Question

Complete and balance the following nuclear equations by supplying the missing particle: (a) \(_{16}^{32} S+{ }_{0}^{1} n \longrightarrow 1_{1} p+?\) (b) \({ }_{4}^{7} \mathrm{Be}+{ }_{-1}^{0}\) (orbital electron) \(\longrightarrow\) ? (c) ? \(\underset{76}{\longrightarrow} \frac{187}{76}+{ }_{-1}^{0}\) (d) \({ }_{42}^{98} \mathrm{Mo}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{0}^{1} \mathrm{n}+?\) (e) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow 135 \mathrm{Xe}+2{ }_{0}^{1} \mathrm{n}+?\)

Short answer

The short answers for the missing particles in each nuclear equation are as follows: (a) \(_{15}^{32}\textrm{P}\) (b) \(_{3}^{7}\textrm{Li}\) (c) \(_{75}^{187}\textrm{Re}\) (d) \(_{43}^{99}\textrm{Tc}\) (e) \(_{38}^{99}\textrm{Sr}\)

Step-by-step solution

(a) Finding the missing particle in equation (a)

In order to balance this nuclear equation, conserve both the atomic mass number (A) and the atomic number (Z) as follows: Atomic mass conservation: \(32 + 1 = 1 + A_{new}\) Atomic number conservation: \(16 + 0 = 1 + Z_{new}\) Solve both these equations to find the unknowns: \(A_{new}= 32\) \(Z_{new}= 15\) Therefore, the missing particle is \(_{15}^{32}\textrm{P}\).

(b) Finding the missing particle in equation (b)

Similar to part (a), balance the atomic mass number (A) and the atomic number (Z): Atomic mass conservation: \(7 + 0 = A_{new}\) Atomic number conservation: \(4 + (-1) = Z_{new}\) Solve both these equations to find the unknowns: \(A_{new}= 7\) \(Z_{new}= 3\) Therefore, the missing particle is \(_{3}^{7}\textrm{Li}\).

(c) Finding the missing particle in equation (c)

Since the atomic mass number and the atomic number of the original particle are unknown, we must balance the equation to determine these values. Atomic mass conservation: \(A = 187 + 0\) Atomic number conservation: \(Z = 76 + (-1)\) Where A and Z are the atomic mass number and atomic number of the unknown particle, respectively. Solve both these equations to find the unknowns: \(A = 187\) \(Z = 75\) Therefore, the missing particle is \(_{75}^{187}\textrm{Re}\).

(d) Finding the missing particle in equation (d)

Balance the atomic mass number (A) and the atomic number (Z): Atomic mass conservation: \(98 + 2 = 1 + A_{new}\) Atomic number conservation: \(42 + 1 = 0 + Z_{new}\) Solve both these equations to find the unknowns: \(A_{new}= 99\) \(Z_{new}= 43\) Therefore, the missing particle is \(_{43}^{99}\textrm{Tc}\).

(e) Finding the missing particle in equation (e)

Balance the atomic mass number (A) and the atomic number (Z): Atomic mass conservation: \(235 + 1 = 135 + 2(1) + A_{new}\) Atomic number conservation: \(92 + 0 = 54 + 2(0) + Z_{new}\) Solve both these equations to find the unknowns: \(A_{new}= 99\) \(Z_{new}= 38\) Therefore, the missing particle is \(_{38}^{99}\textrm{Sr}\).

Key concepts

Atomic Mass Conservation

In nuclear equations, atomic mass conservation is crucial to keep the process balanced. Atomic mass number, denoted as \(A\), represents the sum of protons and neutrons in an atomic nucleus. When completing nuclear reactions, it is essential that the total atomic mass on the reactant side equals that on the product side.
This principle ensures that matter is neither created nor destroyed during the reaction.

For example, consider the reaction in exercise (a):

• The reactant side includes \(^{32}_{16}\textrm{S}\) and \(^1_0\textrm{n}\) giving a total atomic mass of \(32 + 1 = 33\).
• The products include a proton \(^1_1\textrm{p}\) and an unknown particle.

To find the missing particle, we solve for its atomic mass \(A_{new}\) by ensuring that the total mass remains 33, leading to \(32\) plus the mass of the protonthe, confirming the missing particle as \(^{32}_{15}\textrm{P}\). Understanding atomic mass conservation in nuclear reactions helps maintain the balance and validate the reaction process.

Atomic Number Conservation

Another vital concept when balancing nuclear equations is the conservation of the atomic number, represented as \(Z\). The atomic number equals the number of protons in an atom and determines the element. This principle states that the sum of atomic numbers in reactants must equal that in the products. Conservation of charge is achieved by balancing the atomic number, ensuring the same number of protons on both sides.

Take for instance, the equation in part (b):

• We start with \( ^{7}_{4}\textrm{Be}\) and an electron \(^0_{-1}\)
• On the product side, the missing particle needs to ensure continuity.

Solving the conservation equation, \(4 + (-1) = Z_{new}\), gives us \(Z_{new} = 3\), indicating the presence of the particle \(^{7}_{3}\textrm{Li}\).
This straightforward principle of atomic number conservation guarantees that the reaction remains consistent and true to the fundamentals of nuclear physics.

Balancing Nuclear Reactions

Balancing nuclear reactions requires careful attention to detail. It involves ensuring both atomic mass and atomic number are correctly balanced between reactants and products. This balance reflects the accuracy in representing both conserved mass and charge, vital for accurately modeling nuclear processes.

Let's examine part (e) of the exercise:

• We begin with \(^{235}_{92}\textrm{U}\) and a neutron \(^1_0\textrm{n}\).
• On the product side, we identify \(^{135}_{54}\textrm{Xe}\), two neutrons \(2^1_0\textrm{n}\), and an unknown particle.

Applying the conservation principles:

• For atomic mass: \(235 + 1 = 135 + 2(1) + A_{new}\) converts to \(A_{new} = 99\).
• For atomic number: \(92 + 0 = 54 + 2(0) + Z_{new}\), giving \(Z_{new} = 38\).

The missing particle identified using these balances is \(^{99}_{38}\textrm{Sr}\). Focusing on both mass and number keeps nuclear reactions balanced, allowing us to accurately understand atomic interactions and transformations during such processes.