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Initially, \(100 \mathrm{~g}\) of a radioactive material is present. After 3 days, only \(75 \mathrm{~g}\) remains. How much additional time will it take for radioactive decay to reduce the amount present to \(30 \mathrm{~g}\) ?

Short Answer

Expert verified
Answer: It will take an additional 5.40 days.

Step by step solution

01

Understand the exponential decay formula

In a radioactive decay process, the amount of the radioactive material decreases exponentially over time. The formula to represent this process is: $$N(t) = N_0e^{-kt}$$ where: - \(N(t)\) is the remaining amount of the radioactive substance at time t - \(N_0\) is the initial amount of the radioactive substance - \(k\) is the decay constant (a positive value) - \(t\) is the time in days
02

Find the decay constant k using given information

We are given that the initial amount of the radioactive material is 100 g, and after 3 days, 75 g remains. We can plug this information into the exponential decay formula to find the decay constant k. $$75 = 100e^{-3k}$$ Next, we need to solve this equation for k.
03

Solve the equation for k

Divide both sides by 100: $$\frac{75}{100} = e^{-3k}$$ Take the natural logarithm of both sides: $$\ln{\frac{3}{4}} = -3k$$ Now, solve for k: $$k = -\frac{1}{3}\ln{\frac{3}{4}}$$
04

Use the exponential decay formula to find the additional time

Now we want to find the additional time it will take for the amount of the radioactive material to reduce to 30 g. Plug in the values for the remaining amount \(N(t)\), initial amount \(N_0\), and decay constant k into the exponential decay formula: $$30 = 100e^{-\left(-\frac{1}{3}\ln{\frac{3}{4}}\right)t}$$
05

Solve for t

First, divide both sides by 100: $$\frac{3}{10} = e^{\left(-\frac{1}{3}\ln{\frac{3}{4}}\right)t}$$ Take the natural logarithm of both sides: $$\ln{\frac{3}{10}} = -\frac{1}{3}\ln{\frac{3}{4}}t$$ Now, solve for t: $$t = \frac{-3\ln{\frac{3}{10}}}{\ln{\frac{3}{4}}}$$ Calculate t: $$t \approx 5.40 \mathrm{~days}$$
06

State the final answer

It will take an additional 5.40 days for the amount of the radioactive material to reduce from 75 g to 30 g.

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Key Concepts

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

Exponential Decay Formula
The exponential decay formula is critical to understanding phenomena like radioactive decay. In simple terms, this formula describes how quantities decrease over time at a rate proportional to their current value. For example, radioactive substances lose mass by emitting particles, and this loss can be modeled using this formula.

The general form of the exponential decay formula is:
\[ N(t) = N_0e^{-kt} \]
Here,
  • \(N(t)\) represents the amount of the substance remaining after time \(t\),
  • \(N_0\) is the initial amount present,
  • \(k\) is the decay constant, which indicates how quickly the substance decays,
  • and \(t\) is the passage of time.
This powerful equation allows scientists and engineers to predict the future behavior of radioactive materials, among other applications.
Decay Constant
The decay constant, denoted by \(k\), plays a pivotal role in the exponential decay process. It is a value unique to each radioactive substance and determines the speed at which the substance decays. The decay constant is part of the natural exponential function in the decay formula and is always a positive number. When \(k\) is larger, the substance decays more quickly, and conversely, a smaller \(k\) means the substance will decay over a longer period.

Finding \(k\) is often the first step in solving decay problems, as it must be known to calculate the remaining quantity of a substance. In the given problem, we calculated \(k\) using the amounts of the material at two different times. Once determined, this constant is fundamental for predicting how much time will elapse before the substance reaches a certain level of decay.

Importance of the Decay Constant

Understanding the decay constant is not only essential in academic exercises but also in real-world applications like nuclear medicine, where the decay rate of isotopes is crucial for diagnostics and treatment.
Solving Exponential Equations
Solving exponential equations is a common requirement in many scientific fields. These equations, where the unknown appears as an exponent, are solved using logarithms. The natural logarithm, denoted as \(\ln\), is particularly useful because it is the inverse operation of the natural exponential function, \(e^x\).

To solve an exponential equation, it's often necessary to isolate the exponential expression and then apply logarithms to both sides of the equation to extract the variable from the exponent.

Steps for Solving Exponential Equations

  • Isolate the exponential part of the equation.
  • Take the natural logarithm of both sides to remove the base \(e\) from the exponent.
  • Solve for the variable.
  • Check the solution in the original equation.
In our problem, we used these steps to determine the time it would take for the radioactive material to decay to a specific mass. This process of solving exponential equations is essential for accurately predicting and understanding exponential decay in various contexts.

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Most popular questions from this chapter

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ \frac{d y}{d t}=1+y^{2}, \quad y(\pi / 4)=-1 $$

A metal casting is placed in an environment maintained at a constant temperature, \(S_{0}\). Assume the temperature of the casting varies according to Newton's law of cooling. A thermal probe attached to the casting records the temperature \(\theta(t)\) listed. Use this information to determine (a) the initial temperature of the casting. (b) the temperature of the surroundings. $$\theta(t)=80-40 e^{-2 t}{ }^{\circ} \mathrm{F}$$

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(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ (\ln y) y^{\prime}+t=1, \quad y(3)=e $$

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