Problem 43
Question
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?
Step-by-Step Solution
Verified Answer
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).
1Step 1: 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:
2Step 2: 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:
3Step 3: 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.
Key Concepts
Understanding Half-Life in Radioactive DecayRadioactivity and Its Measurable TraitsDecay Constant and its Role in Radioactive DecayExponential Decay and Its Application
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.
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.
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.
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.
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.
Other exercises in this chapter
Problem 41
Some watch dials are coated with a phosphor, like ZnS, and a polymer in which some of the \({ }^{1} \mathrm{H}\) atoms have been replaced by \({ }^{3} \mathrm{H
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It takes \(4 \mathrm{~h} \mathrm{} 39 \mathrm{~min}\) for a \(2.00\)-mg sample of radium-230 to decay to \(0.25 \mathrm{mg}\). What is the half-life of radium-2
View solution Problem 44
How much time is required for a \(6.25-\mathrm{mg}\) sample of \({ }^{51} \mathrm{Cr}\) to decay to \(0.75 \mathrm{mg}\) if it has a half-life of \(27.8\) days?
View solution Problem 46
Cobalt-60, which undergoes beta decay, has a half-life of \(5.26 \mathrm{yr}\). (a) How many beta particles arc emitted in \(600 \mathrm{~s}\) by a \(3.75-\math
View solution