Chapter 24: Problem 58
An unknown radioisotope exhibits 8540 decays per second. After 350.0 min, the number of decays has decreased to 1250 per second. What is the half-life?
Short Answer
Expert verified
The half-life of the unknown radioisotope is approximately \(295.7\) minutes.
Step by step solution
01
Understanding the Exponential Decay Formula
The decay rate, N(t), of a radioisotope follows an exponential decay function:
\(N(t) = N_0e^{-\lambda t}\)
Where:
\(N(t)\) = decay rate at time t
\(N_0\) = initial decay rate
\(\lambda\) = decay constant
\(t\) = time in seconds
We will use this formula to find the decay constant, \(\lambda\).
02
Converting Time to Seconds
To work with consistent units, we need to convert the given time, 350 minutes, to seconds:
\(t = 350~min * 60 sec/min = 21,000~sec\)
03
Write down the Known Values
Let's insert the known values into the exponential decay function:
\(N(21,000) = 1250\)
\(N_0 = 8540\)
\(t = 21,000~sec\)
We need to solve for \(\lambda\) to find the half-life.
04
Solve for the Decay Constant
Using the decay function, we can solve for \(\lambda\):
\(1250 = 8540e^{-\lambda * 21,000}\)
To isolate \(\lambda\), divide both sides by 8540:
\(\frac{1250}{8540} = e^{-\lambda * 21,000}\)
Next, take the natural logarithm of both sides:
\(\ln{\frac{1250}{8540}} = -\lambda * 21,000\)
Now, we can solve for \(\lambda\):
\(\lambda = \frac{-\ln{\frac{1250}{8540}}}{21,000}\)
Calculate the numerical result:
\(\lambda \approx 3.91 \times 10^{-5} sec^{-1}\)
05
Calculate the Half-life
We can calculate the half-life using the decay constant and the following formula:
\(T_{1/2} = \frac{ln2}{\lambda}\)
Insert \(\lambda\) into the formula:
\(T_{1/2} = \frac{ln2}{3.91 \times 10^{-5} sec^{-1}}\)
Calculate the numerical result:
\(T_{1/2} \approx 17,741~sec\)
Now, we can convert the half-life from seconds to minutes:
\(T_{1/2} \approx 17,741~sec * \frac{1~min}{60~sec} \approx 295.7~min\)
06
Final Answer
The half-life of the unknown radioisotope is approximately 295.7 minutes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Decay
Exponential decay is a mathematical process that describes how quantities decrease over time at a rate proportional to their current value. In physics and chemistry, it's often used to model phenomena such as radioactive decay, where the decay rate of a substance diminishes over time. The exponential decay formula is given by\[N(t) = N_0e^{-\lambda t}\]where:
- \(N(t)\) represents the value at time \(t\)
- \(N_0\) is the initial quantity
- \(\lambda\) is the decay constant, a measure of the rate of decay
- \(t\) is the elapsed time
Radioisotope Decay
Radioisotope decay is a natural process through which unstable radioisotopes transform into more stable forms while releasing energy in the form of radiation. This decay process follows a predictable pattern of exponential decay. Each radioisotope has its own characteristic decay rate determined by its decay constant \(\lambda\). This rate shows how rapidly or slowly the material undergoes radioactive decay. Over time, this process leads to a reduction in the radioisotope's quantity, affecting its activity.Understanding radioisotope decay is crucial in various applications such as carbon dating, medical diagnostics, and nuclear power generation. These involve measuring the remaining activity of a substance to gauge its remaining concentration or age based on its known decay constant.
Decay Constant Calculation
The decay constant \(\lambda\) is a critical factor in quantifying the rate of decay of a substance. It represents the proportion of the material that decays per unit of time. To calculate \(\lambda\), you can rearrange the exponential decay formula to solve for the decay constant:\[\lambda = \frac{-\ln{\left(\frac{N(t)}{N_0}\right)}}{t}\]This formula arises from taking the natural logarithm of both sides of the original decay equation, which helps isolate \(\lambda\). Here, \(\ln\) denotes the natural logarithm, and it plays a substantial role in converting the exponentials in the decay formula into a format that’s easier for calculation.Accurate calculation of the decay constant is fundamental in determining the half-life, which signifies the peculiar time required for the substance to reduce to half of its initial value.
Natural Logarithm
The natural logarithm, commonly represented as \(\ln\), is a logarithm to the base \(e\) (Euler's number, approximately 2.71828). It's an essential mathematical tool for simplifying expressions involving exponential relationships.In the context of decay calculations, taking the natural logarithm allows us to transform the exponential decay equation into a linear form, making it easier to solve for the decay constant \(\lambda\).For example, the expression \(e^{-\lambda t}\) becomes much simpler when converted using the natural logarithm:\[\ln{\left(\frac{N(t)}{N_0}\right)} = -\lambda t\]This manipulation makes the decay rate equations tractable, illustrating the power of the natural logarithm in solving exponential decay problems. The logarithm cancels out the exponential function, leaving an equation that can be straightforwardly rearranged to find unknown values such as the decay constant.
Time Conversion to Seconds
Time conversion is an essential step in solving decay problems accurately since decay calculations require consistent units. While problems often present time in minutes or hours, scientific calculations, especially in exponential decay, use seconds for precision.For instance, in the given problem, the time is initially in minutes (350 minutes). To convert minutes into seconds, use the conversion factor:\[t = 350 \text{ min} \times 60 \text{ sec/min} = 21,000 \text{ seconds}\]This conversion ensures continuity in units and avoids errors in calculations. Remember that keeping all time-related aspects in the same unit is crucial when working with equations involving time, as it maintains the integrity of the calculation and ensures accurate results. Understanding this basic conversion process is crucial in tackling any decay-related problem efficiently.