Problem 16 - Chemistry The Molecular Science

Question

Write a balanced nuclear equation for each word statement. (a) Magnesium- 28 undergoes \(\beta\) emission. (b) When uranium- 238 is reacted with carbon-12, four neutrons are emitted and a new element forms. (c) Hydrogen- 2 and helium- 3 react to form helium-4 and another particle. (d) Argon- 38 forms by positron emission. (e) Platinum- 175 forms osmium- 171 by spontaneous radioactive decay.

Step-by-Step Solution

Answer

(a) \( ^{28}_{12}\text{Mg} \rightarrow ^{28}_{13}\text{Al} + \beta^{-} \) (b) \(^{238}_{92}\text{U} + ^{12}_{6}\text{C} \rightarrow ^{246}_{98}\text{Cf} + 4\times ^{1}_{0}\text{n} \) (c) \( ^{2}_{1}\text{H} + ^{3}_{2}\text{He} \rightarrow ^{4}_{2}\text{He} + ^{1}_{1}\text{H} \) (d) \( ^{38}_{19}\text{K} \rightarrow ^{38}_{18}\text{Ar} + \beta^{+} \) (e) \( ^{175}_{78}\text{Pt} \rightarrow ^{171}_{76}\text{Os} + ^{4}_{2}{\text{He}} \).

Step 1: Understand Beta Emission

Magnesium-28 undergoing beta emission means that a neutron in the nucleus is converted into a proton, emitting a beta particle (\( \beta^{-} \), which is an electron) and an antineutrino. The atomic number increases by 1, while the mass number remains the same.

Step 2: Write the Equation for Part (a)

Write the nuclear equation for magnesium-28 (\( ^{28}_{12}\text{Mg} \)) undergoing beta emission. Since beta emission increases the atomic number by 1 without changing the mass, the product is aluminum-28 (\( ^{28}_{13}\text{Al} \)). The equation is: \[ ^{28}_{12}\text{Mg} \rightarrow ^{28}_{13}\text{Al} + \beta^{-} \].

Step 3: Understand Reaction with Carbon and Neutron Emission

For uranium-238 reacting with carbon-12, when four neutrons are emitted, the resulting product is an element with a mass number reduced by the combined mass of the neutrons and carbon.

Step 4: Write the Equation for Part (b)

Uranium-238 (\(^{238}_{92}{\text{U}}\)) reacts with carbon-12 (\(^{12}_{6}{\text{C}}\)). Four neutrons (\(4\times ^{1}_{0}{\text{n}}\)) are emitted, so the element formed will have an atomic number of 98 and a mass number of 246, which is californium-246 (\(^{246}_{98}{\text{Cf}}\)). The equation is: \[ ^{238}_{92}\text{U} + ^{12}_{6}\text{C} \rightarrow ^{246}_{98}\text{Cf} + 4 \times ^{1}_{0}\text{n} \].

Step 5: Understand Fusion Reaction for Hydrogen and Helium

Hydrogen-2 and helium-3 fusing to form helium-4 and another particle needs understanding that the missing mass must be accounted by another produced particle. From typical fusion reactions, the missing particle is often a proton.

Step 6: Write the Equation for Part (c)

The reaction \(^{2}_{1}{\text{H}}\) (deuterium) and \(^{3}_{2}{\text{He}}\) (helium-3) forms \(^{4}_{2}{\text{He}}\) (helium-4) and a proton (\(^{1}_{1}{\text{H}}\)). The equation is: \[ ^{2}_{1}\text{H} + ^{3}_{2}\text{He} \rightarrow ^{4}_{2}\text{He} + ^{1}_{1}\text{H} \].

Step 7: Understand Positron Emission

Positron emission occurs when a proton in the nucleus is converted into a neutron, with the emission of a positron (\( \beta^{+} \)) and a neutrino.

