Question

Complete the following nuclear equations, and identify \(\mathrm{X}\) in each case: (a) \({ }_{12}^{26} \mathrm{Mg}+{ }_{1}^{1} \mathrm{p} \longrightarrow \alpha+\mathrm{X}\) (b) \({ }_{27}^{59} \mathrm{Co}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{27}^{60} \mathrm{Co}+\mathrm{X}\) (c) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{36}^{94} \mathrm{Kr}+{ }_{56}^{139} \mathrm{Ba}+3 \mathrm{X}\) (d) \(\frac{53}{24} \mathrm{Cr}+{ }_{2}^{4} \alpha \longrightarrow{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) \((\mathrm{e}){ }_{8}^{20} \mathrm{O} \longrightarrow{ }_{9}^{20} \mathrm{~F}+\mathrm{X}\)

Short answer

(a) \( _{11}^{23} \mathrm{Na} \), (b) \( _{1}^{1} \mathrm{H} \), (c) \( _{0}^{1} \mathrm{n} \), (d) \( _{26}^{56} \mathrm{Fe} \), (e) \( _{-1}^{0} \beta \).

Step-by-step solution

Analyze Reaction (a)

The nuclear equation is: \[ {}_{12}^{26} \mathrm{Mg} + {}_{1}^{1} \mathrm{p} \rightarrow \alpha + \mathrm{X} \]We know that an alpha particle (\(\alpha\)) is \( _{2}^{4} \mathrm{He} \).Conserve both atomic and mass numbers to find \(\mathrm{X}\).

Solve for X in Reaction (a)

Equation for mass numbers: \[ 26 + 1 = 4 + A \] Thus, \( A = 23 \). Equation for atomic numbers: \[ 12 + 1 = 2 + Z \] Thus, \( Z = 11 \). Therefore, \(\mathrm{X}\) is \( _{11}^{23} \mathrm{Na} \) (sodium).

Analyze Reaction (b)

The nuclear equation is: \[ {}_{27}^{59} \mathrm{Co} + {}_{1}^{2} \mathrm{H} \rightarrow {}_{27}^{60} \mathrm{Co} + \mathrm{X} \]Conserve both atomic and mass numbers to find \(\mathrm{X}\).

Solve for X in Reaction (b)

Equation for mass numbers: \[ 59 + 2 = 60 + A \] Thus, \( A = 1 \).Equation for atomic numbers: \[ 27 + 1 = 27 + Z \] Thus, \( Z = 1 \). Therefore, \(\mathrm{X}\) is \( _{1}^{1} \mathrm{H} \) (a proton).

Analyze Reaction (c)

The nuclear equation is:\[ {}_{92}^{235} \mathrm{U} + {}_{0}^{1} \mathrm{n} \rightarrow {}_{36}^{94} \mathrm{Kr} + {}_{56}^{139} \mathrm{Ba} + 3\mathrm{X} \]Conserve both atomic and mass numbers to find \(\mathrm{X}\).

Solve for X in Reaction (c)

Equation for mass numbers: \[ 235 + 1 = 94 + 139 + 3A \] Thus, \( 3A = 3 \) and \( A = 1 \).Equation for atomic numbers: \[ 92 + 0 = 36 + 56 + 3Z \] Thus, \( 3Z = 0 \) and \( Z = 0 \). Therefore, \(\mathrm{X}\) is \( _{0}^{1} \mathrm{n} \) (a neutron).

Analyze Reaction (d)

The nuclear equation is:\[ \frac{53}{24} \mathrm{Cr} + {}_{2}^{4} \alpha \rightarrow {}_{0}^{1} \mathrm{n} + \mathrm{X} \]Conserve both atomic and mass numbers to find \(\mathrm{X}\).

Solve for X in Reaction (d)

Equation for mass numbers: \[ 53 + 4 = 1 + A \] Thus, \( A = 56 \).Equation for atomic numbers: \[ 24 + 2 = 0 + Z \] Thus, \( Z = 26 \). Therefore, \(\mathrm{X}\) is \( _{26}^{56} \mathrm{Fe} \) (iron).

Analyze Reaction (e)

The nuclear equation is:\[ {}_{8}^{20} \mathrm{O} \rightarrow {}_{9}^{20} \mathrm{~F} + \mathrm{X} \]Conserve both atomic and mass numbers to find \(\mathrm{X}\).

Solve for X in Reaction (e)

Equation for mass numbers: \[ 20 = 20 + A \] Thus, \( A = 0 \).Equation for atomic numbers: \[ 8 = 9 + Z \] Thus, \( Z = -1 \). Therefore, \(\mathrm{X}\) is \( _{-1}^{0} \beta \) (a beta particle, \( \beta^+ \)).

Key concepts

Conservation of Mass Number

In nuclear equations, the conservation of mass number plays a vital role in ensuring that the total number of nucleons (protons and neutrons) remains constant throughout the reaction. This principle states that the sum of mass numbers, which are the superscripts in the element symbol, must be equal on both sides of a nuclear reaction.
When solving nuclear equations, you can think of it like balancing a scale: each side of the reaction must weigh the same in terms of nucleons. For example, if you begin with a mass number of 26 in magnesium and 1 in a proton, the products must collectively have the same mass number total.
This not only helps in identifying missing elements but also ensures the stability and correctness of nuclear reactions.

Conservation of Atomic Number

The conservation of atomic number is another critical principle in nuclear reactions. It ensures that the net charge remains constant from reactants to products, meaning the sum of atomic numbers, which are the subscripts in the element symbol, must be identical on both sides of the equation.
This conservation is pivotal when determining unknown products in nuclear equations. For instance, when a process starts with an atomic number of 12 for magnesium and 1 for a proton, the resulting components' atomic numbers should add up to 13.
This balance not only aids in finding the identity of unknown elements but also maintains the chemical identity throughout the nuclear process. It ensures that the charge and elemental identity are preserved.

Alpha Particle

An alpha particle is a type of nuclear radiation consisting of two protons and two neutrons, just like a helium nucleus but without any electrons. It's symbolically represented as \(_{2}^{4}\mathrm{He}\).
Alpha particles are commonly released in radioactive decay processes where an unstable nucleus expels these helium nuclei to become more stable.
Due to their significant mass and double positive charge, alpha particles have low penetration power and can be stopped by a sheet of paper or skin. However, they can cause significant damage when they interact with living tissues directly, thus necessitating careful handling.
Alpha decay's mathematics involves reducing the atomic number by 2 and the mass number by 4 of the original nucleus, helping us balance the reaction and identify missing products.

Beta Particle

A beta particle is a high-speed electron or positron released from a nucleus during radioactive decay. It is represented typically as \(_{-1}^{0}\beta\).
There are two types of beta decay: beta-minus (\(\beta^-\)) and beta-plus (\(\beta^+\)). In beta-minus decay, a neutron transforms into a proton, and an electron (\(\beta^-\)) is emitted, while in beta-plus decay, a proton becomes a neutron, releasing a positron (\(\beta^+\)).
Beta particles have greater penetration power compared to alpha particles, often passing through paper but stopped by denser materials like aluminum.
In nuclear equations, beta decay plays a significant role in converting one element to another, with changes in the atomic number without altering the mass number, thus enabling complex transmutations in nuclear chemistry.