## Cause of radioactive decay

**Radioactivity** is a nuclear reaction it must be connected with the instability of the nucleus or main cause of radioactive substances. The nucleus of an atom composed of two elementary particles namely protons and neutrons which cause many nuclear reactions.

Since all elements are not radioactive. Thus the ratio of the neutron to the proton or nuclear binding energy is cause for natural radioactivity.

### Nuclear binding energy and stability

Nuclear scientists studied this problem and concluded that stability or instability is connected with the pairing of nuclear spins or nuclear binding energy.

Just as electrons spin around their own axes and just as electron- spin pairing leads to be stable. So also, the nuclear protons and neutrons spin around their own axes, and pairing of spins of neutrons among neutrons, and pairing of spins of protons among protons leads to nuclear stability.

- Nuclei with an even number of protons and even number of neutrons are the most abundant and most stable isotopes of elements. Even number leads to spin pairing, and odd number leads to unpaired spins.
- Nuclei with an even number of neutrons and odd number protons or an odd number of neutrons and even number of protons slightly less stable than even number of neutrons and protons.
- Least stable isotopes are those which have odd numbers of protons and an odd number of neutrons. The nuclear spin pairing is a maximum when odd numbers of both are present.

### Nuclear stability and neutron-proton ratio

The stability of the nucleus of an atom also influenced by the relative numbers of protons and neutrons along with the odd or even number of protons and neutrons.

The neutron/proton or n/p ratio helps to predict the stability of the nucleus of radioactive elements.

The above graph is obtained by plotting the number of neutrons in the nucleus of the stable isotopes against the respective number of protons.

- A study of this figure shows that the actual n/p plot of stable isotopes breaks from hypothetical 1:1 plot around an atomic number 20.
- After atomic number 20, the line rises rather steeply. The number of protons increases inside the nucleus more and more neutrons are needed to minimize the proton-proton binding energy and thereby to add to nuclear stability.

Neutrons serve as binding martial inside the nucleus and the way an unstable nucleus disintegrates will be resided by its position with respect to the actual n/p plot of the nucleus. - The isotope located above this actual n/p plot with too high an n/p ratio, located below the plot it is too low in n/p ratio. In either case, the unstable nucleus should decay so as to approach the actual n/p plot.

Problem

Why gold-197 is non-radioelement but radium-226 radioactive?

Answer

The number of neutron and protons 188 and 79 respectively in an isotope of gold-197.

n/p ratio = 118/79 =1.49.

n/p values for gold-197 less than 1.5. Thus isotope of a gold-197 stable.

The number of neutron and protons 138 and 88 respectively in isotopes radium-226.

n/p ratio = 226/88 = 1.57.

n/p values for radium-226 less than 1.5. Thus isotope of a radium-226 stable.

#### The neutron to proton ratio too high

Isotopes with too many neutrons in the nucleus can attain greater stability if one of the decays of the neutron to the proton in a **nuclear reaction**. Such disintegration leads to the emission of an electron from inside the nucleus.

_{0}n^{1} → _{1}H^{1} + _{-1}e^{0} (electron)

Thus n/p ratio higher than the expected n/p value for stability beta emission will occur.

n/p ratio is high when the mass number of radioactive isotopes greater than the average atomic weight of the element.

#### Radioactive carbon and iodine

n/p ratio for stable isotopes of carbon – 12 is 1.0 with six protons and six neutrons. But for carbon – 14 is 1.3 with eight neutrons and six protons. It can be predicted caron – 14 will be radioactive and will emit beta rays.

_{6}C^{14} → _{7}N^{14} + _{-1}e^{0}

Similarly, the n/p ratio for the stable isotope of iodine-127 is 1.4 with seventy-four neutrons and fifty-three protons. Iodine-133 the n/p ratio equal to 1.5 with eighty neutrons and fifty-three protons and it is a beta emitter.

