Question

Complete each of the following nuclear equations by supplying the missing particle. a. \(? \rightarrow_{84}^{210} \mathrm{Po}+_{2}^{4} \mathrm{He}\) b. \(40 \mathrm{K} \rightarrow_{18}^{40} \mathrm{Ar}+?\) c. \(\frac{137}{57} \mathrm{La}+_{-1} \mathrm{e} \rightarrow ?\)

Short answer

The completed nuclear equations are: a. \(_{82}^{206}\mathrm{Pb} \rightarrow_{84}^{210}\mathrm{Po} + _{2}^{4}\mathrm{He}\) b. \(_{19}^{40}\mathrm{K} \rightarrow_{18}^{40}\mathrm{Ar} + _{-1}^0\mathrm{e}\) c. \(_{57}^{137}\mathrm{La} + _{-1}^0\mathrm{e} \rightarrow_{56}^{137}\mathrm{Ba}\)

Step-by-step solution

Identify given particles

In this equation, we have two known particles: Po (Polonium) with atomic mass number 210 and an atomic number of 84, and He (Helium nuclei) with atomic mass number 4 and atomic number 2. #a. Completing the nuclear equation for Po and He#

Apply conservation of atomic mass number (A) and atomic number (Z)

To find the missing particle, we need to apply the conservation of atomic mass number (A) and atomic number (Z). This means that the total atomic mass number and atomic number before the reaction should be equal to the sum after the reaction. Let the atomic mass number of the missing particle be A1 and the atomic number be Z1. So, we have: \(A1 + 4 = 210\) (Conservation of atomic mass number) \(Z1 + 2 = 84\) (Conservation of atomic number) #a. Completing the nuclear equation for Po and He#

Solve for A1 and Z1

Now we can solve the two equations we found in step 2 for A1 and Z1: \(A1 = 210 - 4 = 206\) \(Z1 = 84 - 2 = 82\) So, the missing particle has an atomic mass number of 206 and an atomic number of 82, which corresponds to Pb (Lead). Therefore, the completed nuclear equation is: \(_{82}^{206}\mathrm{Pb} \rightarrow_{84}^{210}\mathrm{Po} + _{2}^{4}\mathrm{He}\) #b. Completing the nuclear equation for K and Ar#

Identify given particles

In this equation, we have two known particles: K (Potassium) with atomic mass number 40 and an atomic number of 19, and Ar (Argon) with atomic mass number 40 and an atomic number of 18. #b. Completing the nuclear equation for K and Ar#

Apply conservation of atomic mass number (A) and atomic number (Z)

To find the missing particle, we need to apply the conservation of atomic mass number (A) and atomic number (Z). Let the atomic mass number of the missing particle be A2 and the atomic number be Z2. So, we have: \(40 = 40 + A2\) (Conservation of atomic mass number) \(19 = 18 + Z2\) (Conservation of atomic number) #b. Completing the nuclear equation for K and Ar#

Solve for A2 and Z2

Now we can solve the two equations we found in step 2 for A2 and Z2: \(A2 = 40 - 40 = 0\) \(Z2 = 19 - 18 = 1\) So, the missing particle has an atomic mass number of 0 and an atomic number of 1, which corresponds to an electron (e). Therefore, the completed nuclear equation is: \(_{19}^{40}\mathrm{K} \rightarrow_{18}^{40}\mathrm{Ar} + _{-1}^0\mathrm{e}\) #c. Completing the nuclear equation for La and e#

Identify given particles

In this equation, we have two known particles: La (Lanthanum) with an atomic mass number 137 and an atomic number of 57, and e (electron) with atomic mass number 0 and an atomic number of -1. #c. Completing the nuclear equation for La and e#

Apply conservation of atomic mass number (A) and atomic number (Z)

