Chapter 1: Problem 63
Polonium \(210\left(\mathrm{Po}^{210}\right)\) has a half-life of 140 days. (a) If a sample of \(\mathrm{Po}^{210}\) has a mass of 300 micrograms, find \(\mathrm{a}\) formula for the mass after \(t\) days. (b) How long would it take this sample to decay to \(20 \%\) of its original amount?
Short Answer
Expert verified
The formula for mass after \( t \) days is \( M(t) = 300 \left(\frac{1}{2}\right)^{\frac{t}{140}} \). It takes about 324 days to decay to 20% of the original mass.
Step by step solution
01
Understand the Half-Life Formula
The half-life formula to determine the remaining quantity of a radioactive sample is given by the equation: \[ M(t) = M_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]where \( M(t) \) is the remaining mass after time \( t \), \( M_0 \) is the initial mass, and \( T_{1/2} \) is the half-life of the substance. Here, \( T_{1/2} = 140 \) days.
02
Derive the Formula for Mass after t Days
Plug in the given initial mass and half-life into the half-life formula.Starting with \( M_0 = 300 \) micrograms and \( T_{1/2} = 140 \) days:\[ M(t) = 300 \left(\frac{1}{2}\right)^{\frac{t}{140}} \]This is the formula for the mass of \( \mathrm{Po}^{210} \) after \( t \) days.
03
Set Up the Equation for 20% of Original Mass
To find the time it takes for the mass to decay to 20% of its original amount, we start with:\[ M(t) = 0.2 \times M_0 = 0.2 \times 300 \]Thus, \( M(t) = 60 \) micrograms.Substitute this into the decay formula:\[ 60 = 300 \left(\frac{1}{2}\right)^{\frac{t}{140}} \]
04
Solve for Time t
Divide both sides by 300:\[ \frac{60}{300} = \left(\frac{1}{2}\right)^{\frac{t}{140}} \]Simplify to:\[ \frac{1}{5} = \left(\frac{1}{2}\right)^{\frac{t}{140}} \]Taking the natural logarithm on both sides results in:\[ \ln\left(\frac{1}{5}\right) = \frac{t}{140} \ln\left(\frac{1}{2}\right) \]Rearrange to solve for \( t \):\[ t = 140 \times \frac{\ln\left(\frac{1}{5}\right)}{\ln\left(\frac{1}{2}\right)} \]
05
Compute the Time Using the Formula
Calculate \( \ln\left(\frac{1}{5}\right) \approx -1.6094 \) and \( \ln\left(\frac{1}{2}\right) \approx -0.6931 \), then substitute in:\[ t = 140 \times \frac{-1.6094}{-0.6931} \approx 324 \] Thus, it takes approximately 324 days for the sample to decay to 20% of its original mass.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Half-Life
Half-life is a crucial concept in the study of radioactive decay. It refers to the amount of time it takes for half of a radioactive substance to decay into a different material. This is not only applicable to radioactivity but can also be used metaphorically in everyday scenarios where something diminishes by half over regular intervals.
The half-life is inherent to each radioactive isotope. For Polonium-210, this is 140 days. So, every 140 days, only half of the original amount remains. This understanding helps in predicting how long a certain quantity of a radioactive substance will continue to exist before it becomes less significant or nearly non-detectable. It's interesting to note that while the quantity decreases, the concept of half-life implies that the substance never fully disappears through decay alone—a theoretical aspect that fascinates many in the physics community.
Knowing the half-life of a radioisotope allows scientists, archivists, and health professionals to calculate not just how much of a radioactive sample remains after a certain period, but also in scheduling safe handling times and understanding potential environmental impacts.
The half-life is inherent to each radioactive isotope. For Polonium-210, this is 140 days. So, every 140 days, only half of the original amount remains. This understanding helps in predicting how long a certain quantity of a radioactive substance will continue to exist before it becomes less significant or nearly non-detectable. It's interesting to note that while the quantity decreases, the concept of half-life implies that the substance never fully disappears through decay alone—a theoretical aspect that fascinates many in the physics community.
Knowing the half-life of a radioisotope allows scientists, archivists, and health professionals to calculate not just how much of a radioactive sample remains after a certain period, but also in scheduling safe handling times and understanding potential environmental impacts.
Exponential Decay Formula
The exponential decay formula is a mathematical model used to describe the decay process of substances, such as radioactive materials. The formula is given by:
For instance, using Polonium-210 with a half-life of 140 days, we derived the equation \( M(t) = 300 \left(\frac{1}{2}\right)^{\frac{t}{140}} \) to express how the mass changes over time. This is why understanding exponential decay is essential when navigating questions about the remaining life of a radioactive sample or determining its long-term behavior.
- \( M(t) = M_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \)
- \( M(t) \) is the mass remaining after time \( t \),
- \( M_0 \) is the initial mass at time \( t = 0 \), and
- \( T_{1/2} \) is the half-life of the substance.
For instance, using Polonium-210 with a half-life of 140 days, we derived the equation \( M(t) = 300 \left(\frac{1}{2}\right)^{\frac{t}{140}} \) to express how the mass changes over time. This is why understanding exponential decay is essential when navigating questions about the remaining life of a radioactive sample or determining its long-term behavior.
Logarithmic Equations
Logarithmic equations are a powerful tool for solving exponential decay problems, especially when you need to find time-related solutions. When you work with exponential decay, at some point, you'll often need to extract the exponent. This is where the logarithm comes into play.
In our Polonium-210 example, to find out when the sample reduces to 20% of its original mass, we took the natural logarithm of both sides of the equation \( \left(\frac{1}{5}\right) = \left(\frac{1}{2}\right)^{\frac{t}{140}} \). This gives:
Solving the logarithmic equation involves basic operations and often requires a calculator or a logarithm table, as it's based on known values like \( \ln(\frac{1}{2}) \approx -0.6931 \), which helps simplify otherwise complex equations.
In our Polonium-210 example, to find out when the sample reduces to 20% of its original mass, we took the natural logarithm of both sides of the equation \( \left(\frac{1}{5}\right) = \left(\frac{1}{2}\right)^{\frac{t}{140}} \). This gives:
- \( \ln\left(\frac{1}{5}\right) = \frac{t}{140} \ln\left(\frac{1}{2}\right) \)
Solving the logarithmic equation involves basic operations and often requires a calculator or a logarithm table, as it's based on known values like \( \ln(\frac{1}{2}) \approx -0.6931 \), which helps simplify otherwise complex equations.