Problem 11 - Chemistry An Atoms First Approach

Question

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \(\frac{3}{1} \mathrm{H}(\beta)\) b. \(\frac{8}{3} L i(\beta \text { followed by } \alpha)\) c. \(^{7}_{4}\) Be (electron capture)d. \(^{8}_{5} B\) (positron)

Step-by-Step Solution

Answer

The radioactive decay equations for the given nuclides are: a. \[^{3}_{1}\mathrm{H} \rightarrow ^{3}_{2}\mathrm{He} + ^{0}_{-1}\beta\] b. \[^{8}_{3}\mathrm{Li} \rightarrow ^{8}_{4}\mathrm{Be} + ^{0}_{-1}\beta\] \[^{8}_{4}\mathrm{Be} \rightarrow ^{4}_{2}\mathrm{He} + ^{4}_{2}\alpha\] c. \[^{7}_{4}\mathrm{Be} + ^{0}_{-1}e \rightarrow ^{7}_{3}\mathrm{Li}\] d. \[^{8}_{5}\mathrm{B} \rightarrow ^{8}_{4}\mathrm{Be} + ^{0}_{+1}e\]

Step 1: 1. Identify decay mode for Hydrogen-3

: For Hydrogen-3, it's mentioned that the decay mode produces a β particle (beta decay). In beta decay, a neutron transforms into a proton and an electron (beta particle).

Step 2: 2. Write the radioactive decay equation for Hydrogen-3

: Applying the beta decay, we get the following reaction: \[^{3}_{1}\mathrm{H} \rightarrow ^{3}_{2}\mathrm{He} + ^{0}_{-1}\beta\]

Step 3: 3. Identify decay mode for Lithium-8

: For Lithium-8, it's mentioned that the decay mode is beta decay followed by alpha decay. In alpha decay, the nucleus loses an alpha particle, which consists of 2 protons and 2 neutrons.

Step 4: 4. Write the radioactive decay equation for Lithium-8

: Applying both the beta and alpha decay, we get the following reactions: \[^{8}_{3}\mathrm{Li} \rightarrow ^{8}_{4}\mathrm{Be} + ^{0}_{-1}\beta\] \[^{8}_{4}\mathrm{Be} \rightarrow ^{4}_{2}\mathrm{He} + ^{4}_{2}\alpha\]

Step 5: 5. Identify decay mode for Beryllium-7

: For Beryllium-7, it's mentioned that the decay mode is electron capture. In electron capture, an electron is absorbed by the nucleus, and a proton transforms into a neutron.

Step 6: 6. Write the radioactive decay equation for Beryllium-7

: Applying the electron capture, we get the following reaction: \[^{7}_{4}\mathrm{Be} + ^{0}_{-1}e \rightarrow ^{7}_{3}\mathrm{Li}\]

Step 7: 7. Identify decay mode for Boron-8

: For Boron-8, it's mentioned that the decay mode produces a positron. Positron decay is similar to beta decay but differs in that it involves the emission of a positron (an antiparticle of an electron).

Step 8: 8. Write the radioactive decay equation for Boron-8

: Applying the positron decay, we get the following reaction: \[^{8}_{5}\mathrm{B} \rightarrow ^{8}_{4}\mathrm{Be} + ^{0}_{+1}e\]

Key Concepts

Beta Decay

Beta decay is a common mode of radioactive decay in which a nucleus transforms into a different element by emitting a beta particle, which can be an electron or a positron. Specifically in negative beta decay, as in the case of Hydrogen-3 (\textsuperscript{3}H), a neutron in the nucleus is transformed into a proton while releasing an electron (beta particle) and an antineutrino.

The general equation for negative beta decay is: \[ _{Z}^{A}\text{X} \rightarrow _{Z+1}^{A}\text{Y} + _{-1}^{0}\beta + \bar{u}_e \] Where:\begin{itemize} \item \( _{Z}^{A}\text{X} \) is the parent nucleus \item \( _{Z+1}^{A}\text{Y} \) is the daughter nucleus \item \( _{-1}^{0}\beta \) represents the emitted beta particle (electron) \item \( \bar{u}_e \) is the antineutrino\end{itemize} The conservation of charge and mass/atomic number is crucial in completing the equation for the beta decay process.

Alpha Decay

Alpha decay is another form of radioactive decay where an unstable nucleus emits an alpha particle, which consists of two protons and two neutrons (basically a helium-4 nucleus). This type of decay reduces the mass number of the original atom by four and the atomic number by two, thus leading to a new element.

The general equation for alpha decay is: \[ _{Z}^{A}\text{X} \rightarrow _{Z-2}^{A-4}\text{Y} + _{2}^{4}\text{He} \] Where:\begin{itemize} \item \( _{Z}^{A}\text{X} \) is the parent nucleus \item \( _{Z-2}^{A-4}\text{Y} \) is the daughter nucleus \item \( _{2}^{4}\text{He} \) is the emitted alpha particle\end{itemize} Lithium-8 (\textsuperscript{8}Li), as an example, undergoes beta decay to first form Beryllium-8 (\textsuperscript{8}Be), which then undergoes alpha decay, releasing an alpha particle and forming Helium-4 (\textsuperscript{4}He).

Electron Capture

Electron capture is a process in which an inner orbital electron is captured by the nucleus of its own atom, causing a proton to convert into a neutron and thereby changing the element. This typically results in the emission of an X-ray as the electron cloud reorganizes.

The general equation for electron capture is: \[ _{Z}^{A}\text{X} + _{-1}^{0}e \rightarrow _{Z-1}^{A}\text{Y} + u_e \] Where:\begin{itemize} \item \( _{Z}^{A}\text{X} \) is the parent nucleus \item \( _{-1}^{0}e \) is the captured electron \item \( _{Z-1}^{A}\text{Y} \) is the daughter nucleus \item \( u_e \) is the neutrino\end{itemize} Beryllium-7 (\textsuperscript{7}Be) is an example of a nuclide that undergoes electron capture to transform into Lithium-7 (\textsuperscript{7}Li). It is essential to note that the mass number remains the same during this process, but the atomic number decreases by one.

Positron Emission

Positron emission, also known as positive beta decay, involves an unstable nucleus emitting a positron, the antimatter counterpart of an electron. This process occurs when a proton in the nucleus is converted into a neutron, with the emission of a positron and a neutrino. It tends to happen in proton-rich nuclides.

The general equation for positron emission is: \[ _{Z}^{A}\text{X} \rightarrow _{Z-1}^{A}\text{Y} + _{+1}^{0}e + u_e \] Where:\begin{itemize} \item \( _{Z}^{A}\text{X} \) is the parent nucleus \item \( _{Z-1}^{A}\text{Y} \) is the daughter nucleus \item \( _{+1}^{0}e \) is the emitted positron \item \( u_e \) is the neutrino\end{itemize} An example of a nuclide that undergoes positron emission is Boron-8 (\textsuperscript{8}B), resulting in the formation of Beryllium-8 (\textsuperscript{8}Be). In such transformations, the mass number stays the same, but the atomic number decreases by one unit.