Question

Write balanced equations for (a) ${}_{92}^{238}\mathrm{U}(\alpha ,\mathrm{n}{)}_{{}_{94}^{24}}^{24}\mathrm{Pu}$, (b) ${}_7^{14}\mathrm{N}(\alpha ,\mathrm{p}{)}_8^{17}\mathrm{O},(\mathrm{c}){}_{26}^{56}\mathrm{Fe}{(\alpha ,{\beta }^{-})}_{29}^{60}\mathrm{Cu}$.

Short answer

The balanced equations for the given nuclear reactions are: (a) ${}_{92}^{238}\mathrm{U}+{}_2^4\alpha \to {}_{94}^{24}\mathrm{Pu}+{}_0^1\mathrm{n}$ (b) ${}_7^{14}\mathrm{N}+{}_2^4\alpha \to {}_8^{17}\mathrm{O}+{}_1^1\mathrm{p}$ (c) ${}_{26}^{56}\mathrm{Fe}+{}_2^4\alpha \to {}_{29}^{60}\mathrm{Cu}+{\beta }^{-}$

Step-by-step solution

(a) Balanced Equation for U-alpha-n-Pu reaction

For the given reaction: ${}_{92}^{238}\mathrm{U}(\alpha ,\mathrm{n}){}_{94}^{24}\mathrm{Pu}$, let's balance the equation by adding the alpha particle and neutron. ${}_{92}^{238}\mathrm{U}+{}_2^4\alpha \to {}_{94}^{24}\mathrm{Pu}+{}_0^1\mathrm{n}$ Now, let's verify the conservation of protons and neutrons. On both sides of the equation, protons and neutrons should be equal: Left side: Protons (Z) = 92 + 2 = 94 Neutrons (N) = (238 - 92) + (4 - 2) = 146 + 2 = 148 Right side: Protons (Z) = 94 Neutrons (N) = 24 - 94 = 150 - 1 = 148 Since the protons and neutrons are equal on both sides, the equation is balanced.

(b) Balanced Equation for N-alpha-p-O reaction

For the given reaction: ${}_7^{14}\mathrm{N}(\alpha ,\mathrm{p}{)}_8^{17}\mathrm{O}$, let's balance the equation by adding the alpha particle and proton. ${}_7^{14}\mathrm{N}+{}_2^4\alpha \to {}_8^{17}\mathrm{O}+{}_1^1\mathrm{p}$ Now, let's verify the conservation of protons and neutrons: Left side: Protons (Z) = 7 + 2 = 9 Neutrons (N) = (14 - 7) + (4 - 2) = 7 + 2 = 9 Right side: Protons (Z) = 8 + 1 = 9 Neutrons (N) = (17 - 8) = 9 Since the protons and neutrons are equal on both sides, the equation is balanced.

(c) Balanced Equation for Fe-alpha-beta-Cu reaction

For the given reaction: ${}_{26}^{56}\mathrm{Fe}{(\alpha ,{\beta }^{-})}_{29}^{60}\mathrm{Cu}$, let's balance the equation by adding the alpha particle and beta particle. ${}_{26}^{56}\mathrm{Fe}+{}_2^4\alpha \to {}_{29}^{60}\mathrm{Cu}+{\beta }^{-}$ Now, let's verify the conservation of protons and neutrons: Left side: Protons (Z) = 26 + 2 = 28 Neutrons (N) = (56 - 26) + (4 - 2) = 30 + 2 = 32 Right side: Protons (Z) = 29 - 1 = 28 Neutrons (N) = (60 - 29) = 31 + 1 = 32 Since the protons and neutrons are equal on both sides, the equation is balanced.

Key concepts

Nuclear Reactions

Understanding nuclear reactions is fundamental to grasping how elements transform into different elements, which is at the heart of nuclear chemistry. A nuclear reaction involves the change in an atom's nucleus and can lead to the transmutation of one element into another. These reactions are governed by several principles, including the conservation of mass and charge, and feature various types of particles, including alpha and beta particles, protons, and neutrons.

Key types of nuclear reactions include fission, where a nucleus splits into smaller parts, and fusion, where lighter nuclei combine to form a heavier nucleus. Both reactions result in the release or absorption of energy and can be represented by balanced nuclear equations that illustrate the principle of conservation by showing that the number of protons and neutrons remains constant before and after the reaction.

Alpha Particles

Alpha particles are a type of ionizing radiation ejected from the nuclei of certain radioactive substances. An alpha particle is essentially a helium-4 nucleus, consisting of two protons and two neutrons, symbolized as ${}_2^4\text{He}$ or simply $\text{α}$. Owing to their composition, alpha particles are relatively massive and carry a +2 charge. In nuclear equations, the emission of an alpha particle from a nucleus results in a decrease in both atomic number and mass number of the original atom, due to the loss of the two protons and two neutrons.

Alpha particles play a crucial role in nuclear reactions such as radioactive decay, where a heavier nucleus releases an alpha particle to become a lighter element. This is significantly relevant in studying the decay series of heavy elements like uranium or thorium.

Conservation of Mass and Charge

The principle of conservation of mass and charge is vital in balancing nuclear equations. It states that during a nuclear reaction, the total mass and the total charge must be conserved – they should remain unchanged from the reactants to the products.

When balancing nuclear equations, one must ensure that the sum of atomic numbers (protons) and mass numbers (total of protons and neutrons) on the left side of the equation equals that on the right side. This principle explains why in an alpha decay process, the parent nucleus loses two protons and two neutrons – reflected in a decrease of four in mass number and two in atomic number. The conservation of charge is similarly critical; the charge of the nucleus changes based on the particles it emits or captures, and these changes must be accounted for to balance the equation.

Neutron Balance

Neutron balance refers to the conservation of neutrons before and after a nuclear reaction. Neutrons, along with protons, make up the majority of an atom's mass and are fundamental in stabilizing the nucleus. Unlike protons, neutrons do not carry an electrical charge, but they still play a significant role in nuclear reactions.

To achieve neutron balance in nuclear equations, one must count the number of neutrons in both the reactants and products and ensure they are equal. This involves calculating the difference between the mass number and the atomic number for each nuclide involved in the reaction. If neutrons are neither created nor destroyed during a reaction, such as in alpha or proton emissions, this balance is straightforward. However, in reactions involving beta decay, a neutron is transformed into a proton (or vice versa), and an electron or positron is emitted, requiring additional consideration to achieve neutron balance.