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A radjoactive material has a half-life of 2 weeks. After 5 weeks, \(20 \mathrm{~g}\) of the material is seen to remain. How much material was initially present?

Short Answer

Expert verified
Answer: The initial amount of radioactive material was approximately 113.14 grams.

Step by step solution

01

Write down the formula for radioactive decay.

The decay formula for a radioactive material is \(N(t) = N_0 \times 2^{-\frac{t}{t_{1/2}}}\), where \(N(t)\) is the amount of material remaining after time \(t\), \(N_0\) is the initial amount of material, and \(t_{1/2}\) is the half-life. In this case, we are given the half-life, \(t_{1/2} = 2\) weeks, and the amount of material remaining after 5 weeks, \(N(5) = 20\) g. We want to find \(N_0\).
02

Insert the given values into the equation.

We'll plug in the given values into the decay formula: \(20 = N_0 \times 2^{-\frac{5}{2}}\)
03

Solve the equation for \(N_0\).

First, we need to find the value of \(2^{-\frac{5}{2}}\): \(2^{-\frac{5}{2}}=\dfrac{1}{2^{\frac{5}{2}}}=\dfrac{1}{\sqrt{2^5}}=\dfrac{1}{4\sqrt{2}}\) Now, we can plug this value back into the equation: \(20 = N_0 \times \dfrac{1}{4\sqrt{2}}\) To find out \(N_0\), we need to multiply both sides of the equation by \(4\sqrt{2}\): \(20 \times 4\sqrt{2} = N_0\)
04

Calculate the initial amount of material.

Now, we can calculate the value of \(N_0\): \(N_0 = 20 \times 4\sqrt{2} = 80\sqrt{2}\) Which is approximately: \(N_0 \approx 113.14\) Therefore, there were initially about \(113.14\) g of the radioactive material present.

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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 pivotal in understanding radioactive decay. The half-life of a substance is the time required for half of the radioactive material to decay. It functions as a timer of sorts, letting scientists know how quickly a substance will reduce by half.
In practical terms, if you have 100 grams of a material with a half-life of two weeks, in another two weeks, you will have 50 grams. This process continues until the substance transforms into a more stable form. In our exercise, the given half-life is 2 weeks.
Understanding half-life helps in predicting how much of a substance will remain over time. It is a fundamental property of radioactive materials, highlighting their decay pattern. When solving problems involving half-life, it's crucial to identify this period since it feeds directly into the decay formula, impacting how calculations are made.
Exponential Decay
Exponential decay is a mathematical representation of how quantities decrease at a rate proportional to their current value. This concept is neatly defined by an exponential function, and in the case of radioactive decay, it showcases how a material's amount reduces rapidly at first and eventually tapers off.
The formula typically used for this is:
  • \[ N(t) = N_0 \times 2^{-\frac{t}{t_{1/2}}} \]
This equation illustrates how the original amount \( N_0 \) decreases over time \( t \) based on the half-life \( t_{1/2} \).
In practical applications, like the exercise provided, plugging the time (5 weeks in this case) and half-life (2 weeks) into the formula determines the remaining material. Through this calculation, it becomes clear how exponential decay governs the declining pattern of radioactive substances.Importantly, this decay is not linear. The material doesn't decrease by the same quantity each period but instead shrinks to a fraction of its size in a predictable manner.
Initial Amount Calculation
Calculating the initial amount of a substance when given its remaining quantity after a certain time requires reversing the exponential decay model. This step-by-step process helps determine how much material was originally there before decay set in.
In our exercise, we are given:
  • Half-life \( t_{1/2} \): 2 weeks
  • The remaining amount \( N(t) \): 20g after 5 weeks
To find the initial amount \( N_0 \), insert these values into the decay equation. Rearrange to solve for \( N_0 \).
Here's the rearrangement:
  • \[ 20 = N_0 \times 2^{-\frac{5}{2}} \]
  • \[ N_0 = 20 \times 4\sqrt{2} \]
The final computation yields the start quantity: approximately 113.14 grams.
Understanding and performing these calculations demonstrates not only mathematical skills but also a grasp of the underlying scientific principles of radioactive decay. This is particularly useful in fields like chemistry and physics, where such calculations often form the basis for experimental predictions and interpretations.

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Most popular questions from this chapter

Active oxygen and free radicals are believed to be exacerbating factors in causing cell injury and aging in living tissue. \({ }^{1}\) These molecules also accelerate the deterioration of foods. Researchers are therefore interested in understanding the protective role of natural antioxidants. In the study of one such antioxidant (Hsian-tsao leaf gum), the antioxidation activity of the substance has been found to depend on concentration in the following way: $$ \frac{d A(c)}{d c}=k\left[A^{*}-A(c)\right], \quad A(0)=0 . $$ In this equation, the dependent variable \(A\) is a quantitative measure of antioxidant activity at concentration \(c\). The constant \(A^{*}\) represents a limiting or equilibrium value of this activity, and \(k\) is a positive rate constant. (a) Let \(B(c)=A(c)-A^{*}\) and reformulate the given initial value problem in terms of this new dependent variable, \(B\). (b) Solve the new initial value problem for \(B(c)\) and then determine the quantity of interest, \(A(c)\). Does the activity \(A(c)\) ever exceed the value \(A^{*}\) ? (c) Determine the concentration at which \(95 \%\) of the limiting antioxidation activity is achieved. (Your answer is a function of the rate constant \(k\).)

In each exercise, the general solution of the differential equation \(y^{\prime}+p(t) y=g(t)\) is given, where \(C\) is an arbitrary constant. Determine the functions \(p(t)\) and \(g(t)\). \(y(t)=C e^{t^{2}}+2\)

A differential equation of the form $$ y^{\prime}=p_{1}(t)+p_{2}(t) y+p_{3}(t) y^{2} $$ is known as a Riccati equation. \({ }^{5}\) Equations of this form arise when we model onedimensional motion with air resistance; see Section 2.9. In general, this equation is not separable. In certain cases, however (such as in Exercises 24-26), the equation does assume a separable form. Solve the given initial value problem and determine the \(t\)-interval of existence. $$ y^{\prime}=\left(y^{2}+2 y+1\right) \sin t, \quad y(0)=0 $$

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