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. \({ }^{3} \mathrm{H}(\beta)\) b. \({ }_{3}^{8} \mathrm{Li}(\beta\) followed by \(\alpha\) ) c. \({ }_{4}^{7} \mathrm{Be}\) (electron capture) d. \({ }_{5}^{8} \mathrm{~B}\) (positron)

Short answer

a. \({ }^{3} \mathrm{H} \rightarrow { }^{3} \mathrm{He} + e^{-}\) b. \({ }_{3}^{8} \mathrm{Li} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{2}^{4} \mathrm{He} + e^{-}\) c. \(e^{-} + { }_{4}^{7} \mathrm{Be} \rightarrow { }_{3}^{7} \mathrm{Li}\) d. \({ }_{5}^{8} \mathrm{B} \rightarrow { }_{4}^{8} \mathrm{Be} + e^{+}\)

Step-by-step solution

Identify the decay process

Here, we have four different decay processes: a. beta decay for \({ }^{3} \mathrm{H}\) b. beta decay followed by alpha decay for \({ }_{3}^{8} \mathrm{Li}\) c. electron capture for \({ }_{4}^{7} \mathrm{Be}\) d. positron emission for \({ }_{5}^{8} \mathrm{B}\)

Write equation for beta decay

In beta decay, a neutron is converted into a proton, and a beta particle (an electron) is emitted. For decay a: \({ }^{3} \mathrm{H} \rightarrow { }^{3} \mathrm{He} + e^{-}\)

Write equation for beta decay followed by alpha decay

For decay b, we have a two-step process. First, the beta decay occurs, followed by alpha decay. We will write both the equations separately and then combine them: Beta decay: \({ }_{3}^{8} \mathrm{Li} \rightarrow { }_{4}^{8} \mathrm{Be} + e^{-}\) Alpha decay: \({ }_{4}^{8} \mathrm{Be} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{2}^{4} \mathrm{He}\) Combined: \({ }_{3}^{8} \mathrm{Li} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{2}^{4} \mathrm{He} + e^{-}\)

Write equation for electron capture

In electron capture, a proton captures an electron to produce a neutron. For decay c: \(e^{-} + { }_{4}^{7} \mathrm{Be} \rightarrow { }_{3}^{7} \mathrm{Li}\)

Write equation for positron emission

In positron emission, a proton is converted into a neutron, and a positron is emitted. For decay d: \({ }_{5}^{8} \mathrm{B} \rightarrow { }_{4}^{8} \mathrm{Be} + e^{+}\) So, the four decay equations are: a. \({ }^{3} \mathrm{H} \rightarrow { }^{3} \mathrm{He} + e^{-}\) b. \({ }_{3}^{8} \mathrm{Li} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{2}^{4} \mathrm{He} + e^{-}\) c. \(e^{-} + { }_{4}^{7} \mathrm{Be} \rightarrow { }_{3}^{7} \mathrm{Li}\) d. \({ }_{5}^{8} \mathrm{B} \rightarrow { }_{4}^{8} \mathrm{Be} + e^{+}\)

Key concepts

Beta Decay

Beta decay is a common form of radioactive decay where a neutron in an atom's nucleus is transformed into a proton. This process results in the emission of a beta particle, which is essentially an electron. The change increases the atomic number by one, resulting in the formation of a new element. For example, in the decay of tritium ({ }^{3} \text{H}), the equation is:

\( { }^{3} \mathrm{H} \rightarrow { }^{3} \mathrm{He} + e^{-} \)

Here, hydrogen ({ }^{3} \text{H}) decays into helium ({ }^{3} \text{He}), with the release of an electron ({ e^{-} }). Beta decay helps to explain the stability of isotopes and is crucial for understanding nuclear reactions.

• Neutron converts into a proton.
• An electron (beta particle) is released.
• Increases atomic number by 1.

Alpha Decay

Alpha decay involves the emission of an alpha particle from the nucleus of an atom. An alpha particle consists of two protons and two neutrons, equivalent to a helium nucleus. This leads to a decrease in both the atomic number and mass number. In the sequence involving lithium ({ }_{3}^{8} \text{Li}), after a beta decay, it can further decay via alpha decay:

\( { }_{4}^{8} \mathrm{Be} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{2}^{4} \mathrm{He} \)

The transition to helium isotopes highlights the significant loss of mass and energy in alpha decay. This process is typical in heavy nuclei where the atom becomes more stable by emitting an alpha particle.

• Emits an alpha particle (helium nucleus).
• Reduces atomic number by 2 and mass number by 4.
• Common in heavy elements.

Electron Capture

Electron capture is a unique decay process where an inner atomic electron is captured by a proton in the nucleus, transforming it into a neutron. This results in a decrease in the atomic number. In the example with beryllium ({ }_{4}^{7} \text{Be}), the process is described by the equation:

\( e^{-} + { }_{4}^{7} \mathrm{Be} \rightarrow { }_{3}^{7} \mathrm{Li} \)

This transformation is essential for understanding how elements adjust their ratios of protons and neutrons to gain stability. Electron capture is common in proton-rich unstable isotopes.

• An electron is absorbed by the nucleus.
• Proton converts into a neutron.
• Decreases atomic number by 1.

Positron Emission

Positron emission is a decay process that is somewhat the inverse of beta decay. A proton is converted into a neutron, with a positron (the positron is the antimatter counterpart of an electron) being emitted. This decreases the atomic number by one. For boron ({ }_{5}^{8} \text{B}), the process is expressed as:

\( { }_{5}^{8} \mathrm{B} \rightarrow { }_{4}^{8} \mathrm{Be} + e^{+} \)

Positron emission is primarily seen in proton-rich isotopes. It plays a critical role in nuclear medicine, such as in positron emission tomography (PET scans).

• Proton converts into a neutron.
• Emits a positron ({ e^{+} }).
• Decreases atomic number by 1.