Question

Complete each of the following nuclear equations by supplying the missing particle. a. \(\quad_{85}^{196} \mathrm{At} \rightarrow_{2}^{4} \mathrm{He}+?\) b. \(_{84}^{208} \mathrm{Po} \rightarrow_{2}^{4} \mathrm{He}+?\) c. \(_{86}^{210} \mathrm{Rn} \rightarrow_{2}^{4} \mathrm{He}+?\)

Short answer

The missing particles in the given nuclear equations are: a. \(\quad_{85}^{196} \mathrm{At} \rightarrow_{2}^{4} \mathrm{He}+\quad_{83}^{192} \mathrm{Bi}\) (bismuth) b. \(_{84}^{208} \mathrm{Po} \rightarrow_{2}^{4} \mathrm{He}+\quad_{82}^{204} \mathrm{Pb}\) (lead) c. \(_{86}^{210} \mathrm{Rn} \rightarrow_{2}^{4} \mathrm{He}+\quad_{84}^{206} \mathrm{Po}\) (polonium)

Step-by-step solution

Observe the given nuclear equations

First, let's take a look at the given nuclear equations: a. \(\quad_{85}^{196} \mathrm{At} \rightarrow_{2}^{4} \mathrm{He}+?\) b. \(_{84}^{208} \mathrm{Po} \rightarrow_{2}^{4} \mathrm{He}+?\) c. \(_{86}^{210} \mathrm{Rn} \rightarrow_{2}^{4} \mathrm{He}+?\) In each nuclear equation, we have an initial nucleus (on the left) that undergoes alpha decay, emitting an alpha particle (\(_{2}^{4} \mathrm{He}\)) and transforming into an unknown nucleus (denoted by the question mark).

Use conservation laws

The mass number (A) and atomic number (Z) are conserved in nuclear reactions. Therefore, the sum of the mass numbers and the sum of the atomic numbers on the left-hand side of the equation must equal the sum of the mass numbers and the sum of atomic numbers on the right-hand side. For each nuclear equation, we can write the following expressions for the conservation of mass number (A) and atomic number (Z): Mass number: \(A_{initial} = A_{alpha} + A_{unknown}\) Atomic number: \(Z_{initial} = Z_{alpha} + Z_{unknown}\)

Calculate the unknown mass and atomic numbers

Solve for \(A_{unknown}\) and \(Z_{unknown}\) in each equation: a. \(A_{unknown} = A_{initial} - A_{alpha} = 196 - 4 = 192\) \(Z_{unknown} = Z_{initial} - Z_{alpha} = 85 - 2 = 83\) b. \(A_{unknown} = 208 - 4 = 204\) \(Z_{unknown} = 84 - 2 = 82\) c. \(A_{unknown} = 210 - 4 = 206\) \(Z_{unknown} = 86 - 2 = 84\)

Determine the unknown nuclei

Now that we have the mass and atomic numbers for the unknown nuclei, we can identify them using the periodic table or a chart of nuclides: a. \(_{83}^{192} \mathrm{Bi}\) (bismuth) b. \(_{82}^{204} \mathrm{Pb}\) (lead) c. \(_{84}^{206} \mathrm{Po}\) (polonium)

Write the complete nuclear equations

Finally, substitute the unknown nuclei back into the original nuclear equations: a. \(\quad_{85}^{196} \mathrm{At} \rightarrow_{2}^{4} \mathrm{He}+\quad_{83}^{192} \mathrm{Bi}\) b. \(_{84}^{208} \mathrm{Po} \rightarrow_{2}^{4} \mathrm{He}+\quad_{82}^{204} \mathrm{Pb}\) c. \(_{86}^{210} \mathrm{Rn} \rightarrow_{2}^{4} \mathrm{He}+\quad_{84}^{206} \mathrm{Po}\)

Key concepts

Alpha Decay

Alpha decay is a type of radioactive decay where an unstable atomic nucleus emits an alpha particle, which consists of two protons and two neutrons, the same as a helium-4 nucleus (₂⁴He). These two protons and two neutrons are bound together and ejected from the parent atom's nucleus. In the exercise, elements like Astatine (₈₅¹⁹⁶At), Polonium (₈₄²⁰⁸Po), and Radon (₈₆²¹⁰Rn) are shown releasing alpha particles, signifying this kind of decay. After an alpha particle is emitted, the original atomic nucleus loses 4 in mass number and 2 in atomic number, resulting in the formation of a new element.

It's a common misconception that alpha particles only slightly alter the mass of the nucleus they leave behind. However, the emission of an alpha particle significantly changes the structure of the original element, causing it to transform into a completely different element altogether.

Conservation of Mass Number

The conservation of mass number is crucial to understand when looking at nuclear reactions, including alpha decay. The mass number, denoted as A, is the total number of protons and neutrons in a nucleus. This number must remain constant before and after the decay process. When an alpha particle (₂⁴He) is emitted, its mass number of 4 must be subtracted from the mass number of the original element to maintain the balance. That's why, in the solutions, the new elements have a mass number that is exactly 4 less than the starting element. For instance, when Astatine (₈₅¹⁹⁶At) undergoes alpha decay, it transforms into Bismuth (₈₃¹⁹²Bi), implying that the mass number conservation is satisfied (196 = 4 + 192).

Atomic Number Conservation

Similar to the conservation of mass number, the atomic number, represented by Z, must also be conserved in nuclear reactions. The atomic number indicates the element's identity in the Periodic Table, as it is the number of protons in the nucleus of an atom. In alpha decay, the atomic number decreases by two because the alpha particle carries away two protons. The exercise demonstrates this principle; when we subtract the atomic number of the alpha particle (₂) from the initial element, we find the atomic number of the new element. For example, Polonium (₈₄²⁰⁸Po) forms Lead (₈₂²⁰⁴Pb) after emitting an alpha particle, which means that the atomic number has changed according to the equation 84 = 2 + 82, upholding the concept of atomic number conservation.

Periodic Table

The Periodic Table is an organized chart of elements, primarily sorted by increasing atomic number, which represents the number of protons found in the nucleus of an atom. It serves as a powerful tool for understanding and predicting the behaviors of elements. When working with nuclear equations such as those in the exercise, the Periodic Table helps us identify the new element formed after alpha decay. Knowing that an alpha particle has a mass number of 4 and an atomic number of 2, we can use those values to determine the identity of the unknown products in the nuclear equation. By subtracting these numbers from their respective totals in the parent atom, we can locate the new element on the Periodic Table based on the atomic number and mass number arrived upon, as shown in the steps of the solution with elements like Bismuth, Lead, and Polonium.