Step 8: Write the Equation for Part (d)

Argon-38 (\(^{38}_{18}\text{Ar} \)) forms by positron emission, so the original element has one more proton and the same mass number, which is potassium-38 (\(^{38}_{19}\text{K} \)). The equation is: \[ ^{38}_{19}\text{K} \rightarrow ^{38}_{18}\text{Ar} + \beta^{+} \].

Step 9: Understand Spontaneous Decay

For spontaneous radioactive decay, recognize that no external particle initiates the decay. The decay results in the production of a different element that has lost some mass or protons.

Step 10: Write the Equation for Part (e)

Platinum-175 (\(^{175}_{78}{\text{Pt}}\)) forms osmium-171 (\(^{171}_{76}{\text{Os}}\)) through alpha decay, where the mass number decreases by 4 and the atomic number by 2, releasing an alpha particle (\(^{4}_{2}{\text{He}}\)). The equation is: \[ ^{175}_{78}\text{Pt} \rightarrow ^{171}_{76}\text{Os} + ^{4}_{2}\text{He} \].

Key Concepts

Beta Emission

Beta emission is a type of radioactive decay where a neutron in an atom's nucleus is transformed. This transformation results in the neutron becoming a proton. Alongside, a beta particle, which is essentially an electron (\( \beta^{-} \)), and an antineutrino are emitted.
This process increases the atomic number by 1 because of the newly formed proton, but the mass number remains unchanged.
For example, in the beta emission of magnesium-28 (\(^{28}_{12}\text{Mg} \)), it becomes aluminum-28 (\(^{28}_{13}\text{Al} \)). Thus, the nuclear equation is:

• \( ^{28}_{12}\text{Mg} \rightarrow ^{28}_{13}\text{Al} + \beta^{-} \)

Understanding beta emission helps us identify changes in atomic structure during this decay process.

Fusian Reactions

Fusion reactions occur when light nuclei combine to form a heavier nucleus. This process releases a significant amount of energy. A common example involves isotopes of hydrogen and helium.
As seen in the fusion of \(^{2}_{1}\text{H} \) (deuterium) and \(^{3}_{2}\text{He} \) (helium-3) to produce \(^{4}_{2}\text{He} \) (helium-4) and a proton (\(^{1}_{1}\text{H} \)).
The balanced equation is:

• \(^{2}_{1}\text{H} + ^{3}_{2}\text{He} \rightarrow ^{4}_{2}\text{He} + ^{1}_{1}\text{H} \)

This equation shows how mass and energy are conserved in fusion, an essential process powering stars, including our sun.

Positron Emission

Positron emission is a decay mechanism where a proton in a nucleus is converted into a neutron. This results in the emission of a positron (\( \beta^{+} \)) and a neutrino.
Unlike electrons, positrons are positively charged particles. This transformation decreases the atomic number by 1, but the mass number stays the same.
As illustrated by the decay of potassium-38 (\(^{38}_{19}\text{K} \)) to argon-38 (\(^{38}_{18}\text{Ar} \)):

• \(^{38}_{19}\text{K} \rightarrow ^{38}_{18}\text{Ar} + \beta^{+} \)

Positron emission is useful in medical imaging techniques such as PET scans due to its capability to trace biological processes.

Spontaneous Radioactive Decay

Spontaneous radioactive decay occurs naturally in certain unstable isotopes without needing an external trigger. This type of decay rearranges the atomic structure, forming a different element often with a lower mass or fewer protons.
An example of spontaneous decay is the alpha decay of platinum-175 (\(^{175}_{78}\text{Pt} \)) into osmium-171 (\(^{171}_{76}\text{Os} \)). It releases an alpha particle (\(^{4}_{2}\text{He} \)):

• \(^{175}_{78}\text{Pt} \rightarrow ^{171}_{76}\text{Os} + ^{4}_{2}\text{He} \)

Spontaneous decay helps scientists understand the long-term stability and evolution of elements found in nature.