/*! 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 4 Consider a radioactive substance... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 4

Consider a radioactive substance with halflife 10 days. If there are initially \(5000 \mathrm{~g}\) of the substance, how much remains after 365 days?

Short Answer

Expert verified
After 365 days, approximately 6.505 x 10^{-8} grams remain.

Step by step solution

01

Understand the Problem

The exercise involves calculating the remaining amount of a radioactive substance after a given time, given its half-life.
02

Write Down the Half-Life Formula

The amount of a radioactive substance remaining can be calculated using the formula: defining: \(A(t) = A_0 \times (1/2)^{(t/T_{1/2})}\)
03

Identify the Given Values

From the problem, the initial amount \(A_0 = 5000 \text{ grams}\). The half-life \(T_{1/2} = 10 \text{ days}\). Time elapsed \(t = 365 \text{ days}\).
04

Substitute the Values into the Formula

Plug the given values into the formula: \(A(365) = 5000 \times (1/2)^{(365/10)}\)
05

Simplify the Exponent

Calculate the exponent: \(365/10 = 36.5\). So the formula becomes: \(A(365) = 5000 \times (1/2)^{36.5}\)
06

Calculate the Remaining Amount

Use a calculator to compute the value of \((1/2)^{36.5}\) and then multiply by 5000: \(A(365) \approx 5000 \times 1.301 \times 10^{-11} \approx 6.505 \times 10^{-8} \text{ grams}\)
07

Verify the Calculation

Review each step to ensure there are no mistakes in the calculations.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Half-Life
Half-life is the time required for half of the radioactive substance to decay. Imagine you have 100 grams of a substance with a half-life of 5 days. After 5 days, only 50 grams remain.
The half-life doesn't change, no matter how much substance you start with.
  • First period: 100 grams becomes 50 grams.
  • Second period: 50 grams becomes 25 grams.
This pattern continues as time progresses.
Exponential Decay Explained
Radioactive decay follows an exponential trend. This means that the rate of decay is proportional to the amount of substance left.
At the start, decay happens quickly because there's a lot of the substance. Over time, less substance remains, and decay happens more slowly. The decay can be modeled mathematically using exponential functions.
The equation looks like this:
\(A(t) = A_0 \times (\frac{1}{2})^{(\frac{t}{T_{1/2}})}\)
where:
  • \(A(t)\) is the amount of substance left after time \(t\).
  • \(A_0\) is the initial amount of the substance.
  • \(T_{1/2}\) is the half-life of the substance.
What Are Radioactive Substances?
Radioactive substances contain unstable atoms. These atoms lose energy over time by emitting radiation in the form of particles or electromagnetic waves.
  • Common examples include uranium, radon, and carbon-14.
  • These substances can be found naturally or made in labs.
Radioactive decay processes differ based on the type of particle emitted, such as alpha, beta, or gamma decay.
Initial Amount: The Starting Point
The initial amount \((A_0)\) is the quantity of the radioactive substance at the beginning of the observation.
In the given problem, the initial amount is 5000 grams. This value is crucial because all calculations will hinge on this starting value. Without knowing this, we can鈥檛 predict how much will remain later.
Time Elapsed and Its Role
Time elapsed \((t)\) is the time duration over which decay is observed. In this problem, we're looking at 365 days.
The formula incorporates this to allow tracking of decay over any period. Radionuclides can decay over seconds to millions of years, depending on their half-lives.
Putting it All Together: The Decay Formula
This is the formula we use: \(A(t) = A_0 \times (\frac{1}{2})^{(\frac{t}{T_{1/2}})}\).
We calculate ahead by plugging in our known values:
  • \(A_0 = 5000 \text{ grams}\)
  • \(T_{1/2} = 10 \text{ days}\)
  • \(t = 365 \text{ days}\)
So, it becomes: \(A(365) = 5000 \times (\frac{1}{2})^{(\frac{365}{10})}\).
From here, simplify the term inside the exponent: \(\frac{365}{10} = 36.5\).
Now the formula is: \(A(365) = 5000 \times (\frac{1}{2})^{36.5}\).
Finally, use a calculator to find \( (\frac{1}{2})^{36.5} \approx 1.301 \times 10^{-11} \), and multiply by 5000 to get the remaining amount: \(A(365) \approx 6.505 \times 10^{-8} \text{ grams}\).
This exercise illustrates how drastically radioactive substances can decay over a long period.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(Tumor and Organism Growth) Ludwig von Bertalanffy (1901-1972) made valuable contributions to the study of organism growth, including the growth of tumors, based on the relationship between body size of the organism and the metabolic rate. He theorized that weight is directly proportional to volume and that the metabolic rate is proportional to surface area. This indicates that surface area is proportional to \(V^{2 / 3}\) because \(V\) is measured in cubic units and \(S\) in square units. Therefore, Bertalanffy studied the IVP \(d V / d t=a V^{2 / 3}-b V, V(0)=V_{0}\). (a) Solve this initial value problem to show that volume is given by \(V(t)=\) \(\left[\frac{a}{b}-\left(\frac{a}{b}-V_{0}^{1 / 3}\right) e^{-b t / 3}\right]^{3}\). (b) Find \(\lim _{t \rightarrow \infty} V(t)\) and explain what the limit represents. (c) This initial value problem is similar to that involving the Logistic equation \(d V / d t=a V^{2}-b V, V(0)=V_{0}\). Compare the limiting volume between the Logistic equation and the model proposed by Bertalanffy in the cases when \(a / b>1\) and when \(a / b<1\).

If \(k=1 / 4, A(t)=0.25, B(t)=1.75\left(u_{d}-u(t)\right)\), and \(C(t)=70-10 \cos (\pi t / 12)\), determine the value of \(u_{d}\) needed so that the average temperature in the building over a \(24 \mathrm{~h}\) period is \(70^{\circ} \mathrm{F}\).

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?

Suppose that the population of a small town is initially 5000 . Due to the construction of an interstate highway, the population doubles over the next year. If the rate of growth is proportional to the current population, when will the population reach 25,000 ? What is the population after five years?

A certain group of people with initial population 10,000 grows at a rate proportional to the number present. The population doubles in five years. In how many years will the population triple?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.