Question

Complete each of the following nuclear equations by supplying the missing particle. a. \(\quad_{89}^{210} \mathrm{Ac} \rightarrow_{2}^{4} \mathrm{He}+?\) b. \(\quad \frac{131}{53} \mathrm{I} \rightarrow_{54}^{131} \mathrm{Xe}+?\) c. \(_{86}^{210} \mathrm{Rn} \rightarrow_{2}^{4} \mathrm{He}+?\)

Short answer

a. The missing particle is \(_{87}^{206}\mathrm{Th}\) (thorium-206). b. The missing particle is \(_{-1}^{0}\mathrm{e^{-}}\) (an electron). c. The missing particle is \(_{84}^{206}\mathrm{Po}\) (polonium-206).

Step-by-step solution

Check the conservation of mass numbers

On the left side of the equation, we have \(^{210}\mathrm{Ac}\), and on the right side, we have \(^{4}\mathrm{He}\). So, the total mass number on the right side (including the missing particle) should be 210.

Check the conservation of atomic numbers

On the left side of the equation, we have \(_{89}\mathrm{Ac}\), and on the right side, we have \(_{2}\mathrm{He}\). So, the total atomic number on the right side (including the missing particle) should be 89.

Determine the missing particle (mass number and atomic number)

The missing particle must have a mass number of 206 (210 - 4) and an atomic number of 87 (89 - 2). So, the missing particle is \(_{87}^{206}\mathrm{Th}\). #b. Determine the missing particle in I -> Xe + ?#

Check the conservation of mass numbers

On the left side of the equation, we have \(^{131}\mathrm{I}\), and on the right side, we have \(^{131}\mathrm{Xe}\). So, the total mass number on the right side (including the missing particle) should be 131.

Check the conservation of atomic numbers

On the left side of the equation, we have \(_{53}\mathrm{I}\), and on the right side, we have \(_{54}\mathrm{Xe}\). So, the total atomic number on the right side (including the missing particle) should be 53.

Determine the missing particle (mass number and atomic number)

The missing particle must have a mass number of 0 (131 - 131) and an atomic number of -1 (53 - 54). So, the missing particle is \(_{-1}^{0}\mathrm{e^{-}}\), an electron. #c. Determine the missing particle in Rn -> He + ?#

Check the conservation of mass numbers

On the left side of the equation, we have \(^{210}\mathrm{Rn}\), and on the right side, we have \(^{4}\mathrm{He}\). So, the total mass number on the right side (including the missing particle) should be 210.

Check the conservation of atomic numbers

On the left side of the equation, we have \(_{86}\mathrm{Rn}\), and on the right side, we have \(_{2}\mathrm{He}\). So, the total atomic number on the right side (including the missing particle) should be 86.

Determine the missing particle (mass number and atomic number)

The missing particle must have a mass number of 206 (210 - 4) and an atomic number of 84 (86 - 2). So, the missing particle is \(_{84}^{206}\mathrm{Po}\).

Key concepts

Radioactive Decay

Radioactive decay is a process by which an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. This spontaneous transformation can result in a change of an element into a different element or a different isotope of the same element. Three common types of radioactive decay include alpha decay, beta decay, and gamma decay.

In alpha decay, which is depicted in parts a and c of our exercise, a nucleus emits an alpha particle (ucleus{2}{4}{He}), which is essentially a helium nucleus consisting of 2 protons and 2 neutrons. As a result, the original atom loses 4 in mass number and 2 in atomic number, leading to the formation of a new element. Importantly, during radioactive decay, the laws of conservation of mass and atomic number are obeyed, which allows us to predict the resulting products accurately.

Conservation of Mass Number

The conservation of mass number is a fundamental principle used to balance nuclear equations. The mass number, represented by the upper number next to an element's symbol, denotes the total number of protons and neutrons in a nucleus. When a nucleus undergoes radioactive decay, the total mass number of the particles before the decay must equal the total mass number of the particles after decay.

For example, in the solution for part a of the exercise, we see that the decay of Actinium-210 (ucleus{89}{210}{Ac}) results in a mass number of 210. When it emits an alpha particle (ucleus{2}{4}{He}), it leaves behind Thorium-206 (ucleus{87}{206}{Th}) to maintain the mass number, with a simple subtraction (210 - 4 = 206) confirming that the conservation of mass number holds true.

Conservation of Atomic Number

Correspondingly, the conservation of atomic number, also known as the conservation of charge, states that the total charge before and after a nuclear reaction must be equal. The atomic number, indicated by the lower number adjacent to an element's symbol, reveals the number of protons in a nucleus and, thereby, the element's identity.

When working through nuclear equations such as in part b of our exercise, where Iodine-131 (ucleus{53}{131}{I}) undergoes beta decay to become Xenon-131 (ucleus{54}{131}{Xe}), an electron (ucleus{-1}{0}{e^{-}}) is emitted. This process increases the atomic number by 1 while maintaining the mass number, perfectly illustrating the conservation of atomic number. Subtraction comes into play again (53 - 54 = -1), allowing us to identify the emitted particle as an electron, with the change in atomic number revealing the identity of the new element.