Question

Write a balanced nuclear equation showing the bombardment of \({ }_{4}^{9} \mathrm{Be}\) with alpha particles to produce \({ }_{6}^{12} \mathrm{C}\) and a neutron.

Short answer

The balanced nuclear equation for the bombardment of \({ }_{4}^{9}\mathrm{Be}\) with alpha particles to produce \({ }_{6}^{12} \mathrm{C}\) and a neutron is: \begin{equation} ^{9}_{4}\mathrm{Be} + _{2}^{4}\mathrm{He} \rightarrow _{6}^{12}\mathrm{C} + _{0}^{1}\mathrm{n} \end{equation}

Step-by-step solution

Identify the given particles and their mass numbers (A) and atomic numbers (Z)

We are given the following particles: 1. Beryllium-9: Its symbol is \({ }_{4}^{9}\mathrm{Be}\), with a mass number (A) of 9 and atomic number (Z) of 4. 2. Alpha particle: An alpha particle is the same as a helium nucleus, with a symbol of \({ }_{2}^{4}\mathrm{He}\), with a mass number (A) of 4 and atomic number (Z) of 2. 3. Carbon-12: Its symbol is \({ }_{6}^{12}\mathrm{C}\), with a mass number (A) of 12 and atomic number (Z) of 6. 4. Neutron: A neutron has a mass number (A) of 1 and an atomic number (Z) of 0. Its symbol is \({ }_{0}^{1}\mathrm{n}\).

Write the balanced nuclear equation

To write a balanced nuclear equation, we must ensure that the sum of mass numbers and the sum of atomic numbers are conserved on both sides of the equation. So, we have: \begin{equation} ^{9}_{4}\mathrm{Be} + _{2}^{4}\mathrm{He} \rightarrow _{6}^{12}\mathrm{C} + _{0}^{1}\mathrm{n} \end{equation} Now let's check if the equation is balanced.

Verify the conservation of mass number (A) and atomic number (Z)

For the left side of the equation, the sum of the mass numbers is: \begin{equation} A_{\mathrm{Be}} + A_{\mathrm{He}} = 9 + 4 = 13 \end{equation} For the right side of the equation, the sum of the mass numbers is: \begin{equation} A_{\mathrm{C}} + A_{\mathrm{n}} = 12 + 1 = 13 \end{equation} These two sums are equal, so the mass numbers are conserved. For the left side of the equation, the sum of the atomic numbers is: \begin{equation} Z_{\mathrm{Be}} + Z_{\mathrm{He}} = 4 + 2 = 6 \end{equation} For the right side of the equation, the sum of the atomic numbers is: \begin{equation} Z_{\mathrm{C}} + Z_{\mathrm{n}} = 6 + 0 = 6 \end{equation} These two sums are equal, so the atomic numbers are conserved. The conservation of mass number and atomic number is satisfied, so the balanced nuclear equation is: \begin{equation} ^{9}_{4}\mathrm{Be} + _{2}^{4}\mathrm{He} \rightarrow _{6}^{12}\mathrm{C} + _{0}^{1}\mathrm{n} \end{equation}

Key concepts

Balancing Nuclear Equations

Balancing nuclear equations is a vital step in understanding nuclear reactions. It requires you to ensure that both sides of the equation have equal sum totals of mass numbers and atomic numbers. This way, physics' fundamental principles are adhered to. In every nuclear reaction:

• The sum of the mass numbers (A) on one side should be equal to the sum of mass numbers on the other side.
• Similarly, the sum of the atomic numbers (Z) must also balance across the equation.

These nuclear equations are similar to balancing chemical equations, but they focus more on the conservation of nucleons rather than molecules.
To write a balanced nuclear equation, begin by identifying all participating particles' mass and atomic numbers. For example, in the provided exercise, you start with beryllium, an alpha particle, and create carbon and a neutron. Using this information, compose the equation and ensure both sides' numbers balance to reflect the conservation laws precisely.
Following these steps meticulously ensures a comprehensive understanding of nuclear transformations.

Conservation of Mass Number

The conservation of mass number is a crucial concept where the total number of nucleons—protons and neutrons—remains constant in a nuclear reaction. It's important as it ensures matter is neither created nor destroyed in nuclear transformations but simply rearranged.
Let's dissect the initial exercise: When beryllium \(( { }_{4}^{9} ext{Be})\) is bombarded with an alpha particle \(( { }_{2}^{4} ext{He})\), carbon \(( { }_{6}^{12} ext{C})\) and a neutron \(( { }_{0}^{1} ext{n})\) are produced. The total mass number before the reaction \(9 + 4 = 13\) matches the total mass number after the reaction \(12 + 1 = 13\).

• This shows the mass number is conserved because each side equals 13, the initial total sum.

By understanding this principle, analyzing and verifying nuclear reactions become more systematic and logical.

Conservation of Atomic Number

The atomic number conservation principle states that the total atomic number (sum of protons) remains the same on both sides of a nuclear equation. This principle upholds the idea that the identity of the atoms involved doesn't change; rather, they're rearranged.
In the original exercise, let's look at the initial elements: Beryllium has an atomic number of 4, and the alpha particle has 2, for a total of 6 before the reaction. After the reaction, carbon has an atomic number of 6, and the neutron has none (0), resulting in a total of 6.

• With these calculations at hand, it's evident that the sum of the atomic numbers remains unchanged (6 on the left equals 6 on the right), confirming the conservation.

This demonstrates not only the integrity of the nuclear equation but also provides confidence in the reaction's viability, ensuring it aligns with fundamental nuclear physics laws.