Problem 16 - Chemistry

Question

In each of the following radioactive decay processes, supply the missing particle. a. \(^{60} \mathrm{Co} \rightarrow^{60} \mathrm{Ni}+?\) b. \(^{97} \mathrm{Tc}+? \rightarrow^{97} \mathrm{Mo}\) c. \(^{99} \mathrm{Tc} \rightarrow^{99} \mathrm{Ru}+?\) d. \(^{239} \mathrm{Pu} \rightarrow^{235} \mathrm{U}+?\)

Step-by-Step Solution

Answer

a. The missing particle is a beta particle, or an electron: \(^0_{-1}\beta\). b. The missing particle is a positron: \(^0_{+1}e\). c. The missing particle is a beta particle, or an electron: \(^0_{-1}\beta\). d. The missing particle is an alpha particle: \(^4_2\alpha\).

Step 1: Identify given values

In the given radioactive decay process, we have the following information: - The mass of the initial substance, \(^{60}\mathrm{Co}\). - The mass of the decay product, \(^{60}\mathrm{Ni}\).

Step 2: Apply conservation laws

Using the conservation of atomic and mass numbers, we can write the following equations: - Atomic number: 27 = 28 + Z - Mass number: 60 = 60 + A From the equations above, we can calculate A and Z as: - A = 0 - Z = -1 The missing particle is an electron, also known as a beta particle, which is represented by \(^0_{-1}\beta\).

Step 3: Answer#a.

The missing particle is a beta particle, or an electron: \(^0_{-1}\beta\). b. \(^{97} \mathrm{Tc}+? \rightarrow^{97} \mathrm{Mo}\)

Step 1: Identify given values

In the given radioactive decay process, we have the following informations: - The mass of the initial substance, \(^{97}\mathrm{Tc}\). - The mass of the decay product, \(^{97}\mathrm{Mo}\).

Step 2: Apply conservation laws

Now we apply the conservation laws again. - Atomic number: 43 + Z = 42 - Mass number: 97 + A = 97 From the above equations, we get A = 0 and Z = -1. This means the missing particle is a positron, represented by \(^0_{+1}e\).

Step 6: Answer#b.

The missing particle is a positron: \(^0_{+1}e\). c. \(^{99} \mathrm{Tc} \rightarrow^{99} \mathrm{Ru}+?\)

Step 1: Identify given values

In the given radioactive decay process, we have the following informations: - The mass of the starting substance, \(^{99}\mathrm{Tc}\). - The mass of the decay product, \(^{99}\mathrm{Ru}\).

Step 2: Apply conservation laws

We can apply the conservation laws as follows: - Atomic number: 43 = 44 + Z - Mass number: 99 = 99 + A Solving the equations gives A = 0 and Z = -1. The missing particle is a beta particle, or an electron, represented by \(^0_{-1}\beta\).

Step 9: Answer#c.

The missing particle is a beta particle, or an electron: \(^0_{-1}\beta\). d. \(^{239} \mathrm{Pu} \rightarrow^{235} \mathrm{U}+?\)

Step 1: Identify given values

In the given radioactive decay process, we have the following informations: - The mass of the starting substance, \(^{239}\mathrm{Pu}\). - The mass of the decay product, \(^{235}\mathrm{U}\).

Step 2: Apply conservation laws

For this radioactive decay process, we can write the conservation laws as follows: - Atomic number: 94 = 92 + Z - Mass number: 239 = 235 + A The calculated values for A are 4 and Z equals 2. The missing particle is an alpha particle, which is represented by \(^4_2\alpha\).

Step 12: Answer#d.

The missing particle is an alpha particle: \(^4_2\alpha\).

Key Concepts

Conservation Laws

In radioactive decay, conservation laws are like nature's bookkeeping. They ensure that everything balances perfectly. The main players here are the atomic number and the mass number.

When an atom undergoes decay, these two numbers must stay the same before and after the process, just like balancing both sides of an equation.

• The atomic number (Z) represents the number of protons in the nucleus.
• The mass number (A) is the total number of protons and neutrons.

By applying these laws, we can determine missing particles during decay. For example, if a particle has left an atom and we know the atomic and mass numbers, we can find out what that particle was. This understanding is essential to predicting the products of nuclear reactions.

Beta Particle

Beta particles are fascinating products of radioactive decay. They are high-energy, high-speed electrons or positrons ejected from the nucleus.

During a beta decay process, a neutron is transformed into a proton, and this is when a beta particle is emitted.

• The beta particle is represented by (^0_{-1}eta ight), or sometimes (^0_{-1}e ight), indicating it’s an electron.
• It occurs when the atomic number (Z) of a nucleus increases by 1, while the mass number (A) remains unchanged.

This process helps achieve a more stable ratio of protons to neutrons within the nucleus. Beta particles are a crucial aspect of how elements transform from one to another in nature.

Alpha Particle

Alpha particles are another intriguing product of radioactive decay. They are essentially helium nuclei, made up of two protons and two neutrons.

Alpha decay usually happens in heavier elements. It allows the nucleus to shed some excess mass and find a more stable state.

• An alpha particle is represented by (^4_2 ext{He} ight) or (^4_2oldsymbol{ ext{α}} ight), revealing its structure similar to helium.
• During alpha decay, the atomic number falls by 2, and the mass number by 4.

This step is vital in reducing the nucleus’s size and achieving stability. Understanding alpha particles helps in grasping the overall lifecycle of elements in nature.

Positron Emission

Positron emission is an extraordinary form of radioactive decay. A positron is the antimatter counterpart of an electron, with the same mass but a positive charge.

During positron emission, a proton is transformed into a neutron, and a positron is emitted from the nucleus.

• The positron is represented by (^0_{+1}e ight) or sometimes by (^0_{+1}eta^+ ight).
• This process decreases the atomic number by 1, while the mass number stays the same.

Positron emission is crucial in reducing the number of protons in a nucleus, allowing it to reach a more balanced state. It's a fundamental concept in nuclear physics and medical imaging, such as PET scans.