Question

Cesium-137, a beta emitter, has a half-life of \(30 \mathrm{y}\). a. Write the balanced nuclear equation for the beta decay of cesium-137. b. How many milligrams of a \(16-\mathrm{mg}\) sample of cesium- 137 would remain after \(90 \mathrm{y}\) ? c. How many years are required for \(28 \mathrm{mg}\) of cesium- 137 to decay to \(3.5 \mathrm{mg}\) of cesium- \(137 ?\)

Short answer

a. \[ ^{137}_{55}Cs \rightarrow ^{137}_{56}Ba + \beta^- \ \] b. 2 mg c. 90 years

Step-by-step solution

Understanding Beta Decay

Beta decay occurs when a neutron in the nucleus of an atom is converted into a proton and an electron (beta particle). The electron is emitted from the nucleus, and the atomic number of the element increases by one.

Write the Nuclear Equation

For cesium-137, the initial isotope is \textsuperscript{137}_{55}Cs. When it undergoes beta decay, it transforms into \textsuperscript{137}_{56}Ba. The balanced nuclear equation is: \[ ^{137}_{55}Cs \rightarrow ^{137}_{56}Ba + \beta^- \ \]

Calculate the Remaining Amount of Cesium-137

Use the half-life formula: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \] where: * \(N_0\) is the initial amount (16 mg), * \(t\) is the elapsed time (90 years), and * \(t_{1/2}\) is the half-life (30 years).Substitute the values: \[ N(90\text{ y}) = 16 \left( \frac{1}{2} \right)^{\frac{90}{30}} \] This simplifies to: \[ N(90\text{ y}) = 16 \left( \frac{1}{2} \right)^3 \] \[ N(90\text{ y}) = 16 \cdot \frac{1}{8} = 2 \text{ mg} \]

Determine the Time for Decay

To find the time required for 28 mg of cesium-137 to decay to 3.5 mg, use the same half-life formula: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \] Rearrange to solve for \(t\): \[ \frac{N(t)}{N_0} = \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \] \[ \frac{3.5}{28} = \left( \frac{1}{2} \right)^{\frac{t}{30}} \] Simplify: \[ \frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{t}{30}} \] Since \( \frac{1}{8} = \left( \frac{1}{2} \right)^3 \), equate the exponents: \[ 3 = \frac{t}{30} \] \[ t = 90 \text{ years} \]

Key concepts

half-life

In the context of radioactive decay, half-life is the time required for half of the radioactive atoms in a sample to decay. In other words, it is the time it takes for the quantity of the substance to reduce to half its initial amount. This concept is critical for predicting how long a radioactive substance will remain active. For instance, cesium-137 has a half-life of 30 years. This means that every 30 years, the amount of cesium-137 in a sample will halve. The concept of half-life helps in various fields such as nuclear medicine, radiometric dating, and nuclear power.

nuclear equation

A nuclear equation represents the changes that occur within a nucleus during a nuclear reaction or decay process. For cesium-137, which undergoes beta decay, the equation is written to show the transformation. Beta decay occurs when a neutron in the nucleus converts into a proton and an electron. The electron (beta particle) is emitted, while the new proton stays in the nucleus, increasing the atomic number by one. This process can be represented by the nuclear equation: \[ ^{137}_{55}Cs \rightarrow ^{137}_{56}Ba + \beta^- \] Here, cesium-137 (\textsuperscript{137}_{55}Cs) decays into barium-137 (\textsuperscript{137}_{56}Ba), and a beta particle (\( \beta^- \)) is released. The atomic number increases from 55 to 56, while the mass number (137) remains the same. Writing balanced nuclear equations helps us understand the changes in the nucleus and the particles involved.

radioactive decay calculations

Radioactive decay calculations are essential to predict how much of a radioactive substance remains after a given period or to find out how long it takes for a substance to decay to a specific amount. For example, to determine how much cesium-137 remains after 90 years from an initial 16 mg sample, we use the half-life formula: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \] Substituting the given values, \[ N(90 \text{ y}) = 16 \left( \frac{1}{2} \right)^{\frac{90}{30}} = 16 \left( \frac{1}{2} \right)^3 = 2 \text{ mg} \] This shows that only 2 mg of cesium-137 remains after 90 years. Similarly, to find how long it takes for 28 mg to decay to 3.5 mg: \[ \frac{3.5}{28} = \left( \frac{1}{2} \right)^{\frac{t}{30}} \] Simplifying, \[ \frac{1}{8} = \left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{\frac{t}{30}} \] Matching exponents, we get \[ 3 = \frac{t}{30} \, t = 90 \text{ years} \] These calculations help in understanding the decay process better and have practical applications in safety, medicine, and environmental studies.