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Chapter 4: Problem 15

A certain radioactive substance has a half-life of \(38 \mathrm{hr}\). Find how long it takes for \(90 \%\) of the radioactivity to be dissipated.

Short Answer

Expert verified
126.06 hours

Step by step solution

01

Understand the problem

We need to find the time it takes for 90% of a radioactive substance to decay or dissipate. The half-life provides the time it takes for half of the substance to decay.
02

Define the decay formula

Radioactive decay can be modeled by the formula \[ N(t) = N_0 \left(\frac{1}{2}\right)^{t/T} \]where \( N(t) \) is the remaining amount after time \( t \), \( N_0 \) is the initial amount, and \( T \) is the half-life.
03

Set up the equation for 90% dissipation

Since we want 90% of the substance to be gone, only 10% remains. Set \( N(t) = 0.1N_0 \). The equation becomes:\[ 0.1N_0 = N_0 \left(\frac{1}{2}\right)^{t/38} \].
04

Simplify the equation

Divide both sides by \( N_0 \) to eliminate it:\[ 0.1 = \left(\frac{1}{2}\right)^{t/38} \].
05

Solve for t using logarithms

Take the logarithm of both sides to solve for \( t \):\[ \log_{10}(0.1) = \frac{t}{38} \log_{10}\left(\frac{1}{2}\right) \].Solve for \( t \):\[ t = 38 \times \frac{\log_{10}(0.1)}{\log_{10}(0.5)} \].
06

Calculate the result

Calculate \( \log_{10}(0.1) = -1 \) and \( \log_{10}(0.5) \approx -0.301 \). Substituting these values gives:\[ t = 38 \times \frac{-1}{-0.301} \approx 126.06 \] hours.

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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 fundamental to understanding radioactive decay. It is defined as the time required for half of the radioactive atoms in a substance to decay. This means that if you start with a certain quantity of a radioactive substance, after one half-life, only half of it will remain radioactive.

During each subsequent half-life, half of the remaining radioactive atoms will decay. As a result, the amount of active material continuously decreases over time.

  • Half-life helps in predicting how long it takes for a certain percentage of a radioactive substance to decay.
  • It is a fixed property for each radioactive material.
  • The half-life does not depend on the initial amount of the substance. It remains constant regardless of how much material you start with.
Understanding half-life is crucial when solving problems involving radioactive decay, like determining how long it takes for a significant portion of a substance, such as 90%, to dissipate.
Exponential Decay Formula
The mathematical modeling of radioactive decay is elegantly captured by the exponential decay formula. This formula expresses the decay process as an exponential function, reflecting how quickly a substance decreases over time.

The general formula is given by: \[N(t) = N_0 \left(\frac{1}{2}\right)^{t/T}\] Where:
  • \(N(t)\) is the amount of substance remaining after time \(t\).
  • \(N_0\) is the initial amount of the substance.
  • \(T\) represents the half-life of the substance.
The key aspect of the exponential decay formula is the fraction \(\left(\frac{1}{2}\right)\), which demonstrates that the substance is halved every half-life period. This formula is crucial for solving problems related to radioactive decay, as it allows us to predict how much of a substance will remain after any given time.

Applying the exponential decay formula includes substituting the known values, such as half-life and initial amount, to find the decay over a defined period.
Logarithmic Functions
Logarithmic functions are essential when solving exponential decay problems, especially when you want to find the elapsed time required for a substance to decay to a specific amount. These functions help invert the exponential equation, making it easy to solve for time.

When you have an equation of the form \(b^x = y\), the logarithm allows you to solve for \(x\) as follows:\[x = \log_b(y)\] In radioactive decay problems, we often apply logarithms to both sides of the equation to "bring down the exponent." For example, if we want to solve the equation \(0.1 = \left(\frac{1}{2}\right)^{t/38}\), taking the logarithm helps isolate \(t\).

Some essential points about logarithms include:
  • Logarithms convert multiplicative processes (like exponential growth or decay) into additive processes.
  • They help in solving equations where the variable is an exponent.
  • Common bases for logarithms include \(10\) (common logarithm) and \(e\) (natural logarithm).
The use of logarithmic functions is a powerful method in mathematical problem-solving. It simplifies finding how long it takes for a certain percentage of a radioactive substance to decay.

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