Unit 10: Types of Radiation
Understanding the nature of radioactive decay and the particles and energy it releases.
10.1 Radioactivity & Unstable Nuclei
The stability of an atomic nucleus depends on the balance between its protons and neutrons. If a nucleus has too many protons, too many neutrons, or is simply too large, it becomes unstable.
An unstable nucleus will spontaneously change to become more stable. In doing so, it emits energy in the form of particles or electromagnetic waves. This spontaneous emission of radiation from an unstable nucleus is called radioactivity or radioactive decay.
This process is a nuclear reaction, meaning it involves changes within the nucleus itself. This is fundamentally different from a chemical reaction, which only involves the rearrangement of electrons outside the nucleus. As a result of radioactive decay, an atom can transform into an atom of a completely different element.
Solved Examples:
- What makes an atomic nucleus
unstable?
Solution: An imbalance in its proton-to-neutron ratio, or being too large overall. - Define radioactivity.
Solution: The spontaneous emission of radiation from the nucleus of an unstable atom. - What is the main difference between a nuclear reaction and a chemical
reaction?
Solution: A nuclear reaction involves changes inside the nucleus (protons and neutrons), while a chemical reaction only involves the atom's electrons. - What is another term for
radioactivity?
Solution: Radioactive decay. - Why do unstable nuclei undergo radioactive
decay?
Solution: To release energy and transform into a more stable nuclear configuration. - Can a chemical process (like heating) stop an atom from being
radioactive?
Solution: No, because chemical processes affect electrons, whereas radioactivity is a nuclear phenomenon. - Is it possible for an atom of one element to become an atom of another
element?
Solution: Yes, through radioactive decay (a nuclear reaction). - Are all isotopes of an element
radioactive?
Solution: No. For example, carbon-12 is stable, while carbon-14 is radioactive. - What are isotopes?
Solution: Atoms of the same element with the same number of protons but a different number of neutrons. - What is the goal of radioactive
decay?
Solution: To achieve a more stable nucleus.
10.2 Alpha Particles (α) & Alpha Decay
Alpha decay typically occurs in very large nuclei that have too many protons. To increase stability, the nucleus ejects an alpha particle (α).
- An alpha particle consists of two protons and two neutrons.
- It is identical to the nucleus of a helium atom.
- It has a mass number of 4 and a charge of +2.
- Its symbol is written as $^4_2\alpha$ or $^4_2He$.
When a nucleus undergoes alpha decay, its mass number decreases by 4 and its atomic number decreases by 2, resulting in the formation of a new element.
Example: The alpha decay of Uranium-238.
$$^{238}_{92}U \rightarrow ^{234}_{90}Th + ^4_2\alpha$$
Solved Examples:
- What is an alpha particle composed
of?
Solution: Two protons and two neutrons. - What is the charge of an alpha
particle?
Solution: +2. - An alpha particle is the nucleus of which
element?
Solution: Helium. - When a nucleus emits an alpha particle, by how much does its mass number
change?
Solution: It decreases by 4. - By how much does the atomic number change during alpha
decay?
Solution: It decreases by 2. - Radium-226 ($^{226}_{88}Ra$) undergoes alpha decay. What is the daughter
nucleus?
Solution: The new atomic number is 88 - 2 = 86 (Radon, Rn). The new mass number is 226 - 4 = 222. The product is Radon-222 ($^{222}_{86}Rn$). - What type of nuclei typically undergoes alpha
decay?
Solution: Very large, heavy nuclei with too many protons. - Write the symbol for an alpha
particle.
Solution: $^4_2\alpha$ or $^4_2He$. - Is alpha decay a chemical or nuclear
process?
Solution: A nuclear process. - Complete the nuclear equation: $^{210}_{84}Po \rightarrow ? +
^4_2\alpha$.
Solution: The product must have a mass number of 210 - 4 = 206 and an atomic number of 84 - 2 = 82 (Lead, Pb). The product is $^{206}_{82}Pb$.
10.3 Beta Particles (β) & Beta Decay
Beta decay typically occurs in nuclei that have too many neutrons compared to protons. To increase stability, a neutron inside the nucleus is converted into a proton and a high-energy electron. The proton stays in the nucleus, and the electron is ejected at high speed. This ejected electron is a beta particle (β).
- A beta particle is a high-energy electron.
- It has a negligible mass (mass number 0) and a charge of -1.
- Its symbol is written as $^0_{-1}\beta$ or $^0_{-1}e$.
When a nucleus undergoes beta decay, its mass number remains unchanged, but its atomic number increases by 1, resulting in the formation of a new element.
Example: The beta decay of Carbon-14.
$$^{14}_{6}C \rightarrow ^{14}_{7}N + ^0_{-1}\beta$$
Solved Examples:
- What is a beta particle?
Solution: A high-energy electron emitted from the nucleus. - What is the charge of a beta
particle?
Solution: -1. - What nuclear process leads to the emission of a beta
particle?
Solution: A neutron transforms into a proton and an electron. - When a nucleus emits a beta particle, by how much does its mass number
change?
Solution: It does not change (it stays the same). - By how much does the atomic number change during beta
decay?
Solution: It increases by 1. - Cobalt-60 ($^{60}_{27}Co$) undergoes beta decay. What is the daughter
nucleus?
Solution: The new atomic number is 27 + 1 = 28 (Nickel, Ni). The mass number remains 60. The product is Nickel-60 ($^{60}_{28}Ni$). - What type of nuclei typically undergoes beta
decay?
