Question

Putting Concepts Together Potassium ion is present in foods and is an essential nutrient in the human body. One of the naturally occurring isotopes of potassium, potassium- 40 , is radioactive. Potassium-40 has a natural abundance of \(0.0117 \%\) and a half- life \(t_{1 / 2}=1.28 \times 10^{9} \mathrm{yr}\). It undergoes radioactive decay in three ways: \(98.2 \%\) is by electron capture, \(1.35 \%\) is by beta emission, and \(0.49 \%\) is by positron emission. (a) Why should we expect \({ }^{40} \mathrm{~K}\) to be radioactive? (b) Write the nuclear equations for the three modes by which \({ }^{40} \mathrm{~K}\) decays. (c) How many \({ }^{40} \mathrm{~K}^{+}\)ions are present in \(1.00 \mathrm{~g}\) of \(\mathrm{KCl}\) ? (d) How long does it take for \(1.00 \%\) of the \({ }^{40} \mathrm{~K}\) in a sample to undergo radioactive decay? SOLUTION (a) The \({ }^{40} \mathrm{~K}\) nucleus contains 19 protons and 21 neutrons. There are very few stable nuclei with odd numbers of both protons and neutrons (Section 21.2). (b) Electron capture is capture of an inner-shell electron by the nucleus: $$ { }_{19}^{40} \mathrm{~K}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{18}^{40} \mathrm{Ar} $$ Beta emission is loss of a beta particle \((-1 \mathrm{e})\) ) by the nucleus: $$ { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{20}^{40} \mathrm{Ca}+{ }_{-1}^{0} \mathrm{e} $$ Positron emission is loss of a positron \(\left(+{ }_{+}^{0} \mathrm{e}\right)\) by the nucleus: $$ { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{18}^{40} \mathrm{Ar}+{ }_{+1}^{0} \mathrm{e} $$ (c) The total number of \(\mathrm{K}^{+}\)ions in the sample is $$ (1.00 \mathrm{~g} \mathrm{KCl})\left(\frac{1 \mathrm{~mol} \mathrm{KCl}}{74.55 \mathrm{~g} \mathrm{KCl}}\right)\left(\frac{1 \mathrm{~mol} \mathrm{~K}}{1 \mathrm{~mol} \mathrm{KCl}}\right)\left(\frac{6.022 \times 10^{23} \mathrm{~K}^{+}}{1 \mathrm{~mol} \mathrm{~K}^{+}}\right)=8.08 \times 10^{21} \mathrm{~K}^{+} \text {ions } $$ Of these, \(0.0117 \%\) are \({ }^{40} \mathrm{~K}^{+}\)ions: $$ \left(8.08 \times 10^{21} \mathrm{~K}^{+} \text {ions }\right)\left(\frac{0.0117^{40} \mathrm{~K}^{+} \text {ions }}{100^{+} \text {ions }}\right)=9.45 \times 10^{17} \text { potassium-40 ions } $$ (d) The decay constant (the rate constant) for the radioactive decay can be calculated from the half-life, using Equation 21.20: $$ k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{1.28 \times 10^{9} \mathrm{yr}}=\left(5.41 \times 10^{-10}\right) / \mathrm{yr} $$ The rate equation, Equation \(21.19\), then allows us to calculate the time required: $$ \begin{aligned} \ln \frac{N_{t}}{N_{0}} &=-k t \\ \ln \frac{99}{100} &=-\left[\left(5.41 \times 10^{-10}\right) / \mathrm{yr}\right] t \\ -0.01005 &=-\left[\left(5.41 \times 10^{-10}\right) / \mathrm{yr}\right] t \\ t &=\frac{-0.01005}{\left(-5.41 \times 10^{-10}\right) / \mathrm{yr}}=1.86 \times 10^{7} \mathrm{yr} \end{aligned} $$ That is, it would take \(18.6\) million years for just \(1.00 \%\) of the \({ }^{40} \mathrm{~K}\) in a sample to decay.

Short answer

Potassium-40 is radioactive due to its odd numbers of both protons and neutrons. The nuclear equations for its decay processes are: Electron capture: \( { }_{19}^{40} \mathrm{~K} + { }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{18}^{40} \mathrm{Ar} \) Beta emission: \( { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{20}^{40} \mathrm{Ca} + { }_{-1}^{0} \mathrm{e} \) Positron emission: \( { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{18}^{40} \mathrm{Ar} + { }_{+1}^{0} \mathrm{e} \) There are \(9.45 \times 10^{17}\) potassium-40 ions in 1g of KCl. It would take 18.6 million years for 1% of the potassium-40 ions to decay.

Step-by-step solution

Analyze the Radioactivity of Potassium-40

From the exercise, we know that potassium-40 contains 19 protons and 21 neutrons. It is radioactive due to its odd numbers of both protons and neutrons.

