Problem 20 - Chemistry

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. \(^{241} \mathrm{Cm}+? \rightarrow^{241} \mathrm{Am}\)

Step-by-Step Solution

Answer

a. In the given decay process, the missing particle is a beta-minus particle (an electron and an electron antineutrino): \(^{0}_{-1} \mathrm{e}^- + \bar{\nu}_e\). b. In the given decay process, the missing particle is an alpha particle: \(^{4}_{2} \mathrm{He}\). c. In the given decay process, the missing particle is a beta-plus particle (a positron and an electron neutrino): \(^{0}_{1} \mathrm{e}^+ + \nu_e\). d. In the given decay process, the missing particle is an electron in electron capture: \(^{0}_{-1} \mathrm{e}^- + \nu_e\).

Step 1: Identify the elements and their atomic numbers

Ge (Germanium) has an atomic number of 32, and Ga (Gallium) has an atomic number of 31. We can now write the decay process with the atomic numbers included: \[^{73}_{31} \mathrm{Ga} \rightarrow^{73}_{32} \mathrm{Ge} + ?\]

Step 2: Determine the type of decay and missing particle

Since the atomic number has increased by 1, it is a beta-minus decay (\(\beta^-\)), and the missing particle is an electron (e\(^-\)) and an electron antineutrino (\(\bar{\nu}_e\)). So the reaction is: \[^{73}_{31} \mathrm{Ga} \rightarrow^{73}_{32} \mathrm{Ge} + ^{0}_{-1} \mathrm{e}^- + \bar{\nu}_e\] b. \(^{192} \mathrm{Pt} \rightarrow^{188} \mathrm{Os}+?\)

Step 1: Identify the elements and their atomic numbers

Pt (Platinum) has an atomic number of 78, and Os (Osmium) has an atomic number of 76. We can now write the decay process with the atomic numbers included: \[^{192}_{78} \mathrm{Pt} \rightarrow^{188}_{76} \mathrm{Os} + ?\]

Step 2: Determine the type of decay and missing particle

Since the atomic number has decreased by 2, it is an alpha decay, and the missing particle is an alpha particle, which consists of 2 protons and 2 neutrons. So the reaction is: \[^{192}_{78} \mathrm{Pt} \rightarrow^{188}_{76} \mathrm{Os} + ^{4}_{2} \mathrm{He}\] c. \(^{205} \mathrm{Bi} \rightarrow^{205} \mathrm{Pb}+?\)

Step 1: Identify the elements and their atomic numbers

Bi (Bismuth) has an atomic number of 83, and Pb (Lead) has an atomic number of 82. We can now write the decay process with the atomic numbers included: \[^{205}_{83} \mathrm{Bi} \rightarrow^{205}_{82} \mathrm{Pb} + ?\]

Step 2: Determine the type of decay and missing particle

Since the atomic number has decreased by 1, it is a beta-plus decay (\(\beta^+\)), and the missing particle is a positron (e\(^+\)) and an electron neutrino (\(\nu_e\)). So the reaction is: \[^{205}_{83} \mathrm{Bi} \rightarrow^{205}_{82} \mathrm{Pb}+^{0}_{1} \mathrm{e}^+ + \nu_e\] d. \(^{241} \mathrm{Cm}+? \rightarrow^{241} \mathrm{Am}\)

Step 1: Identify the elements and their atomic numbers

Cm (Curium) has an atomic number of 96, and Am (Americium) has an atomic number of 95. We can now write the decay process with the atomic numbers included: \[^{241}_{96} \mathrm{Cm} + ?\rightarrow^{241}_{95} \mathrm{Am}\]

Step 2: Determine the type of decay and missing particle

Since the atomic number has decreased by 1, it is a capture process. The missing particle will be an electron from an inner shell of the atom. This is called electron capture (EC). So the reaction is: \[^{241}_{96} \mathrm{Cm} + ^{0}_{-1} \mathrm{e}^- \rightarrow^{241}_{95} \mathrm{Am} + \nu_e\]

Key Concepts

Beta-minus Decay

In beta-minus decay, a neutron in the nucleus of an atom is transformed into a proton, an electron, and an electron antineutrino. This process is typical in neutron-rich unstable isotopes. As a result, the atomic number increases by one, while the mass number remains unchanged. The electron and electron antineutrino are emitted as radiation.

A simple representation of this decay process is:

• A neutron (\(n ightarrow p + e^- + \bar{u}_e\)) converts to a proton.
• The emitted electron is often referred to as a beta particle (\(e^-\)).
• An antineutrino (\(\bar{u}_e \)) is released following the conversion.

The example given in the exercise where Gallium-73 transforms to Germanium-73 illustrates beta-minus decay:\[^{73}_{31} \mathrm{Ga} \rightarrow ^{73}_{32} \mathrm{Ge} + ^{0}_{-1} \mathrm{e}^- + \bar{u}_e\] Here, the increase in the atomic number from 31 to 32 confirms the occurrence of beta-minus decay.

Alpha Decay

Alpha decay is a type of radioactive decay where an unstable nucleus releases an alpha particle. An alpha particle comprises two protons and two neutrons, making it identical to a helium-4 nucleus. This decay process is common in heavy elements striving to achieve a more stable state.

Key features of alpha decay include:

• The mass number of the parent nucleus decreases by four.
• The atomic number decreases by two.
• The emitted alpha particle can be written as \( ^4_2 \mathrm{He} \) or simply \( \alpha \).

In the exercise, Platinum-192 undergoes alpha decay to produce Osmium-188 and an alpha particle:\[^{192}_{78} \mathrm{Pt} \rightarrow ^{188}_{76} \mathrm{Os} + ^{4}_{2} \mathrm{He}\]This type of decay helps in decreasing the size of the nucleus, making it more stable by reducing the proton-neutron imbalance.

Beta-plus Decay

Beta-plus decay is a radioactive process where a proton is converted into a neutron, releasing a positron and an electron neutrino. This type of decay generally happens in proton-rich unstable isotopes. Beta-plus decay decreases the atomic number by one but leaves the mass number unchanged.

During beta-plus decay:

• A proton (\(p \rightarrow n + e^+ + u_e\)) changes into a neutron.
• The emitted positron is analogous to an electron but with a positive charge, denoted as \(e^+\).
• An electron neutrino (\(u_e\)) is also released.

As evidenced in the exercise with Bismuth-205, it undergoes beta-plus decay to form Lead-205:\[^{205}_{83} \mathrm{Bi} \rightarrow ^{205}_{82} \mathrm{Pb} + ^{0}_{1} \mathrm{e}^+ + u_e\]The decrease in the atomic number from 83 to 82 indicates the occurrence of beta-plus decay, stabilizing the nucleus by converting a proton into a neutron.

Electron Capture

Electron capture is a process where an inner orbital electron is captured by the nucleus, inducing a proton to convert into a neutron. This process is another method for proton-rich unstable isotopes to achieve a balance. Electron capture is typically characterized by a decrease in the atomic number by one while the mass number remains unchanged.

Key points about electron capture include:

• An inner electron (\(e^-\)) is absorbed by the nucleus.
• A proton (\(p + e^- \rightarrow n\)) changes into a neutron.
• An electron neutrino (\(u_e\)) is emitted as a byproduct.

In the provided exercise, Curium-241 undergoes electron capture to transform into Americium-241:\[^{241}_{96} \mathrm{Cm} + ^{0}_{-1} \mathrm{e}^- \rightarrow ^{241}_{95} \mathrm{Am} + u_e\]This process highlights how capturing an electron helps the atom to reduce the atomic number, achieving stability without changing the mass number.