/*! 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 12 The half-life of \({ }^{14} \mat... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 12

The half-life of \({ }^{14} \mathrm{C}\) is 5730 years. If the original amount of \({ }^{14} \mathrm{C}\) in a particular living organism is \(20 \mathrm{~g}\) and that found in a fossil of that organism is \(0.01 \mathrm{~g}\), determine the approximate age of the fossil.

Short Answer

Expert verified
The approximate age of the fossil is 62970 years.

Step by step solution

01

Understand Half-Life Formula

The formula to determine the remaining quantity of a substance after a certain period given its half-life is \[ N(t) = N_0 \times \frac{1}{2}^{\frac{t}{t_{1/2}}} \] where: \(N(t)\) = remaining amount after time t,\(N_0\) = initial amount,\(t\) = time elapsed,\(t_{1/2}\) = half-life.
02

Identify Given Values

Determine the values provided in the problem: Initial amount, \(N_0 = 20 \text{ g}\),Remaining amount, \(N(t) = 0.01 \text{ g}\),Half-life of \({}^{14} \text{C} = 5730 \text{ years}.\)
03

Set Up the Equation

Substitute the given values into the half-life formula: \[ 0.01 = 20 \times \frac{1}{2}^{\frac{t}{5730}} \]
04

Solve for \( t \)

First, isolate the exponential part by dividing both sides by 20: \[ \frac{0.01}{20} = \frac{1}{2}^{\frac{t}{5730}}\] which simplifies to: \[ 0.0005 = \frac{1}{2}^{\frac{t}{5730}} \]Take the natural logarithm of both sides to solve for t: \[ \text{ln}(0.0005) = \text{ln}\big(\frac{1}{2}^{\frac{t}{5730}}\big)\]Using the properties of logarithms, this becomes: \[ \text{ln}(0.0005) = \frac{t}{5730} \times \text{ln}\big(\frac{1}{2}\big)\]
05

Calculate the Time

Solve for t: \[ t = \frac{\text{ln}(0.0005)}{\text{ln}(\frac{1}{2})} \times 5730 \]Use the natural logarithm values: \(\text{ln}(0.0005) 鈮� -7.6009\)\(\text{ln}(0.5) 鈮� -0.6931\)Thus: \[ t 鈮� \frac{-7.6009}{-0.6931} \times 5730 \] which simplifies to: \[ t 鈮� 11 \times 5730 鈮� 62970 \text{ years} \]

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Key Concepts

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

Radioactive Decay
Radioactive decay is a process by which an unstable atomic nucleus loses energy by emitting radiation. This leads to a transformation of the atom into a different element or a different isotope of the same element. The rate of decay is specific to each radioactive isotope and can be described using concepts like half-life.
Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. In the context of radioactive decay, these equations help model how the amount of a radioactive substance decreases over time. For example, the equation \(\frac{dN}{dt} = -\frac{N}{\tau}\) describes the rate of decay, where \(N\) is the quantity of the substance, and \( \tau \) is a constant related to the half-life. This differential equation can be solved to get the decay equation.
Natural Logarithms
Natural logarithms (ln) are logarithms with the base \( e \), where \( e \) is approximately equal to 2.71828. In radioactive decay calculations, natural logarithms are used to solve for time when the decay equation is in exponential form. For instance, to find the age of a fossil, we often take the natural logarithm of both sides of the half-life equation to isolate the variable representing time. This step simplifies solving otherwise complex exponential equations.

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Most popular questions from this chapter

A tank contains \(100 \mathrm{gal}\) of a brine solution in which \(20 \mathrm{lb}\) of salt is initially dissolved. (a) Water (containing no salt) is then allowed to flow into the tank at a rate of \(4 \mathrm{gal} / \mathrm{min}\) and the well-stirred mixture flows out of the tank at an equal rate of \(4 \mathrm{gal} / \mathrm{min}\). Determine the amount of salt \(y(t)\) at any time \(t\). What is the eventual concentration of the brine solution in the tank? (b) If instead of water a brine solution with concentration \(2 \mathrm{lb} /\) gal flows into the tank at a rate of \(4 \mathrm{gal} / \mathrm{min}\), what is the eventual concentration of the brine solution in the tank?

A ball of weight \(4 \mathrm{oz}\) is tossed into the air with an initial velocity of \(64 \mathrm{ft} / \mathrm{s}\). (a) Find the velocity of the object at time \(t\) if the air resistance is equivalent to \(1 / 16\) the instantaneous velocity. (b) When does the ball reach its maximum height?

If the limiting velocity of an object of mass \(m\) that is thrown vertically into the air with an initial velocity \(v_{0} \mathrm{~m} / \mathrm{s}\) is \(-9.8 \mathrm{~m} / \mathrm{s}\), what can be said about the relationship between \(m\) and \(c\) in the force due to air resistance \(F_{R}=c v\) acting on the object?

Suppose that we deposit a sum of money in a money market fund that pays interest at an annual rate \(k\), and let \(S(t)\) represent the value of the investment at time \(t\). If the compounding takes place continuously, then the rate at which the value of the investment changes is the interest rate times the value of the investment, \(d S / d t=k S\). Use this equation to find \(S(t)\) if \(S(0)=S_{0}\).

Suppose that a culture of bacteria has an initial population of \(n=100\). If the population doubles every three days, determine the number of bacteria present after 30 days. How much time is required for the population to reach 4250 in number?
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