Beta Decay: Definition, Types and Examples

Radioactive nuclei behave far differently than normal nuclei. Let’s suppose you take a radioactive phosphorus nucleus and a normal phosphorus chunk in two separate pans; you will notice that after some time, the radioactive phosphorus starts to decay, transitioning into some other element, while the normal phosphorus stays as it was.

This transition does not happen without the emission of certain particles in the process; for example, Uranium-238 a heavy unstable nucleus, upon decaying, transforms into thorium ,emitting an alpha particle in the process, aka, a helium nucleus (₂He⁴). Lighter elements like carbon or nitrogen, who have excess protons or neutrons, upon decaying, emit beta particles, which are essentially nothing but electrons or positrons . Beta decay is found to occur in lighter nuclei. In this article, we will explore beta decay in detail.

Table of Contents

• What is Beta Decay?
• Types of Beta Decay
• Practice Problems of Beta Decay
• FAQs of Beta Decay

What is Beta Decay?

Beta(β) decay is said to have occurred when a positron or electron is emitted from the radioactive nucleus. Positron(₊₁e⁰) is the antimatter (which has the same mass, but opposite charge ) of the electron. The electrons and positrons emitted in the β - decay do not come out directly from the nucleus. They are created during the time of emission, just the same way photons are emitted when an electron transitions from a higher energy level to a lower energy level.

Types of Beta Decay

1) β^- decay 

During a β^- decay, neutron is transformed into a proton, electron and an antineutrino( v^- ). An antineutrino has no rest mass, no charge and a spin quantum number of ± ½ . The reaction can be written as,

 n → p + e^- + v^-

p  → proton

e^- → electron

When a nucleus undergoes β^- decay, its atomic number increases by one, but its mass number remains the same. Consider an element X with atomic number Z and mass number A undergoing β^- decay transforms to another nuclei Y. Then, 

_ZX^A ━^{Beta minus decay}⟶ _z+1Y^A

The energy liberated during a nuclear reaction is called its Q value.

Q value of a β^- decay =[m(_ZX^A)-m(_Z+1Y^A)]c², where c is the velocity of light. 

β^- decay occurs when an element has too many neutrons, i.e the ratio of neutrons to protons is large. A carbon nucleus ₆C¹⁴ decays to ₇N¹⁴ by the reaction shown as follows,

₆C¹⁴ ━━━━━━━⟶ ₇N¹⁴ + e^- + v^-

2) β⁺ decay

During a β⁺ decay, a proton changes to a neutron by emitting a positron and a neutrino(v). A neutrino is the antimatter of an antineutrino.

p ━━━━━━━⟶ n + e⁺ + v

When a nucleus undergoes β⁺ decay, its atomic number decreases by one, but its mass number remains the same. Consider an element X with atomic number Z and mass number A undergoing β⁺ decay transforms to another nuclei Y. Then,

_ZX^A ━^{Beta plus decay}⟶ _z-1Y^A 

Q value of a β⁺ decay of X = 

m_e -  mass of an electron in amu.

β⁺ decay occurs when there are excess protons. In the following reaction, radioactive ₇N¹³ transforms to ₆C¹³ with the emission of a positron(e+ )

₇N¹³  ━━━━━⟶ ₆C¹³ + e⁺ + v

3) Electron capture

Electron capture is analogous to positron emission. It occurs when a nucleus captures one of its own orbital electrons and emits a neutrino. Consider a element _zX^A with atomic number Z and mass number A undergoing an electron capture transforms to an element Y. Then, 

_zX^A + e^-1 ━━━━⟶ Z_-1Y^A + v

₄Be⁷ + e^-1 ━━━━⟶ ₃Li⁷ + v

₁₃Al²⁶ + e^-1 ━━━━⟶ ₁₂Mg²⁶ + v

Practice Problems of Beta Decay

Q1)What comes in the place of the question mark?

₇N¹² ━━━━⟶ ? + v + ₆C¹²

Solution) Since the atomic number decreases by one, it is a + decay. Hence, a positron(e+) should come in the place of the question mark.

∴ ₇N¹² ━━━━⟶ e⁺ + ₆C¹² + v

Q2) Calculate the Q - value of the following decay:

O¹⁹ ━━━━⟶ F¹⁹ + e + v^-

Atomic mass of O¹⁹=19.003576 amu

Atomic mass of F¹⁹=18.998403 amu

Solution) O¹⁹ ━━━━⟶ F¹⁹ + e + v^-

Given, m(O¹⁹)=19.003576 amu

m(F¹⁹)=18.998403 amu

The above reaction is a β^- decay. The Q - value can be calculated as 

Q=[m(O¹⁹)-m(F¹⁹)]c²=(19.003576 -18.998403)

× 931.5 MeV=0.005173931.5 MeV=4.8186 MeV

Q3) Calculate the Q- value of the following reaction 

Al²⁵ ━━━━⟶ e⁺ + Mg²⁵ + v

 m(Al²⁵) = 24.990432 amu

m(Mg²⁵) = 24.985839 amu

m_e = 0.511 MeV

Solution)

The given reaction is a + decay. The Q- value of the reaction can be calculated as 

[m(Al²⁵) - m(Mg²⁵) - 2m_e] × c² = [24.990432 - 24.985839] × 931.5 MeV - [2 × 0.511] MeV

=[4.2784 - 1.022] MeV

=3.2564 MeV

Q4) Complete the following reaction

₁₃Al²⁶ + e^-1  ━━━━⟶ ₁₂Mg²⁶ + ?

Solution)

The following reaction is called electron capture. Neutrino() is emitted. 

₁₃Al²⁶ + e^-1  ━━━━⟶ ₁₂Mg²⁶ + v

Q5) Explain how to calculate Q- value of a β⁺ decay reaction?

Solution) During a β⁺ decay, a proton changes to a neutron by emitting a positron and a neutrino(v). A neutrino is the antimatter of an antineutrino.

p  ━━━━⟶ n + e⁺ + v

_ZX^A ━^{Beta plus decay}⟶ _z-1Y^A

Q value of a β⁺ decay of X = 

me- mass of an electron in amu.

Q6) How to calculate Q^- value of a β^- decay reaction?

Solution) During a β^- decay, neutron is transformed into a proton, electron and an antineutrino( v ). 

 n  ━━━━⟶ p + e^- + v^-

p- proton

e^-- electron

_ZX^A ━^{Beta minus decay}⟶ _z+1Y^A

The energy liberated during a nuclear reaction is called its Q value.

Q value of a β^- decay = where c is the velocity of light. 

FAQs of Beta Decay

Question1. Mass of neutrino and antineutrino is not taken into account while calculating the Q- value of a reaction. Why?

Answer. Mass of a neutrino or antineutrino are very small as compared to other products in a reaction, hence they are neglected. 

Question2. What is the difference between proton and positron?

Answer. Both the proton and a positron have a positive charge. The main difference is that the proton is heavier than a positron. 

Question3. Where does the electron emitted in a beta decay come from?

Answer. During a beta minus decay, an electron is formed when a neutron is converted into a proton, and electron and antineutrino are emitted. 

Question4. Atomic number increases in a beta minus decay. Why?

Answer. During a beta minus decay, the neutron changes to a proton, which leads to an increase in atomic number.

Related Topics to Beta Decay in Physics

• Atomic Mass
• Nuclear Fission
• Mass Energy
• Nuclear Force
• Alpha Decay
• Gamma Decay
• Nuclear Fusion

NCERT Class 12 Physics Chapters