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A tree contains a known percentage \(p_{0}\) of a radioactive substance with half-life \(\tau\). When the tree dies the substance decays and isn't replaced. If the percentage of the substance in the fossilized remains of such a tree is found to be \(p_{1}\), how long has the tree been dead?

Short Answer

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#tag_title#Step 2: Substitute the given values #tag_content#We know that the percentage of the substance in the tree after its death is $p_0$ and after decay is $p_1$. We also know the half-life $\tau$. We can substitute these values in the formula: $$ p_1 = p_0 \cdot \frac{1}{2}^{\frac{t}{\tau}} $$ #tag_title#Step 3: Solve for t #tag_content#To find the time elapsed since the tree died, we need to solve for $t$. First, let's divide both sides by $p_0$: $$ \frac{p_1}{p_0} = \frac{1}{2}^{\frac{t}{\tau}} $$ Now, take the logarithm base 2 of both sides to get rid of the exponent: $$ \log_2 \left(\frac{p_1}{p_0}\right) = \frac{t}{\tau} $$ Finally, multiply both sides by $\tau$ to solve for $t$: $$ t = \tau \cdot \log_2 \left(\frac{p_1}{p_0}\right) $$ This equation gives us the time elapsed since the tree died.

Step by step solution

01

Write down the radioactive decay formula

We will use the radioactive decay formula to find out the time elapsed since the tree died: $$ N(t) = N_0 \cdot \frac{1}{2}^{\frac{t}{\tau}} $$

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Key Concepts

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

half-life calculation
Understanding the concept of half-life is crucial in calculating the age of the tree based on the remaining radioactive substance. Half-life is defined as the time taken for half of the radioactive atoms in a sample to decay. This means, after one half-life period, only 50% of the original substance remains.
When calculating half-life, you are essentially determining how many half-life periods have passed for a sample to decrease to the observed level of radioactive material. To solve a problem of this nature, you will need to compare the initial amount of the substance with the amount remaining to find out how much time has passed.
For instance, if initially, a sample had a radioactivity level of 100 grams and after a certain time, it decreases to 25 grams, two half-lives have passed since 100 to 50 grams is one half-life and from 50 to 25 grams is another half-life. Knowing the half-life of the substance is a key part of these calculations, as it allows you to translate the number of half-lives passed into actual time.
radioactive decay formula
The radioactive decay formula helps us track how a substance reduces over time due to radiative emissions. It embodies the exponential nature of the process. The formula can be expressed as:
  • \( N(t) = N_0 \cdot \frac{1}{2}^{\frac{t}{\tau}} \)
Where:
  • \( N(t) \) is the amount of substance remaining after time \( t \).
  • \( N_0 \) is the initial amount of substance.
  • \( \tau \) is the half-life of the substance.
  • \( t \) is the time since the death of the tree.
This equation expresses the remaining percentage of the radioactive substance as a fraction that diminishes exponentially over time. You measure progress by how many half-life cycles have passed, decreasing the initial substance proportionately. In context, this formula allows you to predict the elapsed time since a living organism, such as a tree, stopped replenishing its radioactive content due to death.
exponential decay
Exponential decay describes how the quantity of a substance decreases at a rate proportional to its current value, leading to a rapid decrease initially, which slows down as time progresses. This is a key feature of things like radioactive decay, making it predictable but not uniform over time.
In exponential decay, each equal increment of time results in the reduction of the substance's amount by a consistent fraction or percentage, not a consistent amount. This creates a curve that drops sharply at first and levels out the longer it goes, never quite reaching zero but coming infinitely close.
Regarding radioactive decay in trees, exponential decay helps us infer how fast the radioactive substance is lost, based on known percentages. For instance, if every year, half of the existing radioactive material decays, then plotting these events creates an exponential curve. Using this concept, scientists can backtrack to estimate the age of objects based on the materials they contain.

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Most popular questions from this chapter

An object with mass \(m\) is given an initial velocity \(v_{0} \leq 0\) in a medium that exerts a resistive force of the form \(a|v| /(1+|v|),\) where \(a\) is positive constant. (a) Set up a differential equation for the speed of the object. (b) Use your favorite numerical method to solve the equation you found in (a), to convince yourself that there's a unique number \(a_{0}\) such that \(\lim _{t \rightarrow \infty} s(t)=\infty\) if \(a \leq a_{0}\) and \(\lim _{t \rightarrow \infty} s(t)\) exists (finite) if \(a>a_{0}\). (We say that \(a_{0}\) is the bifurcation value of \(a\).) Try to find \(a_{0}\) and \(\lim _{t \rightarrow \infty} s(t)\) in the case where \(a>a_{0} .\) HINT: See Exercise \(14 .\)

A super bread dough increases in volume at a rate proportional to the volume \(V\) present. If \(V\) increases by a factor of 10 in 2 hours and \(V(0)=V_{0},\) find \(V\) at any time \(t .\) How long will it take for \(V\) to increase to \(100 \mathrm{~V}_{0}\) ?

Consider the mixing problem of Example 4.2.4 in a tank with infinite capacity, but without the assumption that the mixture is stirred instantly so that the salt is always uniformly distributed throughout the mixture. Assume instead that the distribution approaches uniformity as \(t \rightarrow \infty .\) In this case the differential equation for \(Q\) is of the form $$ Q^{\prime}+\frac{a(t)}{t+100} Q=1 $$ where \(\lim _{t \rightarrow \infty} a(t)=1\) (a) Let \(K(t)\) be the concentration of salt at time \(t\). Assuming that \(Q(0)=Q_{0},\) can you guess the value of \(\lim _{t \rightarrow \infty} K(t) ?\) (b) Use numerical methods to confirm your guess in the these cases: (i) \(a(t)=t /(1+t)\) $$ \text { (ii) } a(t)=1-e^{-t^{2}} $$ (iii) \(a(t)=1+\sin \left(e^{-t}\right)\).

Suppose water is added to a tank at \(10 \mathrm{gal} / \mathrm{min}\), but leaks out at the rate of \(1 / 5 \mathrm{gal} / \mathrm{min}\) for each gallon in the tank. What is the smallest capacity the tank can have if the process is to continue indefinitely?

Find a first order differential equation for the given family of curves. $$ e^{x y}=c y $$

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