Problem 15 - Chemistry: The Central Science in SI Units, Global Edition

Question

Decay of which nucleus will lead to the following products: (a) uranium-235 by alpha decay; (b) aluminium-26 by positron emission; \((\mathbf{c})\) deuterium by alpha decay; (d) yttrium-90 by beta decay?

Step-by-Step Solution

Answer

The parent nuclei for each decay process are: a) Plutonium-238 (\(_{94}^{238}Pu\)) for Uranium-235 by alpha decay, b) Magnesium-26 (\(_{12}^{26}Mg\)) for Aluminium-26 by positron emission, c) No possible parent nucleus for deuterium by alpha decay since deuterium cannot be a product of alpha decay, d) Strontium-90 (\(_{38}^{90}Sr\)) for Yttrium-90 by beta decay.

Step 1: a) Uranium-235 by Alpha decay.

In alpha decay, the parent nucleus emits an alpha particle, which is a helium-4 nucleus (2 protons and 2 neutrons). The general decay equation is: \(A_{p}\rightarrow A_{d}+_{2}^{4}He\) Where \(A_{p}\) is the parent nucleus, and \(A_{d}\) is the daughter nucleus. For the given decay: \(_{92}^{238}X\rightarrow _{92}^{235}U +_{2}^{4}He\) Thus, the parent nucleus is Plutonium-238, \(_{94}^{238}Pu\).

Step 2: b) Aluminium-26 by Positron Emission

In positron emission, the parent nucleus decays by converting a proton into a neutron and emitting a positron, which is represented by \(e^{+}\). The general decay equation is: \(A_{p}\rightarrow A_{d} + e^{+}\) For the given decay: \(_{13}^{26}X\rightarrow _{12}^{26}Al +_{1}^{0}e^{+}\) Thus, the parent nucleus is Magnesium-26, \(_{12}^{26}Mg\).

Step 3: c) Deuterium by Alpha decay

In an alpha decay, the parent nucleus emits an alpha particle, which is a helium-4 nucleus (2 protons and 2 neutrons). The general decay equation is: \(A_{p}\rightarrow A_{d} + _{2}^{4}He\) For the given decay: \(_{z}^{A}X\rightarrow _{1}^{2}H + _{2}^{4}He\) In this case, we see that deuterium has been mentioned as a product instead of a parent nucleus. Deuterium cannot be a product of an alpha decay since the decaying nucleus must have larger mass and atomic numbers than the product nucleus. Therefore, there is no possible parent nucleus for this decay process.

Step 4: d) Yttrium-90 by Beta decay

In a beta decay, the parent nucleus decays by converting a neutron into a proton and emitting a beta particle, which is an electron, represented by \(e^{-}\). The general decay equation is: \(A_{p}\rightarrow A_{d} + e^{-}\) For the given decay: \(_{38}^{90}X\rightarrow _{39}^{90}Y + _{-1}^{0}e^{-}\) Thus, the parent nucleus is Strontium-90, \(_{38}^{90}Sr\).

Key Concepts

Alpha Decay

Alpha decay is a type of nuclear decay where an unstable nucleus emits an alpha particle, consisting of 2 protons and 2 neutrons, similar to a helium-4 nucleus (\(_{2}^{4}He\)). This process reduces the atomic number of the original atom by 2 and its mass number by 4. Consequently, the parent nucleus transforms into a different element known as the daughter nucleus.

• The emitted alpha particle carries away two protons and two neutrons.
• This results in a reduction of the mass and atomic numbers of the originating nucleus.

For instance, in the textbook exercise, uranium-235 undergoes alpha decay. Here, we learned that the parent nucleus was plutonium-238 (\(_{94}^{238}Pu\)). Thus, the reaction can be illustrated as follows:\[ _{94}^{238}Pu \rightarrow _{92}^{235}U + _{2}^{4}He \]In alpha decay scenarios, it is crucial to balance both mass numbers and atomic numbers. This ensures that the overall reaction observes the conservation of mass and charge.

Positron Emission

Positron emission is a fascinating process where a proton within a nucleus is transformed into a neutron. During this transformation, a positron is emitted. A positron is the antimatter counterpart of an electron, symbolized by \(e^{+}\).
In positron emission, the atomic number of the parent nucleus decreases by one while the mass number remains constant.

• The emission results in a decrease in the atomic number.
• The emitted positron carries away positive charge, affecting charge balance.

The textbook exercise example illustrates positron emission through the decay of magnesium-26 (\(_{12}^{26}Mg\)) into aluminium-26 (\(_{13}^{26}Al\)). This can be written as:\[ _{13}^{26}Mg \rightarrow _{12}^{26}Al + _{1}^{0}e^{+} \]Positron emission is employed in medical applications, like positron emission tomography (PET scans), which utilize these decay properties.

Beta Decay

In beta decay, an unstable nucleus undergoes a transformation where a neutron is converted into a proton, with the simultaneous emission of a beta particle, which is essentially an electron denoted by \(e^{-}\). This process increases the atomic number of the element by one, while its mass number remains unchanged.

• A neutron becomes a proton, increasing the atomic number.
• The emitted electron helps preserve charge balance in the reaction.

An instance of beta decay from the textbook exercise is the transition from strontium-90 to yttrium-90. This is shown in the reaction:\[ _{38}^{90}Sr \rightarrow _{39}^{90}Y + _{-1}^{0}e^{-} \]Understanding beta decay is essential in fields such as nuclear physics and geochronology. Beta decay helps scientists date ancient organic materials by observing the ratios of carbon isotopes.