Question

In each of the following radioactive decay processes, supply the missing particle. a. \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+\) ? b. \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+?\) c. \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+?\) d. \({ }^{241} \mathrm{Cm}+? \rightarrow{ }^{241} \mathrm{Am}\)

Short answer

a. \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+ e⁻\) b. \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+ \alpha\) c. \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+ e⁺\) d. \({ }^{241} \mathrm{Cm}+ n \rightarrow{ }^{241} \mathrm{Am}\)

Step-by-step solution

a. Identify the missing particle in \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+\) ?#

We are given the decay process: \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+\) ?. To find the missing particle, we have to ensure the atomic and mass numbers are conserved in the process. Since both Ga and Ge have the same mass number (73), the missing particle must have a mass number of 0. The atomic number of Ga is 31 and the atomic number of Ge is 32. Thus, the missing particle must have an atomic number of -1 to conserve atomic numbers. This particle, with a mass number of 0 and an atomic number of -1, is an electron which is emitted in beta-minus decay (β⁻). So, the missing particle is an electron (e⁻). The decay process is: \({ }^{73} \mathrm{Ga} \rightarrow{ }^{73} \mathrm{Ge}+ e⁻\).

b. Identify the missing particle in \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+?\)#

We are given the decay process: \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+?\). First, let's find the difference in the mass number (∆A) and the atomic number (∆Z) to ensure conservation: ∆A = 192 - 188 = 4 ∆Z = 78 (Pt) - 76 (Os) = 2 The missing particle must have a mass number of 4 and an atomic number of 2 to conserve the mass and atomic numbers. This particle, with a mass number of 4 and an atomic number of 2, is an alpha particle (α). So, the missing particle is an alpha particle. The decay process is: \(^{192} \mathrm{Pt} \rightarrow{ }^{188} \mathrm{Os}+ \alpha\).

c. Identify the missing particle in \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+?\)#

We are given the decay process: \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+?\). Both Bi and Pb have the same mass number (205), so the missing particle must have a mass number of 0. The atomic number of Bi is 83 and the atomic number of Pb is 82. Thus, the missing particle must have an atomic number of 1 to conserve the atomic numbers. This particle, with a mass number of 0 and an atomic number of 1, is a positron which is emitted in beta-plus decay (β⁺). So, the missing particle is a positron (e⁺). The decay process is: \({ }^{205} \mathrm{Bi} \rightarrow{ }^{205} \mathrm{~Pb}+ e⁺\).

d. Identify the missing particle in \({ }^{241} \mathrm{Cm}+? \rightarrow{ }^{241} \mathrm{Am}\)#

We are given the decay process: \({ }^{241} \mathrm{Cm}+? \rightarrow{ }^{241} \mathrm{Am}\). Both Cm and Am have the same mass number (241), so the missing particle must have a mass number of 0. The atomic number of Cm is 96 and the atomic number of Am is 95. Thus, the missing particle must have an atomic number of 1 to conserve atomic numbers. This particle, with a mass number of 0 and an atomic number of 1, is a neutron which is absorbed in neutron capture. So, the missing particle is a neutron (n). The decay process is: \({ }^{241} \mathrm{Cm}+ n \rightarrow{ }^{241} \mathrm{Am}\).

Key concepts

Beta-minus decay

Beta-minus decay is a type of radioactive decay where a neutron in an atomic nucleus is transformed into a proton. One key point of this process is the emission of an electron and an antineutrino.
This emitted electron is known as a beta particle, and it's denoted by the symbol \( e^- \).
The result of beta-minus decay is the formation of a new element with an atomic number that is increased by one, while the mass number remains the same.
For instance, in the given exercise, gallium (\( ^{73} \text{Ga} \)) decays into germanium (\( ^{73} \text{Ge} \)), through the emission of an electron: \[ ^{73} \text{Ga} \rightarrow ^{73} \text{Ge} + e^- \] Key features of beta-minus decay include:

• Conservation of mass number: The sum of mass numbers remains the same before and after the decay.
• Increase in atomic number: The new element has an atomic number increased by one, gaining a positive charge due to the conversion of a neutron to a proton.
• Production of an electron: The emitted electron is a beta particle with negligible mass and negative charge.

Alpha decay

Alpha decay is another form of radioactive decay, characterized by the emission of an alpha particle from the nucleus. An alpha particle consists of 2 protons and 2 neutrons, essentially making it a helium nucleus, denoted by \( \alpha \).
In alpha decay, both the atomic and mass numbers change in the resulting nucleus.
For the decay of platinum (\( ^{192} \text{Pt} \)) into osmium (\( ^{188} \text{Os} \)), the process involves the emission of an alpha particle:\[ ^{192} \text{Pt} \rightarrow ^{188} \text{Os} + \alpha \] Major characteristics of alpha decay are:

• Decrease in mass number: The mass number decreases by 4 due to the loss of 2 protons and 2 neutrons with the alpha particle.
• Decrease in atomic number: The original element becomes a new element with an atomic number decreased by 2.
• Emission of an alpha particle: The alpha particle is relatively heavy, compared to beta particles, and carries a positive charge.

Beta-plus decay

Beta-plus decay is a form of radioactive decay that involves the conversion of a proton into a neutron within a nucleus, resulting in the emission of a positron and a neutrino. This emitted positron, the antimatter counterpart to an electron, is denoted as \( e^+ \).
Consequently, the atomic number decreases by one, while the mass number remains unchanged.
For example, bismuth (\( ^{205} \text{Bi} \)) decays into lead (\( ^{205} \text{Pb} \)) by emitting a positron:\[ ^{205} \text{Bi} \rightarrow ^{205} \text{Pb} + e^+ \] Some key points about beta-plus decay include:

• Conservation of mass number: The mass number before and after the decay remains constant.
• Decrease in atomic number: The new element formed has an atomic number decreased by one due to the conversion of a proton into a neutron.
• Emission of a positron: The positron is a particle with a positive charge and negligible mass, representing antimatter.

Neutron capture

Neutron capture is a nuclear reaction in which an atomic nucleus captures or absorbs a free neutron. This process alters the composition of the nucleus, typically increasing its atomic mass number by one, while the atomic number remains the same.
This is a key difference from decay processes, such as beta or alpha decay, which change both atomic numbers and sometimes mass numbers.In the given exercise, curium (\( ^{241} \text{Cm} \)) captures a neutron to form americium (\( ^{241} \text{Am} \)):\[ ^{241} \text{Cm} + n \rightarrow ^{241} \text{Am} \] Important aspects of neutron capture include:

• Increase in mass number: The mass number increases by one due to the addition of a neutron.
• Atomic number remains the same: As no protons are involved in the reaction, the atomic number is unchanged.
• Building heavier isotopes: Neutron capture often leads to the production of heavier isotopes and can even lead to subsequent radioactive decay.