it can take place. The following table shows the mass of $$^{14}C$$ and $$^{14}N$$ and the average energy of their nucleons:

 m (kg) E/A (MeV) $$^{14}C$$ (parent) $$2,32530\times10^{-26}$$ -7,526 $$^{14}N$$ (daughter) $$2,32527\times10^{-26}$$ -7,483

The data show that the average energy of the $$^{14}N$$ nucleons is $$43 keV$$ higher, but its mass is approx. $$3\times10^{-31} kg$$ smaller than the $$^{14}C$$ isotope's mass. $$157 keV (0,025 pJ)$$ of energy is released from the decay. This anomaly can be observed for negative beta decaying isotopes near the bottom of the energy valley (e.g. $$^{131}I$$). In such cases, each nucleon enters a less bound (less stable) state, but the "community", i.e. the nucleus, moves into a more favourable position.