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Chapter 3: Problem 6

The radioactive isotope of lead, \(\mathrm{Pb}-209\), decays at a rate proportional to the amount present at any time and has a half-life of \(3.3\) hours. If 1 gram of lead is present initially, how long will it take for \(90 \%\) of the lead to decay?

Short Answer

Expert verified
It takes approximately 10.96 hours for 90% to decay.

Step by step solution

01

Understand the Problem

We want to find the time it takes for 90% of the radioactive isotope \( \mathrm{Pb}-209 \) to decay. Given the half-life \( T_{1/2} = 3.3 \) hours and the initial amount \( A_0 = 1 \) gram, we need to determine when the remaining quantity is 10% of the original amount.
02

Set Up the Decay Equation

The decay of a radioactive isotope can be described by the equation \( A(t) = A_0 e^{-kt} \), where \( A(t) \) is the amount remaining at time \( t \), \( A_0 \) is the initial amount, and \( k \) is the decay constant.
03

Calculate the Decay Constant

Using the half-life formula \( T_{1/2} = \frac{\ln(2)}{k} \), we can solve for \( k \). Given \( T_{1/2} = 3.3 \), we have \( k = \frac{\ln(2)}{3.3} \).
04

Substitute Values

Substitute \( k \) into the decay equation. Also, since 90% decay means only 10% remains, set \( A(t) = 0.1A_0 \). Substitute \( A_0 = 1 \) gram, so \( A(t) = 0.1 \). The equation is now \( 0.1 = 1 e^{-k t} \).
05

Solve for Time \( t \)

Rearrange \( 0.1 = e^{-kt} \) to solve for \( t \): \( e^{-kt} = 0.1 \). Taking the natural log of both sides gives \( -kt = \ln(0.1) \). Solve for \( t \): \( t = -\frac{\ln(0.1)}{k} \).
06

Substitution and Calculation

Substitute \( k = \frac{\ln(2)}{3.3} \) into \( t = -\frac{\ln(0.1)}{k} \). Calculate \( t \) using a calculator: \( t \approx 10.96 \).

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Key Concepts

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

Half-Life
The concept of half-life is essential in understanding radioactive decay. Half-life is the time required for half of the radioactive isotope in a sample to decay.
For example, if you start with a certain amount of radioactive material, after one half-life, you will have half of it remaining.
After two half-lives, only a quarter of the original amount will be left. This means each half-life reduces the remaining quantity by half, no matter how much you start with.
  • The half-life for lead-209 (\(\mathrm{Pb}-209 \)) is 3.3 hours.
  • This half-life is specific to each isotope and does not change unless the isotope is altered.
  • It allows us to predict how quickly a given isotope will decay over time.
Half-life is crucial in many scientific fields including geology, archaeology, and health sciences, as it allows us to approximate the timeline of decay.
Decay Constant
The decay constant, denoted by the symbol \( k \), is a crucial part of the radioactive decay equation. It relates to how quickly a substance undergoes decay.
A higher decay constant means the substance decays more rapidly.
  • The decay constant \( k \) is calculated using the formula: \( k = \frac{\ln(2)}{T_{1/2}} \).
  • \(\ln(2)\) is the natural logarithm of 2, approximately equal to 0.693.
  • For lead-209, the half-life \( (T_{1/2}) \) is 3.3 hours, so \( k = \frac{0.693}{3.3} \).
  • The decay constant is essential for setting up the exponential decay equation.
Understanding the decay constant gives a clear idea about the rate of decay and quantifies how fast or slow the radioactive process is.
Exponential Decay Equation
The exponential decay equation is a powerful tool used to model radioactive decay. It can predict the remaining amount of a substance at any time \( t \). The equation: \[ A(t) = A_0 e^{-kt} \] is instrumental in scientific calculations.
  • \(A(t)\) is the amount of the substance remaining at time \( t \).
  • \(A_0\) is the initial amount of the substance.
  • \(e\) is the base of the natural logarithm, approximately 2.718.
  • \(k\) is the decay constant, determining the rate of decay.
Using the exponential decay equation, we can find out how long it takes for a substance to reach a certain remaining quantity.
For example, to determine when 90% of lead-209 has decayed, we set \( A(t) = 0.1 A_0 \) and solve for \( t \). This equation captures the unique property of exponential decay, where the rate of change is proportional to the current value. It helps in precise calculations for half-life and decay constant relationships.

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Most popular questions from this chapter

Mr. Jones puts two cups of coffee on the breakfast table at the same time. He immediately pours cream into his coffee from a pitcher that was sitting on the table for a long time. He then reads the morning paper for 5 minutes before taking his first sip. Mrs. Jones arrives at the table 5 minutes after the cups were set down, adds cream to her coffee, and takes a sip. Assume that Mr, and Mrs. Jones add exactly the same amount of cream to their cups. Discuss who drinks the hotter cup of coffee. Defend your assertion with sound mathematics.

\(d y / d x=f(x, y)\) is the differential equation of one family, then the differential equation for the orthogonal trajectories of this family is \(d y / d x=-1 / f(x, y)\). (a) Find a differential equation for the family \(y=-x-1+c_{1} e^{x}\). (b) Find the orthogonal trajectories for the family in part (a).

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