Radioactivity – Types of Radioactive Decay

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.

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Alpha Radioactive Decay

An alpha particle is a helium nucleus ⁴₂He. Whenever a nucleus goes through alpha decay, it transforms into a different nucleus by emitting an alpha particle. For example, when ²³⁸₉₂U undergoes alpha-decay, it transforms into ²³⁴₉₀Th.

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He … (1)

Now, ⁴₂He 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_ZX nucleus to ^A-4_Z-2X nucleus is expressed as follows,

^A_ZX → ^A-4_Z-2X + ⁴₂He … (2)

where ^A_ZX is the parent nucleus and ^A-4_Z-2X is the daughter nucleus. It is important to note that the alpha decay of ²³⁸₉₂U can occur without an external source of energy. This is because of the total mass of the decay products (²³⁴₉₀Th and ⁴₂He) < the mass of the original ²³⁸₉₂U

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) c² … (3)

This energy is shared between the daughter nucleus, ^A-4_Z-2X and the alpha particle, ⁴₂He 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,

³²₁₅P → ³²₁₆S + 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,

²²₁₁Na → ²²₁₀Ne + 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,

⁶⁰₂₇Co undergoes a beta decay and transforms into ⁶⁰₂₈Ni. The daughter nucleus (⁶⁰₂₈Ni) 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:

• ²²⁶₈₈Ra
• ²²⁰₈₆Rn

Where,

226. m (²²⁶₈₈Ra) = 226.02540u
227. m (²²⁰₈₆Rn) = 220.01137u

• m (²²²₈₆Rn) = 222.01750u

216. m (²¹⁶₈₄Po) = 216.00189u

Answer:

(a) After emitting an alpha particle (helium nucleus), the mass number of ²²⁶₈₈Ra reduces to 222 (226 – 4) and the atomic number reduces to 86 (88 – 2).Therefore, we have

²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He

Now, Q value of the emitted alpha particle is,
Q = (m_x – m_y – m_He) c² … from equation (3) above)
= {m (²²⁶₈₈Ra) – m (²²²₈₆Rn) – m (⁴₂He)} c²

We know that,

226. m (²²⁶₈₈Ra) = 226.02540u
227. m (²²²₈₆Rn) = 222.01750u

• m (⁴₂He) = 4.002603u

Therefore,
Q-value = [226.02540u – 222.01750u – 4.002603u] c² = 0.005297uc²

We know that, 1u = 931.5 MeV/c². 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 ²²⁰₈₆Rn reduces to 216 (220 – 4) and the atomic number reduces to 84 (86 – 2).Therefore, we have

²²⁰₈₆Rn → ²¹⁶₈₄Po + ⁴₂He

Now, Q value of the emitted alpha particle is,
Q = (m_x – m_y – m_He) c² … from equation (3) above)
= {m (²²⁰₈₆Rn) – m (²¹⁶₈₄Po) – m (⁴₂He)} c²

We know that,

220. m (²²⁰₈₆Rn) = 220.01137u
221. m (²¹⁶₈₄Po) = 216.00189u

• m (⁴₂He) = 4.002603u

Therefore,
Q-value = [220.01137u – 216.00189u – 4.002603u] c² = 0.006877uc²

We know that, 1u = 931.5 MeV/c². 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.