When we move from the heavy nuclei region to the region in the middle region of the plot, we find that there will be an increase in the overall binding energy (The energy with which nucleons bind in the nucleus) and hence the release of energy.
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This indicates that energy is released when a heavy nucleus breaks into two roughly equal fragments. This process is called nuclear fission.
We know that an element consists of a heavy nucleus which is unstable by nature. That’s why it gets disintegrated into two daughter nuclei to become stable.
So, the process by which element disintegrates itself without being forced by any external agent to do is called the radioactivity, or the radioactive decay.
Hence, radioactivity is a property of a heavy nucleus.
The Beta-decay process is the process of emission of an electron or positron from a radioactive nucleus.
In this process, a parent nucleus emits electrons or beta particles while disintegrating itself into two daughter nuclei.
The mass number of daughter nuclei remains the same because the mass of the electron is negligibly low, but the atomic number increases by one.
Let’s take a beta decay example:
Let's say we have 9091Th234.
The atomic number (Z) of Thorium is 91 and the mass number (A) is 234.
It undergoes the beta decay:
9091Th234 → 91Pa234 + -1e0 (electron or the β-particle)
Here, one electron is released.
The mass number of daughter nucleus = 234 - 0 = 234 remained the same and the atomic number (Z) or the charge number = 90 + 1 = 91, got incremented by 1.
We get a daughter nucleus as 91Pa234 .
In general form, the equation is: zXA → z+1YA + -1e0 + Q
Here, Q is the energy released during this process.
The beta decay produces a beta particle, which is a high-speed electron or positron. The mass of a beta particle is 〜1/2000 amu or atomic mass units.
Most of the stable elements have a certain balance between the number of neutrons and protons and if this balance gets disturbed, or whenever there is an excess in the number of neutrons or the number of protons, then the particle which is in excess gets transformed to the other type of particle.
So, in the β-decay process, either the neutron gets converted to a proton, or a proton is converted to a neutron. If a nucleus is formed with more neutrons than needed for stability, a neutron will convert itself into a proton to move towards stability, and the same happens with excess protons. Such transformations occur because of weak forces operating within neutrons or protons.
So, β-decay occurs in two forms, that is:
Beta plus decay, and
Beta minus decay
Let’s understand them one by one:
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In this process, excess protons inside the nucleus get converted into a neutron, releasing a positron and an electron neutrino (ve).
p → n + e+ + ve
Here, a positron is similar to an electron in all aspects, except that it has +e charge, instead of - e. So, in a β+ decay, β+ is used for positron.
In the year 1931, Pauli postulated that a β-particle is accompanied by another particle with zero rest mass and a zero charge called neutrino or ve.
This ve is very similar to an electron.
The value of energy emitted (Q) in this process is negative.
When a neutron is converted into a proton, an electron and a new particle named antineutrino (v’) are created and emitted from the nucleus. Here, β- is used for electrons.
n → p + e- + v’
Antineutrino is an antimatter particle, the counterpart of neutrino. As the mass neutron is greater than the combined mass of proton and electron.
Therefore, the value of energy emitted (Q) in this process is positive.
Fermi carried forward these suggestions in his theory of beta decay.
His formalism is based on the fact that β-decay is similar to the situation where a proton is created at the time of nuclear de-excitation.
He assumed that interaction responsible for β-decay is very weak, so he went beyond the conventional theory to hypothesize a new force that was extremely weak in comparison to electromagnetism. So that perturbation of quantum mechanics can be applied.
He used the result of Dirac’s time-dependent perturbation theory. According to this, the transition probability per unit time is given by:
λ = 2π/ħ |Hif|2ρ(E)...(1)
Here, ρ(E) is the density of the final states, i.e. number of final states in a particular energy interval. Where Hif is the matrix element of perturbation interaction given by,
Hif = ∫ψf* H ψi dて
Where ψf* = Final state wave function,
H = Dimensionless matrix element,
ψi = Initial state wave function, and
dて = Volume element.
The final state wavefunction must include not only the nucleus but also e and v.
Therefore, ψ*f = ψ*fN ψe* ψv*
Hif = g ∫ [ψ*fN ψe* ψv*] H ψi
Here, g = Fermi coupling constant whose value is 0.9 x 10-4 MeVfm3.
It determines the strength of the interaction.
1. What is the Equation for Beta Decay?
In the β-decay process, a neutron is converted into a proton, where the atomic number of the element increases by 1. At the same time, the mass number remains unchanged.
We can write the general equation for beta decay as:
ZXA → z+1YA + 00v’ +-01e
2. Write a Beta Decay Example.
For example, β-decay of the C-14 element looks like this:
6C14 → 7N14 + 00v’ + -01e
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The atomic number Z increments to Z+1 to give nitrogen.
3. What Can Stop Beta Decay?
Beta-rays have high penetrating power that they can pass a several feel distance in the air and penetrate the skin. Due to their greater penetrating power, the materials like a thin sheet of metal where these metals may include antimony (Sb), tin (Sn), bismuth (Bi), tungsten (W) or other elements and other materials like a sheet of plastic or wooden block can block these rays.
4. Does Beta Decay Release Energy?
Yes, beta decay releases energy.
Therefore, beta particles emit kinetic energy ranging from 0 to Q.
Typically, Q is around 1 MeV, but its value may range from a few keV to a few tens of MeV.