Table of contents

1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
3. Ionic Compounds2h 18m
4. Molecular Compounds2h 18m
5. Classification & Balancing of Chemical Reactions3h 17m
6. Chemical Reactions & Quantities2h 27m
7. Energy, Rate and Equilibrium3h 46m
8. Gases, Liquids and Solids3h 25m
9. Solutions4h 10m
10. Acids and Bases3h 29m
11. Nuclear Chemistry56m
BONUS: Lab Techniques and Procedures1h 38m
BONUS: Mathematical Operations and Functions47m
12. Introduction to Organic Chemistry1h 34m
13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
14. Compounds with Oxygen or Sulfur1h 6m
15. Aldehydes and Ketones1h 1m
16. Carboxylic Acids and Their Derivatives1h 11m
17. Amines39m
18. Amino Acids and Proteins1h 51m
19. Enzymes1h 37m
20. Carbohydrates1h 41m
21. The Generation of Biochemical Energy2h 8m
22. Carbohydrate Metabolism2h 22m
23. Lipids2h 26m
24. Lipid Metabolism1h 45m
25. Protein and Amino Acid Metabolism1h 37m
26. Nucleic Acids and Protein Synthesis2h 54m

11. Nuclear Chemistry

Alpha Decay

GOB Chemistry

11. Nuclear Chemistry

Alpha Decay

Previous problemNext problem

1:41 minutes

Problem 19d

Textbook Question

Complete each of the following nuclear equations and describe the type of radiation:
d. ? → ²³⁴₉₀Th + ⁴₂He

Verified step by step guidance

Step 1: Understand the problem. The given nuclear equation involves a missing reactant that undergoes a nuclear reaction to produce thorium-234 (²³⁴₉₀Th) and a helium nucleus (⁴₂He). The helium nucleus is also known as an alpha particle, indicating that this is an alpha decay process.

Step 2: Recall the law of conservation of mass and atomic numbers. In a nuclear reaction, the sum of the mass numbers (superscripts) and the sum of the atomic numbers (subscripts) must be equal on both sides of the equation.

Step 3: Write the equation with placeholders for the missing reactant. Let the missing reactant be represented as ²³⁸₉₂X, where ²³⁸ is the mass number and ₉₂ is the atomic number. The equation becomes: ²³⁸₉₂X → ²³⁴₉₀Th + ⁴₂He.

Step 4: Verify the conservation of mass numbers. The mass number of the reactant (²³⁸) must equal the sum of the mass numbers of the products (²³⁴ + ⁴). This confirms that the mass number of the missing reactant is ²³⁸.

Step 5: Verify the conservation of atomic numbers. The atomic number of the reactant (₉₂) must equal the sum of the atomic numbers of the products (₉₀ + ₂). This confirms that the atomic number of the missing reactant is ₉₂, identifying it as uranium-238 (²³⁸₉₂U). The type of radiation is alpha radiation.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Nuclear Decay

Nuclear decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This can occur in various forms, including alpha, beta, and gamma decay. In the context of the given equation, the decay process involves the transformation of one element into another, resulting in the emission of particles.

Alpha Particle

An alpha particle is a type of radiation consisting of two protons and two neutrons, essentially a helium nucleus. It is emitted during alpha decay, a common form of nuclear decay for heavy elements. The presence of an alpha particle in the equation indicates that the original nucleus is losing mass and charge, transforming into a new element.

Balancing Nuclear Equations

Balancing nuclear equations involves ensuring that the total number of protons and neutrons is conserved during a nuclear reaction. This means that the sum of the atomic numbers and mass numbers on both sides of the equation must be equal. In the provided equation, identifying the missing particle requires applying this principle to maintain balance.

Watch next

Master Alpha Decay Concept 1 with a bite sized video explanation from Jules

Start learning