Solution: Nuclei with a high neutron-to-proton ratio (too many neutrons). - Write the symbol for a beta particle.
Solution: $^0_{-1}\beta$ or $^0_{-1}e$. - Why does the atomic number increase in beta
decay?
Solution: Because a neutron is converted into a proton, increasing the total number of protons in the nucleus by one. - Complete the nuclear equation: $^{131}_{53}I \rightarrow ? +
^0_{-1}\beta$.
Solution: The product must have a mass number of 131 and an atomic number of 53 + 1 = 54 (Xenon, Xe). The product is $^{131}_{54}Xe$.
10.4 Gamma Rays (γ)
Often, after a nucleus has undergone alpha or beta decay, it is left in an excited, high-energy state. To return to its stable, ground state, the nucleus releases the excess energy in the form of a high-energy electromagnetic wave called a gamma ray (γ).
- Gamma rays are a form of high-energy electromagnetic radiation, similar to X-rays but more energetic.
- They have no mass and no charge.
- Their symbol is $^0_0\gamma$.
The emission of a gamma ray does not change the mass number or the atomic number of the nucleus. Therefore, it does not change the identity of the element. It is simply a way for an excited nucleus to lose energy. Gamma emission almost always accompanies alpha and beta decay.
Solved Examples:
- What is a gamma ray?
Solution: A high-energy photon of electromagnetic radiation emitted from an excited nucleus. - What is the mass and charge of a gamma
ray?
Solution: Zero mass and zero charge. - Why are gamma rays emitted during radioactive
decay?
Solution: To release excess energy from a nucleus that has been left in an excited state after alpha or beta decay. - Does gamma emission change the atomic number of an
atom?
Solution: No. - Does gamma emission change the mass number of an
atom?
Solution: No. - What type of radiation is a gamma ray similar
to?
Solution: X-rays, but with higher energy. - Can gamma decay occur on its own?
Solution: No, it almost always accompanies alpha or beta decay. - Write the symbol for a gamma ray.
Solution: $^0_0\gamma$. - If a nucleus emits a gamma ray, does it become a different
element?
Solution: No, because the number of protons does not change. - What is the purpose of gamma emission for a
nucleus?
Solution: To transition from a high-energy (excited) state to a lower-energy (ground) state.
10.5 Balancing Nuclear Equations
A nuclear equation represents a radioactive decay process. Unlike chemical equations, nuclear equations must explicitly show the mass numbers and atomic numbers of all particles involved.
Balancing a nuclear equation relies on two principles of conservation:
- Conservation of Mass Number: The sum of the mass numbers on the reactant side must equal the sum of the mass numbers on the product side.
- Conservation of Atomic Number (Charge): The sum of the atomic numbers on the reactant side must equal the sum of the atomic numbers on the product side.
Example: Find the missing particle in the decay of Polonium-212.
$$^{212}_{84}Po \rightarrow ^{208}_{82}Pb + ?$$
- Balance Mass Number: 212 = 208 + A. So, A = 4.
- Balance Atomic Number: 84 = 82 + Z. So, Z = 2.
- The missing particle has a mass number of 4 and an atomic number of 2. This is an alpha particle, $^4_2\alpha$.
Solved Examples:
- What two quantities must be conserved when balancing a nuclear
equation?
Solution: Mass number and atomic number. - Complete the equation: $^{234}_{90}Th \rightarrow ^{234}_{91}Pa +
?$
Solution: Mass number is conserved (234 = 234 + 0). Atomic number must be conserved (90 = 91 + Z), so Z = -1. The particle is a beta particle, $^0_{-1}\beta$. - Identify the parent nucleus: $? \rightarrow ^{14}_{7}N +
^0_{-1}e$.
Solution: Mass number: A = 14 + 0 = 14. Atomic number: Z = 7 + (-1) = 6. The parent nucleus is Carbon-14, $^{14}_{6}C$. - What is the product of the alpha decay of Radon-222
($^{222}_{86}Rn$)?
Solution: It emits $^4_2\alpha$. The product will have mass number 222 - 4 = 218 and atomic number 86 - 2 = 84 (Polonium, Po). The product is $^{218}_{84}Po$. - Is the following equation balanced? $^{137}_{55}Cs \rightarrow ^{137}_{56}Ba
+ ^0_{-1}\beta$.
Solution: Yes. Mass numbers: 137 = 137 + 0. Atomic numbers: 55 = 56 + (-1). - What does the subscript in nuclear notation
represent?
Solution: The atomic number (or charge). - What does the superscript represent?
Solution: The mass number. - Complete the equation: $^{239}_{94}Pu \rightarrow ? +
^4_2He$.
Solution: Mass number: 239 - 4 = 235. Atomic number: 94 - 2 = 92 (Uranium, U). The product is $^{235}_{92}U$. - What type of decay is represented by the equation $^{40}_{19}K \rightarrow
^{40}_{20}Ca + ^0_{-1}e$?
Solution: Beta decay. - Balance the equation: $^{27}_{13}Al + ^4_2\alpha \rightarrow ? +
^1_0n$.
Solution: Reactant mass numbers: 27 + 4 = 31. Product mass numbers: A + 1. So A = 30. Reactant atomic numbers: 13 + 2 = 15. Product atomic numbers: Z + 0. So Z = 15 (Phosphorus, P). The product is $^{30}_{15}P$.