_{53}I^{133} → _{54}Xe^{133} + _{-1}e^{0}

### Electron capture nuclear reaction

A nucleus deficient in neutrons will tend to attain nuclear stability by converting one of its protons to a neutron and this will be achieved either by the emission of a position or by the capture of an electron.

_{1}H^{1} → _{0}n^{1} + _{+1}e^{0} (positron)

Positron emission occurs with light radioactive isotopes of the elements of a low atomic number. Nitrogen-13 with n/p ratio of 0.86 disintegrate by positron emission.

_{7}N^{13} → _{6}C^{13} + _{+1}e^{0}

Similarly, iodine-121 has n/p ratio 1.3 which is below the n/p ratio of 1.4, a stable isotope of iodine. Thus iodine expectedly decays by positron emission.

_{53}I^{121} → _{52}Te^{121} + _{+1}e^{0}

Problem

Which of the following elements are beta emitter and why?

- Isotopes of carbon-12 and carbon-14.
- Isotopes of iodine-127 and iodine-133.

Answer

The n/p ratio for stable carbon-12 is 1.0(6n + 6p) but that for carbon-14 is 1.3 (8n + 6p). It will predict that carbon -14 will be radioactive and will emit beta rays.

_{6}C^{14} → _{7}N^{14} + _{-1}e^{0} (beta ray)

Similarly the n/p ratio of the stable Iodine-127 is 1.4 (74n + 53p). For Iodine-133 the n/p ratio equals 1.5 (80n + 53p) and it is again predicted to the beta emitter.

_{53}I^{127} → _{54}Xe^{137} + _{-1}e^{0}

### Examples of nuclear reaction

Orbital electron capture nuclear reaction occurs with too light isotopes of radioactive elements of relatively high atomic numbers. The nucleus captures an electron from the nearest orbital. Orbital electron capture changes one proton to a neutron cause natural radioactivity.

_{37}Rb^{82} + _{-1}e^{0} → _{36}Kr^{82}

_{79}Au^{194} + _{-1}e^{0} → _{78}Pt^{194}

Heaviest nucleus, the total proton-proton repulsion large. The binding effect of the neutron is not enough to lead a stable nonradioactive isotope. For such a nucleus, alpha particle emission common mode of disintegration.

_{90}Th^{232} → _{88}Ra^{228} + _{2}He^{4}

_{92}U^{235} → _{90}Th^{231} + _{2}He^{4}

92U238 → 90Th234 + 2He4

The species thorium-232, thorium-234, uranium-235, and uranium-238 disintegrate to produce more stable isotopes of lead-82.

### Measurement of radioactivity

The rate at which a radioactive sample disintegrates can be determined by counting the number of particles emitted in a given time. The number of disintegration per second expressed the term natural radioactivity.

One gram of radium undergoes about 3.7 × 10^{10} disintegrations per second. The quantity of 3.7 × 10^{10} disintegrations per second is called curie, which is the older unit of radioactivity.

3.7 × 10^{10} curie = 3.7 × 10^{7} millicurie

= 3.7 × 10^{7} microcurie

Millicurie and microcurie respectively correspond to 3.7 × 10^{7} and 3.7 × 10^{4} disintegrations per second.

On this basis, the cause natural radioactivity of radium 1 curie per gram. Phosphorus-32, a beta – emitter, has an activity of 50 millicuries per gram.

This means that for every gram of phosphorus-32 in some material containing this species, there are 50 × 3.7 × 10^{7} disintegrations taking place per second.

#### How to convert becquerel to curie?

SI unit of radioactivity is Becquerel or simply Bq. Becquerel has expressed one disintegration per second.

3.7 × 10^{10} disintegrates per second = 1 curie.

1 disintegration per second = 1 Bq

∴ 3.7 × 10^{10} Bq = 1 curie

Rutherford or simply Rd is the practical unit of radioactivity.

Problem

What is the radioactivity of this element in curie if the radioactive isotopes have x number of disintegration per second?

Answer

3.7 × 10^{10} disintegrations per second = 1 curie.

Thus x number of disintegration per second = x/(3.7 × 10^{10}) curie.