Chemistry Practice Problems Radioactive Decay Practice Problems Solution: Nuclear reactions involve a change in the nucleus ...

Solution: Nuclear reactions involve a change in the nucleus of an atom. For this reason, nuclear equations show the mass number and atomic number of each species. Nuclear equations must be balanced in both mass number (mass balance) and atomic number (charge balance). For example, consider the equation for the decay of carbon-14:146C → 147N + 0-1 ePart A. Write the identity of the missing nucleus for the following nuclear decay reaction:23290Th → 22888Ra + ?Express your answer as an isotope.Part B. Write the identity of the missing nucleus for the following nuclear decay reaction:189F → 01 e +?Express your answer as an isotope.Part C. Write the identity of the missing nucleus for the following nuclear decay reaction:? → 5927Co + 0-1 eExpress your answer as an isotope.

Problem

Nuclear reactions involve a change in the nucleus of an atom. For this reason, nuclear equations show the mass number and atomic number of each species. Nuclear equations must be balanced in both mass number (mass balance) and atomic number (charge balance). For example, consider the equation for the decay of carbon-14:

146C → 147N + 0-1 e

Part A. Write the identity of the missing nucleus for the following nuclear decay reaction:

23290Th → 22888Ra + ?

Express your answer as an isotope.

Part B. Write the identity of the missing nucleus for the following nuclear decay reaction:

189F → 0e +?

Express your answer as an isotope.

Part C. Write the identity of the missing nucleus for the following nuclear decay reaction:

? → 5927Co + 0-1 e

Express your answer as an isotope.

Solution

We’re being asked to complete the given partial decay reactions


To do so, we need to analyze the product that forms during the decay of the initial nuclide.


Recall that in a nuclear reaction, the number of protons and neutrons is affected and the identity of the element changes


The different types of radioactive decay are:

• Alpha decay: forms an alpha particle (42α, atomic mass = 4, atomic number = 2)

• Beta decay: forms a beta particle (0–1β, atomic mass = 0, atomic number = –1). The beta particle appears in the product side.

• Gamma emission: forms a gamma particle (00γ, atomic mass = 0, atomic number = 0)

• Positron emission: forms a positron particle (01e, atomic mass = 0, atomic number = 1)

• Electron capture: the initial nuclide captures an electron (0–1e, atomic mass = 0, atomic number = –1). The electron appears in the reactant side.


The nuclear equation must be balanced with the same total atomic mass and atomic number on both sides.

Step 1: Balance the atomic mass on both sides.

Step 2: Balance the atomic number on both sides.

Step 3: Identify the particle formed.


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