Question

The radioactive isotope \({ }^{100} \mathrm{~T}_{c}\) decays to form the stable isotope \({ }^{100}\) Mo. (a) There are two possible pathways for this decay. Write balanced equations for both. (b) Only one of the pathways is observed. Calculate the energy released by both pathways, and explain why only one is observed. Relevant masses are: \({ }^{100} \mathrm{~T} c(99.907657 \mathrm{u}),{ }^{100} \mathrm{Mo}\) \((99.90748 \mathrm{u}), \mathrm{e}^{-}(0.0005486 \mathrm{u})\)

Short answer

Beta decay ( \[ {}^{100}Tc \to {}^{100}Mo + e^{-} + \bar{\nu} \]) is observed, as it releases more energy than electron capture.

Step-by-step solution

Understanding Radioactive Decay

Radioactive decay involves the transformation of a radioactive isotope into a more stable form. The decay process can follow different pathways, and our task is to determine these pathways for the isotope \({}^{100}Tc\). The most common decay processes for isotopes include beta decay and electron capture.

Writing Balanced Equations for Decay Pathways

For \({}^{100}Tc\), the two possible decay pathways are beta decay and electron capture. 1. **Beta Decay:** In beta decay, a neutron in the nucleus of \({}^{100}Tc\) is transformed into a proton, emitting an electron (\(e^{-}\)) and an antineutrino (\(u\)): \\[ {}^{100}Tc \to {}^{100}Mo + e^{-} + \bar{u} \]2. **Electron Capture:** In electron capture, an inner orbital electron is captured by the nucleus, combining with a proton to form a neutron, releasing a neutrino (\(u\)): \\[ {}^{100}Tc + e^{-} \to {}^{100}Mo + u \].

Calculating Energy Released in Beta Decay

Energy released in beta decay can be calculated using the mass difference between the parent and product isotopes. The energy released, \(Q\), is given by:\[ Q = (m_{Tc} - m_{Mo} - m_{e^{-}}) \times 931.5 \text{ MeV/u} \]Substituting the given masses:\[ Q = (99.907657u - 99.90748u - 0.0005486u) \times 931.5 \text{ MeV/u} = 0.594 ext{ MeV} \].

Calculating Energy Released in Electron Capture

Energy released during electron capture involves only the difference in mass between the parent and product isotopes, as there is no emission of electron mass. Thus, the energy released, \(Q\), is:\[ Q = (m_{Tc} - m_{Mo}) \times 931.5 \text{ MeV/u} \]Substituting the masses:\[ Q = (99.907657u - 99.90748u) \times 931.5 \text{ MeV/u} = 0.165 ext{ MeV} \].

Determining the Observed Pathway

The pathway that is more commonly observed is typically the one that releases more energy. In this case, beta decay releases 0.594 MeV, which is greater than the 0.165 MeV released by electron capture. Therefore, beta decay is the observed pathway for \({}^{100}Tc\) decay.

Key concepts

Beta Decay

Beta decay is one of the most common forms of radioactive decay. It occurs when a neutron in the nucleus of an unstable atom, such as \(^{100}Tc\), transforms into a proton. This transformation is accompanied by the emission of a beta particle, which is essentially an electron, and an antineutrino. The equation representing this process for \(^{100}Tc\) is:\[ {}^{100}Tc \to {}^{100}Mo + e^{-} + \bar{u} \]

• Neutron transforms into a proton
• An electron (\(e^-\)) is emitted
• An antineutrino (\(\bar{u}\)) is also emitted

This process increases the atomic number of the element, transforming \(^{100}Tc\) into the more stable \(^{100}Mo\). Beta decay leads to elements moving up one place in the periodic table while maintaining the same mass number. Beta decay is favored if it results in a more substantial release of energy, making it commonly observed.

Electron Capture

Electron capture is another pathway for radioactive decay. Unlike beta decay, electron capture does not emit a particle like a beta particle. Instead, the nucleus of an atom like \(^{100}Tc\) captures one of its inner orbit electrons, combining it with a proton to form a neutron. This process releases a neutrino:\[ {}^{100}Tc + e^{-} \to {}^{100}Mo + u \]

• Proton combines with an electron to form a neutron
• A neutrino (\(u\)) is emitted

This process reduces the atomic number but keeps the mass number unchanged. Electron capture is generally less common in cases where other pathways, like beta decay, release more energy. In this exercise, electron capture releases less energy, making it less likely to be the observed decay pathway.

Energy Calculation

Energy calculations in decay processes are crucial to understanding which pathway occurs more frequently. The energy released during these decays, known as the Q value, depends on the mass difference between the parent and the daughter isotopes.
For **beta decay**:\[ Q = (m_{Tc} - m_{Mo} - m_{e^{-}}) \times 931.5 \text{ MeV/u} \]Using the given masses, this calculation results in an energy release of 0.594 MeV.
For **electron capture**:\[ Q = (m_{Tc} - m_{Mo}) \times 931.5 \text{ MeV/u} \]The energy release is 0.165 MeV.

• Greater energy release favors the observed pathway
• Beta decay releases more energy than electron capture in this case

Thus, the larger energy release in beta decay makes it the preferred and observed pathway for \(^{100}Tc\) decay.

Isotope Transformation

Isotope transformation refers to the change in the number of protons or neutrons within an atom's nucleus during radioactive decay. This transformation leads to the formation of a different isotope or element.
In **beta decay**, a neutron changing into a proton transforms \(^{100}Tc\) into \(^{100}Mo\), effectively increasing the atomic number by one while keeping the mass number constant.
In **electron capture**, an electron combines with a proton to form a neutron. This transformation reduces the atomic number by one without changing the mass number, also transforming \(^{100}Tc\) into \(^{100}Mo\).

• Changes involve proton to neutron conversion and vice versa
• Mass number remains the same as protons and neutrons are interchanged

During these transformations, the properties of the material, like its stability and type, can change. Understanding isotope transformation is essential for predicting the behavior of radioactive substances and their subsequent effects.