Question

Write balanced nuclear equations for the bombardment of (a) Fe-54 with an alpha particle to produce another nucleus and two protons. (b) Mo-96 with deuterium \(\left({ }_{1}^{2} \mathrm{H}\right)\) to produce a neutron and another nucleus. (c) Ar-40 with an unknown particle to produce \(\mathrm{K}-43\) and a proton. (d) a nucleus with a neutron to produce a proton and \(\mathrm{P}-31 .\)

Short answer

In summary, the balanced nuclear equations for each scenario are: a) _{26}^{54}Fe + _2^4He → _{26}^{56}Fe + 2(_1^1H) b) _{42}^{96}Mo + _1^2H → _{43}^{97}Tc + _0^1n c) _{18}^{40}Ar + _2^4He → _{19}^{43}K + _1^1H d) _{16}^{31}S + _0^1n → _1^1H + _{15}^{31}P

Step-by-step solution

a) Fe-54 + alpha particle => unknown nucleus + two protons

Write the balanced nuclear equation as follows, ensuring the mass numbers and atomic numbers balance on both sides of the equation: \(_{26}^{54}\mathrm{Fe} + _2^4\mathrm{He} \rightarrow _Z^A\mathrm{X} + 2(_1^1\mathrm{H})\) The mass numbers must add up, so: \(54 + 4 = A + 2(1) \Rightarrow A = 56\) The atomic numbers must add up, so: \(26 + 2 = Z + 2(1) \Rightarrow Z = 26\) Hence, the unknown nucleus is \(_{26}^{56}\mathrm{Fe}\) and we can write the balanced nuclear equation as: \(_{26}^{54}\mathrm{Fe} + _2^4\mathrm{He} \rightarrow _{26}^{56}\mathrm{Fe} + 2(_1^1\mathrm{H})\)

b) Mo-96 + deuterium => unknown nucleus + neutron

Write the balanced nuclear equation as follows, ensuring the mass numbers and atomic numbers balance on both sides of the equation: \(_{42}^{96}\mathrm{Mo} + _1^2\mathrm{H} \rightarrow _Z^A\mathrm{X} + _0^1\mathrm{n}\) The mass numbers must add up, so: \(96 + 2 = A + 1 \Rightarrow A = 97\) The atomic numbers must add up, so: \(42 + 1 = Z + 0 \Rightarrow Z = 43\) Hence, the unknown nucleus is \(_{43}^{97}\mathrm{Tc}\) and we can write the balanced nuclear equation as: \(_{42}^{96}\mathrm{Mo} + _1^2\mathrm{H} \rightarrow _{43}^{97}\mathrm{Tc} + _0^1\mathrm{n}\)

c) Ar-40 + unknown particle => K-43 + proton

Write the balanced nuclear equation as follows, ensuring the mass numbers and atomic numbers balance on both sides of the equation: \(_{18}^{40}\mathrm{Ar} + _Z^A\mathrm{X} \rightarrow _{19}^{43}\mathrm{K} + _1^1\mathrm{H}\) The mass numbers must add up, so: \(40 + A = 43 + 1 \Rightarrow A = 4\) The atomic numbers must add up, so: \(18 + Z = 19 + 1 \Rightarrow Z = 2\) Hence, the unknown particle is \(_2^4\mathrm{He}\) (an alpha particle), and we can write the balanced nuclear equation as: \(_{18}^{40}\mathrm{Ar} + _2^4\mathrm{He} \rightarrow _{19}^{43}\mathrm{K} + _1^1\mathrm{H}\)

d) Unknown nucleus + neutron => proton + P-31

Write the balanced nuclear equation as follows, ensuring the mass numbers and atomic numbers balance on both sides of the equation: \(_Z^A\mathrm{X} + _0^1\mathrm{n} \rightarrow _1^1\mathrm{H} + _{15}^{31}\mathrm{P}\) The mass numbers must add up, so: \(A + 1 = 1 + 31 \Rightarrow A = 31\) The atomic numbers must add up, so: \(Z + 0 = 1 + 15 \Rightarrow Z = 16\) Hence, the unknown nucleus is \(_{16}^{31}\mathrm{S}\), and we can write the balanced nuclear equation as: \(_{16}^{31}\mathrm{S} + _0^1\mathrm{n} \rightarrow _1^1\mathrm{H} + _{15}^{31}\mathrm{P}\)

Key concepts

Nuclear Chemistry

Nuclear chemistry involves the study of the chemical and physical properties of elements as influenced by changes in the structure of the atomic nucleus. This field of chemistry explores various processes such as nuclear transmutation and radioactive decay, which are fundamental to understanding how elements are transformed.

One application of nuclear chemistry is the balancing of nuclear equations, which is critical for the conservation of mass and charge during a nuclear reaction. Balancing nuclear equations ensures that the number of protons and neutrons (which are counted by the atomic number and mass number, respectively) are the same on both sides of the equation. This is essential to understanding how new elements or isotopes are formed and is demonstrated in the exercise where we balance equations for reactions involving Fe-54, Mo-96, Ar-40, and an unknown nucleus.

Radioactive Decay

Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. Three common types of decay include alpha decay, beta decay, and gamma decay. Alpha decay, for instance, is the process by which an unstable nucleus emits an alpha particle, consisting of two protons and two neutrons.

Our exercises touch upon this concept when an alpha particle is emitted or absorbed in a nuclear reaction. For example, when Ar-40 reacts with an unknown particle to produce K-43 and a proton, the balancing process reveals that the unknown particle is actually an alpha particle, showing that alpha radiation was part of the nuclear transformation.

Nuclear Reaction

A nuclear reaction involves a change in an atom's nucleus and usually results in the change of one element into another. During a nuclear reaction, the nuclei of atoms can split apart in a process known as fission, or they can crash together and join in a process called fusion. The nuclear equations detailed in the solution section depict different types of nuclear reactions.

For instance, the bombardment of Fe-54 with an alpha particle to produce another nucleus and two protons is a type of nuclear reaction that is harnessed in various technologies, ranging from energy production in stars to synthetic element creation in laboratories. The exercise that the students worked on involves deducing products and reactants in nuclear reactions to create balanced nuclear reaction equations, which illustrate the principle of mass and charge conservation.