To find the missing particle, we need to apply the conservation of atomic mass number (A) and atomic number (Z). Let the atomic mass number of the missing particle be A3 and the atomic number be Z3. So, we have: \(137 = A3\) (Conservation of atomic mass number) \(57 - 1 = Z3\) (Conservation of atomic number) #c. Completing the nuclear equation for La and e#

Solve for A3 and Z3

Now we can solve the two equations found in step 2 for A3 and Z3: \(A3 = 137\) \(Z3 = 57 - 1 = 56\) So, the missing particle has an atomic mass number of 137 and an atomic number of 56, which corresponds to Ba (Barium). Therefore, the completed nuclear equation is: \(_{57}^{137}\mathrm{La} + _{-1}^0\mathrm{e} \rightarrow_{56}^{137}\mathrm{Ba}\)

Key concepts

Atomic Mass Number

The atomic mass number, also known simply as the mass number, is crucial in understanding nuclear equations. It is the total count of protons and neutrons in the nucleus of an atom. Essentially, it tells you about the _bulkiness_ of an atom as protons and neutrons make up the vast majority of an atom's mass.
The notation for atomic mass number in nuclear equations is denoted by the letter \(A\). In the equation \(_{82}^{206}\mathrm{Pb}\), \(A\) is **206**.

• Key Point: Atomic mass number determines the isotope of an element—same protons, different neutrons.
• Mass Number Formula: \(A = Z + N\) where \(Z\) is the atomic number (protons), and \(N\) is the number of neutrons.

In nuclear reactions, the sum of the atomic mass numbers on the reactant side must equal the sum on the product side. This rule assures that we aren't mysteriously losing or gaining 'chunks' of nuclear mass during reactions. Understanding this allows you to solve for unknowns in nuclear equations, as seen in the exercises above.

Atomic Number

The atomic number, denoted as \(Z\), is a fundamental characteristic of an element. It represents the number of protons in the nucleus, which also determines the chemical identity of the element. For instance, hydrogen has an atomic number of **1**, helium **2**, and so on.

The atomic number is what sets one element apart from another. In our nuclear equations:

• \(_{84}^{210}\mathrm{Po}\) has an atomic number of 84, identifying it as polonium.
• \(_{57}^{137}\mathrm{La}\) has an atomic number of 57, identifying it as lanthanum.

Conservation of atomic number is key in nuclear equations. Just like with mass numbers, there must be an equal number of protons (or \(Z\)) before and after the reaction.
For example: In the step \(40 K \rightarrow_{18}^{40} \mathrm{Ar} + ?\), knowing the atomic numbers helps us deduce the missing particle, ensuring all charges are balanced across the equation. And since the change from \(K\) to \(Ar\) involves a decrease of an atomic number by 1, it implies the emission of a particle carrying a positive charge reduction such as an electron \(e^-\).
Such notions allow further exploration in nuclear chemistry, including predicting reaction outcomes and balancing equations.

Conservation of Mass and Charge

Conservation laws are the guardians of nuclear reactions, ensuring balance and clarity. In the context of nuclear chemistry, we focus here on the conservation of mass and charge.

• **Mass Conservation**: The total mass number (sum of protons and neutrons) must remain constant before and after a nuclear reaction. This is because protons and neutrons are not created or destroyed in these exchanges; they only rearrange into different atoms or isotopes.
• **Charge Conservation**: The total charge, depicted by the atomic numbers, also remains unchanged throughout a nuclear reaction. Consequently, the sum of atomic numbers in the reactants will equal the sum in the products.

These conservation principles steer us through nuclear equations like a compass, ensuring accuracy.
For example, in solving: \( \_? \rightarrow_{84}^{210} \mathrm{Po}+_{2}^{4} \mathrm{He} \), you equate the total atomic numbers and mass numbers to deduce the missing particle \((_{82}^{206}\mathrm{Pb})\). This technique enriches your understanding of complex nuclear transformations and helps decode the history of reactions happened inside core depths, such as stars and nuclear reactors.