/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Suppose you find a piece of anci... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose you find a piece of ancient pottery and find that the glaze contains radium, a radioactive element that decays to radon and has a half-life of 1,620 years. There could not have been any radon in the glaze when the pottery was being fired, but now it contains three atoms of radon for each atom of radium. How old is the pottery?

Short Answer

Expert verified
The pottery is 3,240 years old.

Step by step solution

01

Understand the Problem

We need to determine the age of the pottery given that radium decays into radon, and currently, there are three atoms of radon for each atom of radium. The half-life of radium is 1,620 years.
02

Set Up the Decay Formula

The decay of radium can be modeled by the exponential decay formula: \[ N(t) = N_0 \times \frac{1}{2}^{t / T} \]where:- \(N(t)\) is the number of radium atoms at time \(t\)- \(N_0\) is the initial number of radium atoms- \(T\) is the half-life (1,620 years)
03

Determine the Radium Decay Relationship

Given there are three radon atoms for each radium atom now, we express this ratio as: \[ N_0 - N(t) = 3 \times N(t) \]where \(N_0 - N(t)\) is the number of radium atoms that have decayed into radon atoms.
04

Solve for the Present Radium Amount

Re-organize the equation to isolate \(N(t)\): \[ N_0 - N(t) = 3N(t) \implies N_0 = 4N(t) \]Thus, the present amount of radium is one-fourth of the initial amount: \[ N(t) = \frac{N_0}{4} \]
05

Find the Time Elapsed

Substitute \(N(t) = \frac{N_0}{4}\) into the decay formula to solve for \(t\):\[ \frac{N_0}{4} = N_0 \times \frac{1}{2}^{t / 1620} \]Divide both sides by \(N_0\):\[ \frac{1}{4} = \frac{1}{2}^{t / 1620} \]Taking the natural logarithm on both sides:\[ \text{ln} \frac{1}{4} = \text{ln} \frac{1}{2}^{t / 1620} \]Simplify using logarithm rules:\[ \text{ln} \frac{1}{4} = \frac{t}{1620} \text{ln} \frac{1}{2} \]
06

Calculate the Exact Age

Solve for \(t\):\[ t = \frac{1620 \times \text{ln} \frac{1}{4}}{\text{ln} \frac{1}{2}} \]Since \( \text{ln} \frac{1}{4} = -2 \times \text{ln} 2 \) and \( \text{ln} \frac{1}{2} = -\text{ln} 2 \):\[ t = \frac{1620 \times (-2 \text{ln} 2)}{-\text{ln} 2} \]\[ t = 1620 \times 2 = 3240 \text{ years} \]

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Key Concepts

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

half-life
Radioactive decay is a random process where unstable atomic nuclei lose energy. The **half-life** of a radioactive substance is the time it takes for half of the radioactive atoms in a sample to decay. In our pottery example, radium has a half-life of 1,620 years. This means it will take 1,620 years for half of the radium atoms to decay into radon.

The half-life formula is:
\[ N(t) = N_0 \times \frac{1}{2}^{t / T} \]
Where:
  • \(N(t)\) is the number of radium atoms remaining at time \(t\)
  • \(N_0\) is the initial number of radium atoms
  • \(T\) is the half-life (1,620 years in this case)

This formula shows how the remaining number of atoms decreases exponentially over time.


You should note that during each half-life, only a fixed fraction (half) of the substance decays, regardless of the initial amount. This is what defines the exponential nature of radioactive decay.
exponential decay
The term **exponential decay** describes a process in which the amount of a substance decreases rapidly at first, but then more slowly over time. It’s characterized by the property that in each equal time interval, the substance reduces by the same fraction. The equation for exponential decay we used in this problem is:
\[ N(t) = N_0 \times \frac{1}{2}^{t / T} \]
In this equation, the base of the exponent (\( \frac{1}{2} \)) indicates that we are halving the remaining amount after each period \(T\) (the half-life).


**Visualizing Exponential Decay:**
Imagine you have 100 atoms to start with. After one half-life (1,620 years for radium), you'd have 50 atoms left. After another 1,620 years, you'd have 25 atoms left, then 12.5, and so on.


This kind of decay is contrasted with linear decay, where the amount decreases by a fixed quantity each interval. With exponential decay, the speed of decay slows down as fewer and fewer atoms remain.
**Practical Application:** Radioactive dating, like in this exercise, makes use of this concept to determine the age of archaeological finds by measuring the ratios of remaining radioactive material and its decay products.
natural logarithm
The **natural logarithm** (ln) is a logarithm to the base \(e\), where \(e\) is approximately 2.71828. It’s often used in calculations involving exponential growth or decay due to its unique properties that simplify these types of equations.

In this problem, we used the natural logarithm to solve for time \(t\):
  1. We started with the decay equation:
    \[ \frac{1}{4} = \frac{1}{2}^{t / 1620} \]
  2. Took the natural logarithm of both sides:
    \[ \text{ln} \frac{1}{4} = \text{ln} \frac{1}{2}^{t / 1620} \]
  3. Applied the logarithm power rule, which states \( \text{ln}(a^b) = b \text{ln}(a) \):
    \[ \text{ln} \frac{1}{4} = \frac{t}{1620} \text{ln} \frac{1}{2} \]

By using the properties of logarithms, we simplified the equation, making it easier to solve for \(t\).

The natural logarithm helps convert exponential equations into linear ones, which are easier to manipulate algebraically. In general, natural logarithms are extremely useful when dealing with problems related to exponential processes, such as radioactive decay, population growth, and certain financial calculations.

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