Question

Write balanced nuclear equations for the following processes: (a) radon-198 undergoes alpha emission; (b) thorium-234 undergoes beta emission; (c) copper-61 undergoes positron emission; (d) silver-106 undergoes electron capture.

Short answer

(a) \(^{198}\textrm{Rn} \rightarrow ^{194}\textrm{Po} + ^4_2\textrm{He}\) (b) \(^{234}\textrm{Th} \rightarrow ^{234}\textrm{Pa} + ^0_{-1}\textrm{e}\) (c) \(^{61}\textrm{Cu} \rightarrow ^{61}\textrm{Ni} + ^0_{+1}\textrm{e}\) (d) \(^{106}\textrm{Ag} + ^0_{-1}\textrm{e} \rightarrow ^{106}\textrm{Pd}\)

Step-by-step solution

(a) Radon-198 undergoes alpha emission

In alpha emission, an alpha particle (a helium nucleus) is emitted. An alpha particle has a mass number of 4 and an atomic number of 2. We are given that the initial nucleus is Radon-198 (\(^{198}\textrm{Rn}\)), which has a mass number of 198 and an atomic number of 86. To conserve the mass number and atomic number in the equation, we subtract the values of the alpha particle from the initial nucleus: Mass number: 198 - 4 = 194 Atomic number: 86 - 2 = 84 This product nucleus has mass number 194 and atomic number 84. The element with an atomic number of 84 is Polonium (Po). Therefore, the balanced nuclear equation for this process can be written as: \[^{198}\textrm{Rn} \rightarrow ^{194}\textrm{Po} + ^4_2\textrm{He}\]

(b) Thorium-234 undergoes beta emission

In beta emission, a beta particle (an electron) is emitted. A beta particle has a mass number of 0 and an atomic number of -1. We are given that the initial nucleus is Thorium-234 (\(^{234}\textrm{Th}\)), which has a mass number of 234 and an atomic number of 90. Mass number remains the same after the beta emission, and the atomic number increases by 1: Mass number: 234 Atomic number: 90 + 1 = 91 This product nucleus has mass number 234 and atomic number 91. The element with an atomic number of 91 is Protactinium (Pa). Therefore, the balanced nuclear equation for this process can be written as: \[^{234}\textrm{Th} \rightarrow ^{234}\textrm{Pa} + ^0_{-1}\textrm{e}\]

(c) Copper-61 undergoes positron emission

In positron emission, a positron (an anti-electron) is emitted. A positron has a mass number of 0 and an atomic number of +1. We are given that the initial nucleus is Copper-61 (\(^{61}\textrm{Cu}\)), which has a mass number of 61 and an atomic number of 29. Mass number remains the same after the positron emission, and the atomic number decreases by 1: Mass number: 61 Atomic number: 29 - 1 = 28 This product nucleus has mass number 61 and atomic number 28. The element with an atomic number of 28 is Nickel (Ni). Therefore, the balanced nuclear equation for this process can be written as: \[^{61}\textrm{Cu} \rightarrow ^{61}\textrm{Ni} + ^0_{+1}\textrm{e}\]

(d) Silver-106 undergoes electron capture

In electron capture, an electron from an atom's inner shell is captured by the nucleus. The captured electron has a mass number of 0 and an atomic number of -1. We are given that the initial nucleus is Silver-106 (\(^{106}\textrm{Ag}\)), which has a mass number of 106 and an atomic number of 47. Mass number remains the same after electron capture, and the atomic number decreases by 1: Mass number: 106 Atomic number: 47 - 1 = 46 This product nucleus has mass number 106 and atomic number 46. The element with an atomic number of 46 is Palladium (Pd). Therefore, the balanced nuclear equation for this process can be written as: \[^{106}\textrm{Ag} + ^0_{-1}\textrm{e} \rightarrow ^{106}\textrm{Pd}\]

Key concepts

Alpha Emission

Alpha emission is a type of radioactive decay where an unstable nucleus releases an alpha particle. An alpha particle consists of two protons and two neutrons—the same as a helium nucleus. This emission results in the reduction of the atomic mass and the atomic number of the original atom.

• The mass number decreases by 4 units because two protons and two neutrons are removed.
• The atomic number decreases by 2 units, shifting the element two places back on the periodic table.

In the exercise, when radon-198 undergoes alpha emission, it produces polonium-194:
The equation is written as:\[^{198}\text{Rn} \rightarrow ^{194}\text{Po} + ^4_2\text{He}\]This shows how radon becomes a new element, polonium, after emitting an alpha particle.

Beta Emission

Beta emission occurs when a neutron in the nucleus is transformed into a proton, emitting a beta particle, which is essentially an electron. As a result, the nucleus gains one proton but loses a neutron, leading to:

• The mass number staying the same since the overall nucleon count doesn't change.
• The atomic number increases by 1, moving the element one place forward on the periodic table.

In the problem, thorium-234 undergoes beta emission to form protactinium-234:
The equation can be expressed as:\[^{234}\text{Th} \rightarrow ^{234}\text{Pa} + ^0_{-1}\text{e}\]This transformation shows how a neutron's conversion into a proton turns thorium into protactinium.

Positron Emission

Positron emission involves a proton in the nucleus transforming into a neutron. During this process, a positron is emitted—a particle with the mass of an electron but with a positive charge:

• The mass number remains unchanged as the overall count of particles in the nucleus doesn't vary.
• The atomic number decreases by 1, moving the element one place back on the periodic table.

In the example, copper-61 releases a positron to form nickel-61:
This is represented as:\[^{61}\text{Cu} \rightarrow ^{61}\text{Ni} + ^0_{+1}\text{e}\]Thus, copper becomes nickel following the emission of a positron.

Electron Capture

Electron capture is a process where an inner orbital electron is captured by the nucleus. This electron combines with a proton to form a neutron, reducing the proton count:

• The mass number remains unchanged since no nucleons are added or removed.
• The atomic number decreases by 1 because the number of protons decreases.

In the given exercise, silver-106 captures an electron and transforms into palladium-106:
This nuclear reaction is illustrated as:\[^{106}\text{Ag} + ^0_{-1}\text{e} \rightarrow ^{106}\text{Pd}\]So, silver loses a proton and becomes palladium after capturing an electron.