Write Nuclear Equations for Decay Processes

There are three decay processes for potassium-40: electron capture, beta emission, and positron emission. We are supposed to write the nuclear equations for these processes: Electron capture: \[ { }_{19}^{40} \mathrm{~K} + { }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{18}^{40} \mathrm{Ar} \] Beta emission: \[ { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{20}^{40} \mathrm{Ca} + { }_{-1}^{0} \mathrm{e} \] Positron emission: \[ { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{18}^{40} \mathrm{Ar} + { }_{+1}^{0} \mathrm{e} \]

Calculate the Number of Potassium-40 Ions in 1g of KCl

We can use stoichiometry to find the total number of potassium ions and then find the number of potassium-40 ions (given the natural abundance of 0.0117%): \[ (1.00 \mathrm{~g} \mathrm{KCl})\left(\frac{1 \mathrm{~mol} \mathrm{KCl}}{74.55 \mathrm{~g} \mathrm{KCl}}\right)\left(\frac{1 \mathrm{~mol} \mathrm{~K}}{1 \mathrm{~mol} \mathrm{KCl}}\right)\left(\frac{6.022 \times 10^{23} \mathrm{~K}^{+}}{1 \mathrm{~mol} \mathrm{~K}^{+}}\right)=8.08 \times 10^{21} \mathrm{~K}^{+} \text {ions } \] From this total number, we find the potassium-40 ion count: \[ \left(8.08 \times 10^{21} \mathrm{~K}^{+} \text {ions }\right)\left(\frac{0.0117^{40} \mathrm{~K}^{+} \text {ions }}{100^{+} \text {ions }}\right)=9.45 \times 10^{17} \text { potassium-40 ions } \]

Calculate the Time Required for 1% of Potassium-40 to Decay

We first find the decay constant (k) from the given half-life: \[ k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{1.28 \times 10^{9} \mathrm{yr}}=\left(5.41 \times 10^{-10}\right) / \mathrm{yr} \] Now, we use the radioactive decay equation to find the required time for 1% of potassium-40 to decay: \[ \begin{aligned} \ln \frac{N_{t}}{N_{0}} &=-k t \\\ \ln \frac{99}{100} &=-\left[\left(5.41 \times 10^{-10}\right) / \mathrm{yr}\right] t \\\ -0.01005 &=-\left[\left(5.41 \times 10^{-10}\right) / \mathrm{yr}\right] t \\\ t &=\frac{-0.01005}{\left(-5.41 \times 10^{-10}\right) / \mathrm{yr}}=1.86 \times 10^{7} \mathrm{yr} \end{aligned} \] The time it would take for 1% of the potassium-40 ions to decay is 18.6 million years.

Key concepts

Potassium-40

Potassium-40, or \(^{40}K\), is a naturally occurring isotope of potassium. It's an element that's crucial in our diet due to its role in maintaining fluid and electrolyte balance. Despite its importance, some potassium atoms are radioactive, with potassium-40 being one such isotope.
Radioactive isotopes have unstable nuclei that decay over time. Potassium-40's instability arises from its composition of 19 protons and 21 neutrons. This imbalance in protons and neutrons often results in radioactivity because stable nuclei rarely have odd numbers of both. Potassium-40's decay practices make it unique in isotopic studies.

• It is present in low amounts in potassium sources.
• It naturally decays, providing insight into various geological and archaeological dating techniques.

Understanding radioactive isotopes like potassium-40 helps scientists in fields like geology, where it's used for dating rocks.

Nuclear Equations

Nuclear equations represent the changes happening in an atom's nucleus during radioactive decay. Potassium-40 decays via three processes: electron capture, beta emission, and positron emission. Let's understand these processes better.
**Electron Capture**
This occurs when an inner-shell electron combines with a proton in the nucleus, transforming it into a neutron:
\[^{40}_{19}\text{K} + ^{0}_{-1}\text{e} \rightarrow ^{40}_{18}\text{Ar}\]
**Beta Emission**
This process involves the expulsion of a beta particle (electron) from the nucleus, converting a neutron to a proton:
\[^{40}_{19}\text{K} \rightarrow ^{40}_{20}\text{Ca} + ^{0}_{-1}\text{e}\]
**Positron Emission**
Here, a positron is emitted, which is the antimatter counterpart of an electron, converting a proton to a neutron:
\[^{40}_{19}\text{K} \rightarrow ^{40}_{18}\text{Ar} + ^{0}_{+1}\text{e}\]
Each of these equations shows how the nucleus changes and the new elements formed.

Half-Life

The half-life of an isotope is the time it takes for half of a given sample to decay. For potassium-40, the half-life is 1.28 billion years. This extremely long half-life means potassium-40 is incredibly stable over shorter human timescales.
Understanding half-life is vital for calculations involving radioactive decay. The half-life tells us not only about the stability of the isotope but also helps in measuring the age of objects.

• In forensic science, it helps in dating specimens.
• In geology, it's useful for determining the age of rocks and fossils.

To calculate how long it takes for a percentage of an isotope to decay, scientists use the decay constant derived from the half-life, which enables precise predictions about radioactive decay rates.

Natural Abundance

Natural abundance refers to the relative amount of an isotope present in a natural setting or sample. Potassium-40's natural abundance is just 0.0117%, a small fraction of the total potassium in the environment. Despite this, it contributes significantly to radiogenic heat in the Earth's interior and plays a role in natural radioactivity.
The calculation of natural abundance helps us determine how much of an isotope, like potassium-40, is present in compounds such as potassium chloride (KCl).

• This involves using stoichiometry to find the total number of ions.
• Subsequently, we multiply by the isotope's fractional abundance.

This way, chemists can measure the precise amounts of isotopes for various practical applications, from food analysis to soil sampling.