In this article, we will look at the three types of radioactive decay namely, alpha, beta, and gamma decay. We will try to understand how these particles are emitted and its effects on the emitting nucleus.

### Suggested Videos

## Alpha Radioactive Decay

An alpha particle is a helium nucleus ^{4}_{2}He. Whenever a nucleus goes through alpha decay, it transforms into a different nucleus by emitting an alpha particle. For example, when ^{238}_{92}U undergoes alpha-decay, it transforms into ^{234}_{90}Th.

^{238}_{92}U → ^{234}_{90}Th + ^{4}_{2}He … (1)

Now, ^{4}_{2}He contains two protons and two neutrons. Hence, after emission, the mass number of the emitting nucleus reduces by four and the atomic number reduces by two. Therefore, the transformation of ^{A}_{Z}X nucleus to ^{A-4}_{Z-2}X nucleus is expressed as follows,

^{A}_{Z}X → ^{A-4}_{Z-2}X + ^{4}_{2}He … (2)

where ^{A}_{Z}X is the parent nucleus and ^{A-4}_{Z-2}X is the daughter nucleus. It is important to note that the alpha decay of ^{238}_{92}U can occur without an external source of energy. This is because of the total mass of the decay products (^{234}_{90}Th and ^{4}_{2}He) < the mass of the original ^{238}_{92}U

Or, the total mass-energy of the decay products is less than that of the original nuclide. This brings us to the concept of ‘*Q value of the process*’ or ‘*Disintegration energy*’ which is the difference between the initial and final mass-energy of the decay products. For an alpha decay, the Q value is expressed as,

Q = (m_{X} – m_{Y} – m_{He}) c^{2} … (3)

This energy is shared between the daughter nucleus, ^{A-4}_{Z-2}X and the alpha particle, ^{4}_{2}He in the form of kinetic energy. Also, alpha decay obeys the radioactive laws.

**Browse more Topics under Nuclei**

- Atomic Mass and Composition of Nucleus
- Mass-Energy and Nuclear Binding Energy
- Nuclear Force
- Nuclear Energy – Nuclear Fusion
- Nuclear Energy – Nuclear Fission
- Radioactivity – Law of Radioactive Decay

## Beta Radioactive Decay

Beta decay is when a nucleus decays spontaneously by emitting an electron or a positron. This is also a spontaneous process, like the alpha decay, with a definite disintegration energy and half-life. And, it follows the radioactive laws. A Beta decay can be a beta minus or a beta plus decay.

*In a Beta minus (*β^{−}*) decay*, an electron is emitted by the nucleus. For example,

^{32}_{15}P → ^{32}_{16}S + e^{–} + v^{–} … (4)

where v- is an antineutrino, a neutral particle with little or no mass. Also, T_{1/2} = 14.3 days.

*In a Beta plus decay*, a positron is emitted by the nucleus. For example,

^{22}_{11}Na → ^{22}_{10}Ne + e^{+} + v … (5)

where v is a neutrino, a neutral particle with little or no mass. Also, T_{1/2} = 2.6 years. The neutrinos and antineutrinos are emitted from the nucleus along with the positron or electron during the beta decay process. Neutrinos interact very weakly with matter. Hence, they were undetected for a very long time.

Further, in a Beta minus decay, a neutron transforms into a proton within the nucleus:

n → p + e^{–} + ν^{–} … (6)

Also, in a Beta plus decay, a proton transforms into a neutron (inside the nucleus):

p → n + e^{+} + ν … (7)

Hence, we can see that the mass number (A) of the emitting nuclide does not change. As shown in equations (6) and (7), either a proton transforms into a neutron or vice versa.

## Gamma Radioactive Decay

We know that atoms have energy levels. Similarly, a nucleus has energy levels too. When a nucleus is in an excited state, it can transition to a lower energy state by emitting an electromagnetic radiation. Further, the difference between the energy states in a nucleus is in MeV. Hence, the photons emitted by the nuclei have MeV energies and called Gamma rays.

After an alpha or beta emission, most radionuclides leave the daughter nucleus in an excited state. This daughter nucleus reaches the ground state by emitting one or multiple gamma rays. For example,

^{60}_{27}Co undergoes a beta decay and transforms into ^{60}_{28}Ni. The daughter nucleus (^{60}_{28}Ni) is in its excited state. This excited nucleus reaches the ground state by the emission of two gamma rays having energies of 1.17 MeV and 1.33 MeV. The energy level diagram shown below depicts this process.

## Solved Examples for You

Question: Find the Q value and kinetic energy of the emitted alpha particle in:

^{226}_{88}Ra^{220}_{86}Rn

Where,

- m (
^{226}_{88}Ra) = 226.02540u - m (
^{220}_{86}Rn) = 220.01137u

- m (
^{222}_{86}Rn) = 222.01750u

- m (
^{216}_{84}Po) = 216.00189u

Answer:

**(a)** After emitting an alpha particle (helium nucleus), the mass number of ^{226}_{88}Ra reduces to 222 (226 – 4) and the atomic number reduces to 86 (88 – 2).Therefore, we have

^{226}_{88}Ra → ^{222}_{86}Rn + ^{4}_{2}He

Now, Q value of the emitted alpha particle is,

Q = (m_{x} – m_{y} – m_{He}) c^{2} … from equation (3) above)

= {m (^{226}_{88}Ra) – m (^{222}_{86}Rn) – m (^{4}_{2}He)} c^{2}

We know that,

- m (
^{226}_{88}Ra) = 226.02540u - m (
^{222}_{86}Rn) = 222.01750u

- m (
^{4}_{2}He) = 4.002603u

Therefore,

Q-value = [226.02540u – 222.01750u – 4.002603u] c^{2} = 0.005297uc^{2}

We know that, 1u = 931.5 MeV/c^{2}. Hence,

Q = 0.005297 x 931.5 ~ 4.94 MeV

And, the Kinetic Energy of the alpha particle,

= {(Mass number after decay) / (Mass number before decay)} x Q

= (222/226) x 4.94 = 4.85 MeV.

**(b) **After emitting an alpha particle (helium nucleus), the mass number of ^{220}_{86}Rn reduces to 216 (220 – 4) and the atomic number reduces to 84 (86 – 2).Therefore, we have

^{220}_{86}Rn → ^{216}_{84}Po + ^{4}_{2}He

Now, Q value of the emitted alpha particle is,

Q = (m_{x} – m_{y} – m_{He}) c^{2} … from equation (3) above)

= {m (^{220}_{86}Rn) – m (^{216}_{84}Po) – m (^{4}_{2}He)} c^{2}

We know that,

- m (
^{220}_{86}Rn) = 220.01137u - m (
^{216}_{84}Po) = 216.00189u

- m (
^{4}_{2}He) = 4.002603u

Therefore,

Q-value = [220.01137u – 216.00189u – 4.002603u] c^{2} = 0.006877uc^{2}

We know that, 1u = 931.5 MeV/c^{2}. Hence,

Q = 0.006877 x 931.5 ~ 6.41 MeV

And, the Kinetic Energy of the alpha particle,

= {(Mass number after decay) / (Mass number before decay)} x Q

= (216/220) x 6.41 = 6.29 MeV.