Problem 76 - Chemistry: An Atoms-Focused Approach

Question

What nuclide is produced in the core of a collapsing giant star by each of the following reactions? a. \({ }_{29}^{65} \mathrm{Cu}+3{ }_{0}^{1} \mathrm{n} \rightarrow ?+{ }_{-1}^{0} \boldsymbol{\beta}\) b. \({ }_{30}^{68} \mathrm{Zn}+2{ }_{0}^{1} \mathrm{n} \rightarrow ?+{ }_{-1}^{0} \beta\) c. \({ }_{38}^{88} \mathrm{Sr}+{ }_{0}^{1} \mathrm{n} \rightarrow ?+{ }_{-1}^{0} \beta\)

Step-by-Step Solution

Answer

a. \({ }_{29}^{65} \mathrm{Cu}+3{ }_{0}^{1} \mathrm{n} \rightarrow ?+{ }_{-1}^{0} \boldsymbol{\beta}\) b. \({ }_{30}^{68} \mathrm{Zn}+2{ }_{0}^{1} \mathrm{n} \rightarrow ?+{ }_{-1}^{0} \beta\) c. \({ }_{38}^{88} \mathrm{Sr}+{ }_{0}^{1} \mathrm{n} \rightarrow ?+{ }_{-1}^{0} \beta\) Answer: The nuclides produced in the core of the collapsing giant star are: a. \({ }_{28}^{68} \mathrm{Ni}\) b. \({ }_{29}^{70} \mathrm{Cu}\) c. \({ }_{37}^{89} \mathrm{Rb}\)

Step 1: Analyze the First Reaction#a.

We are given the reaction: \({ }_{29}^{65} \mathrm{Cu}+3{ }_{0}^{1} \mathrm{n} \rightarrow ?+{ }_{-1}^{0} \boldsymbol{\beta}\) We will first add the reactants' mass and atomic numbers to calculate the resulting product's mass and atomic numbers.

Step 2: Calculate Nuclide Numbers for the First Reaction#a.

The sum of mass numbers (A): \(65 + 3 \times 1 = 68\) The sum of atomic numbers (Z): \(29 + 3 \times 0 = 29\) Now, we need to account for the emitted beta particle, which means a decrease in the atomic number by 1: Z = 29 - 1 = 28 The resulting nuclide is thus \({ }_{28}^{68} \mathrm{X}\).

Step 3: Identify the First Reaction's Product Nuclide#a.

The resulting nuclide has an atomic number of 28, corresponding to the element nickel (Ni). Therefore, the product nuclide of the first reaction is \({ }_{28}^{68} \mathrm{Ni}\).

Step 4: Analyze the Second Reaction#b.

We are given the reaction: \({ }_{30}^{68} \mathrm{Zn}+2{ }_{0}^{1} \mathrm{n} \rightarrow ?+{ }_{-1}^{0} \beta\)

Step 5: Calculate Nuclide Numbers for the Second Reaction#b.

The sum of mass numbers (A): \(68 + 2 \times 1 = 70\) The sum of atomic numbers (Z): \(30 + 2 \times 0 = 30\) Now, we need to account for the emitted beta particle, which means a decrease in the atomic number by 1: Z = 30 - 1 = 29 The resulting nuclide is thus \({ }_{29}^{70} \mathrm{X}\).

Step 6: Identify the Second Reaction's Product Nuclide#b.

The resulting nuclide has an atomic number of 29, corresponding to the element copper (Cu). Therefore, the product nuclide of the second reaction is \({ }_{29}^{70} \mathrm{Cu}\).

Step 7: Analyze the Third Reaction#c.

We are given the reaction: \({ }_{38}^{88} \mathrm{Sr}+{ }_{0}^{1} \mathrm{n} \rightarrow ?+{ }_{-1}^{0} \beta\)

Step 8: Calculate Nuclide Numbers for the Third Reaction#c.

The sum of mass numbers (A): \(88 + 1 = 89\) The sum of atomic numbers (Z): \(38 + 0 = 38\) Now, we need to account for the emitted beta particle, which means a decrease in the atomic number by 1: Z = 38 - 1 = 37 The resulting nuclide is thus \({ }_{37}^{89} \mathrm{X}\).

Step 9: Identify the Third Reaction's Product Nuclide#c.

The resulting nuclide has an atomic number of 37, corresponding to the element rubidium (Rb). Therefore, the product nuclide of the third reaction is \({ }_{37}^{89} \mathrm{Rb}\). To conclude, the nuclides produced in the core of the collapsing giant star are: a. \({ }_{28}^{68} \mathrm{Ni}\) b. \({ }_{29}^{70} \mathrm{Cu}\) c. \({ }_{37}^{89} \mathrm{Rb}\)

Key Concepts

Beta Decay

Beta decay is a process through which an unstable atomic nucleus loses energy. This process involves the emission of a beta particle, which can be an electron (\(\beta^-\)) or a positron (\(\beta^+\)). In the context of reactions like the ones in our exercise, we're focusing on \(\beta^-\) decay. This happens when a neutron in the nucleus is transformed into a proton while emitting an electron and an antineutrino. The result is an atom with a higher atomic number because one neutron becomes a proton.

• Imagine that within the nucleus, one of the neutrons changes its identity to become a proton.
• An electron (beta particle) is then ejected at high speed.
• This transformation results in a change in the atomic number, thus creating a different element.

Understanding beta decay is crucial for predicting the products of nuclear reactions involving neutron capture and beta emission. It allows us to calculate how the composition of the nucleus changes, leading to the formation of a new element.

Nuclide Production

Nuclide production can be understood as the creation of a different nucleus from a nuclear reaction. When discussing nuclear reactions like those in collapsing stars, protons, neutrons, and beta particles play key roles. Each interaction in the nucleus leads to the creation of a new nuclide, impacting the mass and atomic numbers.
In our exercise:

• Additional neutrons are captured by the nucleus, increasing the mass number but leaving the atomic number unaffected.
• The inclusion of a beta decay step then transforms this configuration by lowering the atomic number by 1 for each beta particle emitted.

The result is a different element with a new mass number and atomic number. Knowing the starting element, the number of neutrons absorbed, and the beta particle emissions lets us predict the end product—this is important for understanding the chemical makeup following stellar events.

Atomic Number Calculations

Atomic number calculations are fundamental in identifying elements involved in nuclear reactions. The atomic number, symbolized by \(Z\), represents the number of protons in an atom and identifies the element itself.
During a reaction such as those seen in our exercise:

• Calculating the atomic number involves summing the atomic numbers of all reactants.
• In beta decay, we reduce the calculated atomic number by 1 for each beta particle emitted. This reflects the transformation of a neutron into a proton.
• For instance, in our first reaction: the atomic number initially is the atomic number of copper, 29, and the resultant atomic number is 28 after accounting for the beta emission.

This calculation step-by-step helps in identifying the new element (or nuclide) produced, like nickel in the first case. Thus, atomic number calculations are essential for predicting the new element formed in nuclear reactions.