5.0 Radioactivity

The phenomenon of the spontaneous disintegration of an unstable nuclide into other nuclide's to achieve stability is called radioactivity.

When an unstable nucleus or radioactive element disintegrates, high-energy photons are released.

These particles and photons are collectively known as rays.

Three kinds of rays are produced by naturally occurring radioactivity i.e. $\alpha {\text{, }}\beta $ and $\gamma $ rays.

All these rays have different penetrating power. $$\alpha {\text{ - rays < }}\beta {\text{ - rays < }}\gamma {\text{ - rays}}$$

All these disintegration processes that produce these rays must obey laws of physics called conservation laws (as these processes deal with charge, energy, momentum etc. that should be conserved in any process). These are-

All these physical entities must be conserved when a nucleus disintegrates into smaller nuclear fragments and accompanying $\alpha {\text{ - , }}\beta {\text{ - or }}\gamma {\text{ - rays}}{\text{.}}$

It was then seen that these rays contain charge particles. This can be depicted in a simple experiment.

A piece of radioactive material is placed at the bottom of the of a narrow hole in a lead cylinder.

The cylinder is located within an evacuated chamber as illustrated in Fig. 10 .

A magnetic field is directed into the plane of the screen and a photographic plate is positioned to the right of the hole. Three nods of spots appear on the developed plate, which is associated with the radioactivity of the nuclei in the material.

Since moving particles are deflected by a magnetic field only when they are charged.

This experiment reveals that two types of radioactive rays $\alpha $ and $\beta $ consists of charged particles, while the $\gamma$ ray does not.

Note: The above experiment shows that,

1. $\beta $ particles are negatively charged particle as they are attracted towards the positive plate

2. $\alpha $ particles are positively charged particle as they are attracted towards the negative plate

3. $\gamma$ particles are neutral as they goes un-deviated

When a nucleus disintegrates and produces $\alpha$ rays, it is said to undergo $\alpha$ decay.

Nuclear reaction is given by, $${\text{ }}_Z^AX \to _{Z - 2}^{A - 4}Y + _2^4He + Q$$

where, $$\begin{equation} \begin{aligned} \Rightarrow Q = \Delta m{c^2} \\ \Rightarrow Q = \left[ {\left( {{M_X} - Z{m_e}} \right) - \left\{ {\left( {{M_Y} - \left( {Z - 2} \right){m_e}} \right) + \left( {{M_{He}} - 2{m_e}} \right)} \right\}} \right]{c^2} \\ \Rightarrow Q = \left[ {{M_X} - {M_Y} - {M_{He}}} \right]{c^2} \\\end{aligned} \end{equation} $$

When a nucleus disintegrates and produces $\beta $ , it is said to undergo $\beta $ decay.

As these rays are deflected in a direction opposite to the direction of $\alpha$ rays which are positively charged.

So, they must consist of negatively charged particles.

The experiment shows that $\beta $ particles are electrons.

There is two type of beta decay.

1. ${\beta ^ - }$ decay

2. ${\beta ^ + }$ decay

${\beta ^ - }$ decay occurs when a neutron in the unstable nucleus decays into a proton and an electron.

The electron is usually fast moving and escapes from the atom, leaving behind a positively charged atom.

So, an anti-neutrino is released in this process.

It is a charge less and massless quantity.

$$\begin{equation} \begin{aligned} {\text{Nuclear reaction}} \mapsto {\text{ }}_Z^AX \to _{Z - 2}^{A - 4}Y + _2^4He + \bar v + Q \\ {\text{where }}Q = \Delta m{c^2} = \left[ {\left( {{M_X} - Z{m_e}} \right) - \left\{ {\left( {{M_Y} - \left( {Z - 2} \right){m_e}} \right) + \left( {{M_{He}} - 2{m_e}} \right)} \right\}} \right]{c^2} \\ \quad \quad \quad \quad \quad \quad \quad \quad = \left[ {{M_X} - {M_Y} - {M_{He}}} \right]{c^2} \\\end{aligned} \end{equation} $$

${\beta ^ - }$ decay occurs when a proton in the unstable nucleus decays into a neutron and a positron.

Positron does not exist within the nucleus but rather is generated during the process.

A neutrino is released in this process.

It is also a charge less and massless quantity.

$$\begin{equation} \begin{aligned} {\text{Nuclear reaction}} \mapsto {\text{ }}_Z^AX \to _{Z - 1}^AY + _1^0e + Q \\ {\text{where }}Q = \Delta m{c^2} = \left[ {\left( {{M_X} - Z{m_e}} \right) - \left\{ {\left( {{M_Y} - \left( {Z - 1} \right){m_e}} \right) + {m_e}} \right\}} \right]{c^2} \\ \quad \quad \quad \quad \quad \quad \quad \quad = \left[ {{M_X} - {M_Y} + 2{m_e}} \right]{c^2} \\\end{aligned} \end{equation} $$

Very often a nucleus that undergoes radioactive decay is left in an excited energy state.

