Question

Complete the following nuclear equations, and identify \(\mathrm{X}\) in each case: (a) \({ }_{53}^{135} \mathrm{I} \longrightarrow{ }_{54}^{135} \mathrm{Xe}+\mathrm{X}\) (b) \({ }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{-1}^{0} \beta+\mathrm{X}\) (c) \({ }_{27}^{59} \mathrm{Co}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{25}^{56} \mathrm{Mn}+\mathrm{X}\) (d) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{40}^{99} \mathrm{Zr}+{ }_{52}^{135} \mathrm{Te}+2 \mathrm{X}\)

Short answer

(a) \({ }_{-1}^{0} \beta\); (b) \({ }_{18}^{40} \mathrm{Ar}\); (c) \({ }_{2}^{4} \mathrm{He}\); (d) \({ }_{0}^{1} \mathrm{n}\).

Step-by-step solution

Analyze Equation (a)

In the equation \({ }_{53}^{135} \mathrm{I} \longrightarrow { }_{54}^{135} \mathrm{Xe}+\mathrm{X}\), iodine (I) has atomic number 53 and xenon (Xe) has atomic number 54. The atomic number increases by 1. This indicates that a beta particle \(\beta^-\) (an electron, \({ }_{-1}^{0} \beta\)) is emitted. Thus, \(\mathrm{X} = { }_{-1}^{0} \beta\).

Analyze Equation (b)

In the equation \({ }_{19}^{40} \mathrm{K} \longrightarrow { }_{-1}^{0} \beta+\mathrm{X}\), potassium (K) decays to produce a beta particle \({ }_{-1}^{0} \beta\). Potassium's atomic number is 19, decreasing by one gives 18, and its mass number is unchanged. Therefore \(\mathrm{X} = { }_{18}^{40} \mathrm{Ar}\) (argon).

Analyze Equation (c)

The equation \({ }_{27}^{59} \mathrm{Co}+{ }_{0}^{1} \mathrm{n} \longrightarrow { }_{25}^{56} \mathrm{Mn}+\mathrm{X}\) involves cobalt (Co) with an added neutron \({ }_{0}^{1} \mathrm{n}\). The Mn product is three neutrons lighter in mass ( 59 + 1 - 56) and two protons lower in atomic number. Thus, \(\mathrm{X} = { }_{2}^{4} \mathrm{He}\) (an alpha particle).

Analyze Equation (d)

The equation \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow { }_{40}^{99} \mathrm{Zr}+{ }_{52}^{135} \mathrm{Te}+2 \mathrm{X}\) needs the identification of \(\mathrm{X}\). Uranium (U) reacts with a neutron, and products Zr and Te account for a total atomic mass of 234 (99 + 135) and atomic number of 92 (40 + 52). The neutron adds one mass unit, making 236. To balance, \(2\mathrm{X}\) must provide 2 more atomic masses and account for the neutron leakage; each \(\mathrm{X} = { }_{0}^{1} \mathrm{n}\).

Key concepts

Beta Decay

Beta decay is an essential concept in nuclear physics where an unstable nucleus transforms into a more stable one. During beta decay, a neutron is converted into a proton, an electron (beta particle), and an antineutrino. The electron is expelled from the nucleus, which increases the atomic number by one but leaves the mass number unchanged.

• In a typical beta decay equation, the parent atom transitions into a daughter atom plus a beta particle.
• For example, in the equation \({ }_{53}^{135} \mathrm{I} \rightarrow { }_{54}^{135} \mathrm{Xe} + { }_{-1}^{0} \beta\), iodine undergoes beta decay to form xenon.

When solving such nuclear equations, pay attention to the atomic numbers to identify the new element formed. Remember that the mass number remains constant during beta decay because the emitted electron has negligible mass.

Alpha Particle

An alpha particle is equivalent to a helium nucleus, comprising two protons and two neutrons. Its notation is \({ }_{2}^{4} \mathrm{He}\), and it's a common product of nuclear reactions or decay processes.
Alpha decay occurs when a nucleus ejects an alpha particle, reducing both its atomic number and mass number. This leads to the formation of a different, more stable element with a lower atomic and mass number.

• During alpha decay, the atomic number decreases by 2, and the mass number decreases by 4.
• In the equation \({ }_{27}^{59} \mathrm{Co} + { }_{0}^{1} \mathrm{n} \rightarrow { }_{25}^{56} \mathrm{Mn} + { }_{2}^{4} \mathrm{He}\), cobalt, after interacting with a neutron, emits an alpha particle, resulting in manganese.

Alpha particles are positively charged and have considerable mass, thus they have limited penetration power and are often stopped by just a sheet of paper or skin.

Neutron Emission

Neutron emission is a process where a free neutron is ejected from the nucleus. This kind of decay often accompanies nuclear fission and serves as a mechanism to relieve nuclear instability.
Neutrons, being neutral, are not directly influenced by electromagnetic forces, which allows them to interact deeply with other atoms. This quality makes neutron emission significant in chain reactions, such as those seen in nuclear reactors.

• In the fission equation \({ }_{92}^{235} \mathrm{U} + { }_{0}^{1} \mathrm{n} \rightarrow { }_{40}^{99} \mathrm{Zr} + { }_{52}^{135} \mathrm{Te} + 2 \times { }_{0}^{1} \mathrm{n}\), uranium splits to produce zirconium and tellurium with two additional free neutrons.
• The emitted neutrons can further propagate the fission process when absorbed by other uranium nuclei.

Neutron emission plays a crucial role in sustaining nuclear chain reactions by contributing to the sequence of reactions required for energy release.

Radioactive Decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation. There are several types of decay, each involving different particles and leading to changes in the nucleus.
The most common forms of decay include:

• Alpha Decay: Emission of an alpha particle reduces the atomic number by 2 and the mass number by 4.
• Beta Decay: A neutron transfers into a proton, expelling a beta particle, increasing the atomic number by 1.
• Gamma Decay: Emission of electromagnetic gamma radiation, usually without a change in atomic or mass numbers.

In radioactive decay, the instability drives the process as the atom seeks a stable configuration of protons and neutrons. Knowing the type of decay helps predict the identity of the new element formed and the changes in atomic structure.
Understanding radioactive decay is pivotal in fields such as nuclear energy, medical treatments, and radiometric dating.