Question

Each of the following nuclides is known to undergo radioactive decay by production of a \(\beta\) particle, \({ }_{-1}^{0} \mathrm{e}\). Write a balanced nuclear equation for each process. a. \({ }_{74}^{188} \mathrm{~W}\) b. \({ }_{19}^{40} \mathrm{~K}\) c. \({ }^{198} \mathrm{Au}\)

Short answer

The balanced nuclear equations for the provided nuclides undergoing beta decay are: a. \( {}_{74}^{188}\mathrm{W} \longrightarrow {}_{75}^{188}\mathrm{Re} + {}_{-1}^{0}\mathrm{e}\) b. \( {}_{19}^{40}\mathrm{K} \longrightarrow {}_{20}^{40}\mathrm{Ca} + {}_{-1}^{0}\mathrm{e}\) c. \( {}_{79}^{198}\mathrm{Au} \longrightarrow {}_{80}^{198}\mathrm{Hg} + {}_{-1}^{0}\mathrm{e}\)

Step-by-step solution

a. Balancing the nuclear equation for W

In the beta decay of the tungsten isotope (\( {}_{74}^{188}\mathrm{W} \)), a neutron changes into a proton and emits a beta particle (\({}_{-1}^{0}\mathrm{e}\)): $${}_{74}^{188}\mathrm{W} \longrightarrow {}_{75}^{A}\mathrm{X} + {}_{-1}^{0}\mathrm{e}$$ Here, we need to find A and X. Since the mass number is conserved, we have \(A = 188\). As a neutron is turned into a proton and the atomic number represents the number of protons, we have \(74 + 1 = 75\). So, the element with atomic number 75 is element Re (Rhenium). The balanced nuclear equation is: $${}_{74}^{188}\mathrm{W} \longrightarrow {}_{75}^{188}\mathrm{Re} + {}_{-1}^{0}\mathrm{e}$$

b. Balancing the nuclear equation for K

In the beta decay of the potassium isotope (\( {}_{19}^{40}\mathrm{K} \)), a neutron changes into a proton and emits a beta particle (\({}_{-1}^{0}\mathrm{e}\)): $${}_{19}^{40}\mathrm{K} \longrightarrow {}_{20}^{A}\mathrm{X} + {}_{-1}^{0}\mathrm{e}$$ We need to find A and X. Since the mass number is conserved, we have \(A = 40\). As a neutron is turned into a proton and the atomic number represents the number of protons, we have \(19 + 1 = 20\). So, the element with atomic number 20 is element Ca (Calcium). The balanced nuclear equation is: $${}_{19}^{40}\mathrm{K} \longrightarrow {}_{20}^{40}\mathrm{Ca} + {}_{-1}^{0}\mathrm{e}$$

c. Balancing the nuclear equation for Au

In the beta decay of the gold isotope (\( {}^{198}\mathrm{Au} \)), a neutron changes into a proton and emits a beta particle (\({}_{-1}^{0}\mathrm{e}\)): $$\,_{79}^{198}\mathrm{Au} \longrightarrow {}_{80}^{A}\mathrm{X} + {}_{-1}^{0}\mathrm{e}$$ We need to find A and X. Since the mass number is conserved, we have \(A = 198\). As a neutron is turned into a proton and the atomic number represents the number of protons, we have \(79 + 1 = 80\). So, the element with atomic number 80 is element Hg (Mercury). The balanced nuclear equation is: $$\,_{79}^{198}\mathrm{Au} \longrightarrow {}_{80}^{198}\mathrm{Hg} + {}_{-1}^{0}\mathrm{e}$$

Key concepts

Radioactive Decay

Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. This emission of energy results in the transformation of the original unstable element into a more stable one. Understanding radioactive decay is crucial in nuclear chemistry as it explains how elements change over time.
For students, it's important to note that radioactive decay can occur in different forms, such as alpha, beta, or gamma decay. In each type, different particles or energy forms are emitted from the decay of the nucleus.

• Alpha decay involves the emission of alpha particles, which are composed of 2 protons and 2 neutrons.
• Beta decay occurs when a neutron in the nucleus transforms into a proton (as discussed in the exercise), emitting a beta particle.
• Gamma decay involves the release of energy in the form of gamma rays.

These processes help isotopes move towards stability by reaching a lower energy state, which often involves a change in the number of protons, neutrons, or energy levels within the nucleus.

Beta Decay

Beta decay is a fascinating form of radioactive decay where a neutron in the atomic nucleus transforms into a proton, accompanied by the emission of a beta particle, denoted as \({}_{-1}^{0} \mathrm{e}\). This particle is essentially an electron or its antimatter counterpart, the positron.
In beta decay, the number of protons in the nucleus increases by one while the number of neutrons decreases by one. This transformation is crucial because:

• It changes the identity of the element, as the atomic number (which defines an element) increases.
• The mass number (total protons + neutrons) remains unchanged since a neutron is merely converting into a proton.
• Beta decay helps unstable isotopes become more stable and closer to the ideal neutron-to-proton ratio.

Understanding beta decay helps explain many natural radioactive processes and nuclear reactions, including those used in medical imaging and carbon dating.

Nuclear Equations

Nuclear equations are a simplified way to represent the nuclear reactions and changes occurring within an atom's nucleus. Writing these equations involves ensuring the conservation of both mass and atomic numbers.
When working with nuclear equations, it's essential to remember:

• The sum of the mass numbers (protons + neutrons) on both sides of the equation should be equal.
• Similarly, the total atomic number (protons) should remain balanced on both sides.
• New elements are often produced in the process, as is seen in beta decay where the atomic number increases.

In nuclear equations, symbols are used to represent different elements and particles. These equations help scientists and students predict product isotopes formed after nuclear reactions, like the transformation of tungsten (\({}_{74}^{188}\mathrm{W}\)) to rhenium (\({}_{75}^{188}\mathrm{Re}\)) or potassium (\({}_{19}^{40}\mathrm{K}\)) to calcium (\({}_{20}^{40}\mathrm{Ca}\)). Learning to interpret and write nuclear equations is fundamental for anyone studying nuclear chemistry.