The nucleus can undergo a second decay to attain a lower energy state, by emitting one or more high energy photons.

The photons emitted in such a process are called gamma rays.

$$\begin{equation} \begin{aligned} {\text{Nuclear reaction}} \mapsto {\text{ }}_Z^AX \to _{Z - 1}^A{Y^*} + _1^0e + v \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad _{Z - 1}^A{Y^*} \to _{Z - 1}^AY + \gamma \\\end{aligned} \end{equation} $$

where "*" represents an excited nucleus.

Gamma decay occurs in the second equation.

Sometimes electron from the $K$-shell is captured by the nucleus and this electron along with a proton is converted into a neutron. This known as $K$-capture.

$${\text{Nuclear reaction}} \mapsto {\text{ }}_Z^AX + _{ - 1}^0e \to _{Z - 1}^AY$$

Question 10. Consider the beta decay

$${}^{198}Au \to {}^{198}H{g^*} + {\beta ^ - } + \overline v $$

where ${}^{198}H{g^*}$ represents a mercury nucleus in an excited state at energy 1.088 $Mev$ above the ground state. What can be the maximum kinetic energy of the electron emitted?

The atomic mass of ${}^{198}Au$ is 197.968233 $u$ and that of ${}^{198}Hg$ is 197.966760 $u$.

If the product nucleus ${}^{198}Hg$ is formed in its ground state, the kinetic energy available to the electron and the anti neutrino is $$Q = \left[ {m\left( {{}^{198}Au} \right) - m\left( {{}^{198}Hg} \right)} \right]{c^2}$$

As ${}^{198}H{g^*}$ has $1.088\ MeV$ more energy than ${}^{198}Hg$ in ground state. The kinetic energy actually available is

$$\begin{equation} \begin{aligned} Q = \left[ {m\left( {{}^{198}Au} \right) - m\left( {{}^{198}Hg} \right)} \right]{c^2} - 1.088{\text{ Mev}} \\ \quad = \left( {197.968233{\text{ u}} - 197.966760{\text{ u}}} \right)\left( {931.5\frac{{{\text{Mev}}}}{{\text{u}}}} \right) - 1.088{\text{ Mev}} \\ \quad = 1.3686{\text{ Mev}} - 1.088{\text{ Mev}} = 0.2806{\text{ Mev}} \\\end{aligned} \end{equation} $$

This is also the maximum possible kinetic energy of the electron emitted.

Question 11. What is the wavelength of the 0.186 $MeV$ gamma ray photon emitted by radium ${}_{88}^{226}Ra$?

The photon energy is the difference between two nuclear energy levels.

Relation between the energy level separation $\Delta E$ and the frequency $f$ of the photon is given by, $$\begin{equation} \begin{aligned} {E_i} - {E_f} = hf \\ \Delta E = hf \\\end{aligned} \end{equation} $$.

Since $$f\lambda = c$$,

The wavelength of the photon is, $$\lambda = \frac{{hc}}{{\Delta E}}$$.

First, we must convert the photon energy into Joule $(J)$.

$$\begin{equation} \begin{aligned} \Delta E = \left( {0.186 \times {{10}^6}eV} \right)\left( {\frac{{1.60 \times {{10}^{ - 19}}}}{{1eV}}} \right)\quad \;\, \\ \Delta E = 2.98 \times {10^{ - 14}}J \\\end{aligned} \end{equation} $$

Therefore, the wavelength of the proton is,

$$\begin{equation} \begin{aligned} \lambda = \frac{{hc}}{{\Delta E}} \\ \lambda = \frac{{\left( {6.63 \times {{10}^{ - 34}}Js} \right)\left( {3.00 \times {{10}^8}m{s^{ - 1}}} \right)}}{{2.98 \times {{10}^{ - 14}}J}}\quad \quad \quad \; \\ \lambda = 6.67 \times {10^{ - 12}}m \\\end{aligned} \end{equation} $$

Stable nuclei have the $\left( {\frac{N}{P}} \right)$ ratio either equal to one or more than 1.

The ratio is nearly equal to 1 for light nuclei up to calcium $(Ca)$.

It increases up to 1.6 for heavy nuclei.

A nucleus undergo a type of decay event depending upon its $\left( {\frac{N}{P}} \right)$ ratio.

If it is greater than 1 then it may undergo ${\beta ^ - }$ decay.

Here are the possible cases,