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Cobalt-60 is a strong gamma emitter that has a half-life of \(5.26 \mathrm{yr}\). The cobalt- 60 in a radiotherapy unit must be replaced when its radioactivity falls to \(75 \%\) of the original sample. If an original sample was purchased in June 2013, when will it be necessary to replace the cobalt-60?

Short Answer

Expert verified
The Cobalt-60 will need to be replaced in June 2015, as its radioactivity will have fallen to 75% of its original value after approximately 1.954 years (rounded to 2 years).

Step by step solution

01

Decay constant calculation

\[\lambda = \frac{ln2}{5.26} \] Calculate the decay constant: \[\lambda \approx 0.1315 \,\text{per year}\] #Step 2: Calculate the time elapsed for the radioactivity to fall to 75%# In this step, we need to find the time t when the remaining activity \(N(t)\) is 75% of the initial activity. Using the radioactive decay formula: \[N(t) = N_0 e^{-\lambda t}\] Let's solve for t: \[0.75 N_0 = N_0 e^{-0.1315 t}\] Rearranging the formula to isolate t:
02

Find the time t

\[\frac{0.75 N_0}{N_0} = e^{-0.1315 t}\] Divide both sides by \(N_0\) and calculate t: \[0.75 = e^{-0.1315 t}\] Take the natural logarithm of both sides: \[ln(0.75) = -0.1315 t\] Now, divide by \(-0.1315\) to find the value of t: \[t = \frac{ln(0.75)}{-0.1315}\] Calculate the value of t:
03

Time elapsed calculation

\[t \approx 1.954 \,\text{years}\] Since the original sample was purchased in June 2013, it will need to be replaced in: June 2013 + 1.954 years (rounded to 2 years) #Solution# Cobalt-60 will be necessary to replace in June 2015.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Half-Life in Radioactive Decay
In the world of physics, half-life is a term commonly associated with radioactive substances. It refers to the time required for half of the radioactive atoms present in a sample to decay. This concept is pivotal in many fields, including geology, archaeology, and medicine.

Take, for example, Cobalt-60; it is known to have a half-life of 5.26 years. This means that every 5.26 years, the strength of the radioactive emission, which we rely on for processes like radiotherapy, diminishes by half. When its activity falls to a certain level, it no longer remains effective for its intended use, indicating a need for replacement. The half-life provides a predictable pattern of decay, allowing us to calculate when replacement is due with great precision.
Radioactivity and Its Measurable Traits
The phenomenon of radioactivity is observed when unstable nuclei release energy to gain more stable configurations. This process occurs naturally in elements like uranium or artificially induced isotopes, such as Cobalt-60, used in medical treatments. Radioactivity can be hazardous but also useful, often harnessed in medical diagnostics and treatments.

To manage this powerful force, it's vital to measure radioactivity in terms of the number of disintegrations per unit time — commonly measured in becquerels or curies. In practice, knowing the rate of decay helps in determining the usability span of a radioactive source, such as how long Cobalt-60 can function effectively in a radiotherapy unit before its activity falls below a desired threshold.
Decay Constant and its Role in Radioactive Decay
At the core of understanding radioactive decay lies the decay constant, represented by the symbol \(\lambda\). This constant is crucial because it depicts the rate at which a radioactive substance decays. In mathematical terms, it quantifies the probability of atom disintegration per unit time.

To put it into perspective for Cobalt-60, using the formula \(\lambda = \frac{\ln2}{\text{half-life}}\), we calculate its decay constant. The smaller the value of the decay constant, the slower the rate of radioactive decay, and conversely, a larger value equates to a faster decay. This constant is used in calculations to predict how much time will pass before a sample's radioactivity drops to a specific level, which is essential for planning in medical and industrial applications.
Exponential Decay and Its Application
The concept of exponential decay is often a challenging but important one to grasp. It describes how quantities decrease over time at a rate proportionate to their value. In the case of radioactivity, this implies that the number of undecayed atoms at any given moment directly influences how quickly the atoms continue to decay.

For radioactive substances, the decay rate follows an exponential function, represented mathematically as \(N(t) = N_0 e^{-\lambda t}\), where \(N_0\) is the initial quantity of substance, \(\lambda\) is the decay constant, and \(t\) is the time elapsed. Applying this formula gives us an effective tool to determine how long it will take for a substance like Cobalt-60 to reduce to a certain level of radioactivity, ensuring patient safety and effective treatment.

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Most popular questions from this chapter

A \(26.00-\mathrm{g}\) sample of water containing tritium, \({ }^{3} \mathrm{H}\), emits \(1.50 \times 10^{3}\) beta particles per second. Tritium is a weak beta emitter with a half-life of \(12.3 \mathrm{yr}\). What fraction of all the hydrogen in the water sample is tritium?

How much energy must be supplied to break a single \({ }^{21} \mathrm{Ne}\) nucleus into separated protons and neutrons if the nucleus has a mass of \(20.98846\) amu? What is the nuclear binding energy for \(1 \mathrm{~mol}\) of \({ }^{21} \mathrm{Ne}\) ?

According to current regulations, the maximum permissible dose of strontium- 90 in the body of an adult is \(1 \mu \mathrm{Ci}\left(1 \times 10^{-6} \mathrm{Ci}\right)\). Using the relationship rate \(=k N\), calculate the number of atoms of strontium- 90 to which this dose corresponds. To what mass of strontium- 90 does this correspond? The half-life for strontium- 90 is \(28.8 \mathrm{yr}\).

What particle is produced during the following decay processes: (a) sodium-24 decays to magnesium-24; (b) mercury-188 decays to gold-188; (c) iodine-122 decays to xenon-122; (d) plutonium-242 decays to uranium-238?

The atomic masses of hydrogen-2 (deuterium), helium-4, and lithium-6 are \(2.014102 \mathrm{amu}_{2} 4.002602 \mathrm{amu}\), and \(6.0151228\) amu, respectively. For cach isotope, calculate (a) the nuclear mass, (b) the nuclear binding energy, (c) the nuclear binding energy per nucleon. (d) Which of these three isotopes has the largest nuclear binding energy per nucleon? Does this agrec with the trends plotted in Figure 21.12?

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