/*! 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 The activity of a certain radion... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 4

The activity of a certain radionuclide decreases to 15 percent o its original value in \(10 \mathrm{~d}\). Find its half-life.

Short Answer

Expert verified
The half-life is approximately 3.56 days.

Step by step solution

01

Understanding the Problem

The activity of a radionuclide decreases to 15% of its original value in 10 days. We need to find the half-life, which is the time taken for the activity to reduce to 50% of its initial value.
02

Setting up the Exponential Decay Formula

The decay of a radionuclide follows the exponential decay law: \[ A(t) = A_0 e^{-kt} \] where \( A(t) \) is the activity at time \( t \), \( A_0 \) is the initial activity, and \( k \) is the decay constant.
03

Applying Given Conditions

We know that after 10 days, the activity is 15% of the original, so \[ A(10) = 0.15A_0 \] Substitute into the decay formula: \[ 0.15A_0 = A_0 e^{-10k} \] Cancel \( A_0 \) from both sides: \[ 0.15 = e^{-10k} \]
04

Solving for the Decay Constant \( k \)

Take the natural logarithm of both sides to solve for \( k \): \[ \ \ln(0.15) = -10k \] Simplifying: \[ k = -\frac{\ln(0.15)}{10} \]
05

Finding the Half-Life

The half-life \( T_{1/2} \) is the time when the activity is half of the original, described by \[ 0.5 = e^{-kT_{1/2}} \] Again, take the natural logarithm: \[ \ln(0.5) = -kT_{1/2} \] Solve for \( T_{1/2} \): \[ T_{1/2} = \frac{\ln(2)}{k} \]
06

Calculating the Half-Life

Substitute \( k = -\frac{\ln(0.15)}{10} \) into the equation for \( T_{1/2} \): \[ T_{1/2} = \frac{\ln(2)}{-\frac{\ln(0.15)}{10}} \] Simplify this to find the numeric value of the half-life.

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.

Exponential Decay
Exponential decay is a mathematical concept that describes the process by which a quantity decreases at a rate proportional to its current value. In the context of radioactive decay, it models how unstable atoms, or radionuclides, lose their activity over time. This type of decay is characterized by a consistent percentage loss over equal time periods.
To describe this mathematically, we use the formula:
  • Initial Activity (\(A_0\)): This represents the starting value or the original activity of the radionuclide.
  • Time (\(t\)): The period over which the decay is being measured.
  • Decay Constant (\(k\)): A constant that represents the rate at which the radionuclide decays.
The overall formula is: \[ A(t) = A_0 e^{-kt} \]From this, we'll see that the activity decreases exponentially as time increases, which means it will never truly reach zero, just get closer and closer to it.
Radionuclide
A radionuclide is an atom with an unstable nucleus that becomes more stable by emitting radiation in the form of particles or energy. This process is what we call radioactive decay. Radionuclides can be found naturally in the environment, such as uranium and radon, or be artificially created in labs.
Radionuclides are significant in various fields:
  • Medical Use: They are used in treatment and diagnostics, such as in PET scans and cancer radiotherapy.
  • Energy Production: Nuclear reactions in power plants use radionuclides to generate energy.
  • Research: Understanding decay processes helps scientists in dating ancient artifacts through radiometric techniques.
When studying radionuclides, one of the key aspects is understanding their half-life, which allows us to quantify how long it takes for half of the substance to decay. This is crucial for both practical applications and theoretical understanding of radioactive decay.
Decay Constant
The decay constant, often represented by \(k\), is a crucial parameter in the study of exponential decay, especially in the field of radioactive decay. It quantifies the rate at which a particular radioactive substance decays over time. The larger the decay constant, the more rapidly the substance undergoes decay.
To find the decay constant, we can rewrite the equation for exponential decay:\[ A(t) = A_0 e^{-kt} \]From this equation:
  • The decay constant \(k\) aligns with the speed of the decay process.
  • By knowing the value of \(k\), you can estimate how quickly the activity of a radionuclide will decrease.
  • It is involved in calculating the half-life \(T_{1/2}\) using the following relation: \[ T_{1/2} = \frac{\ln(2)}{k} \]
This shows us that by determining the decay constant, we can establish how long it will take for half of the radionuclide to decay, giving a clearer picture of its stability and how it behaves over time.

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

Positron emission resembles electron emission in all respects except that the shapes of their respective energy spectta are different: there are many low- energy electrons emitted, but few low-energy positrons. Thus the average electron energy in beta decay is about \(0.3 \mathrm{KE}_{\max }\), whereas the aversge positron energy is about \(0.4 \mathrm{KE}_{\text {nox }}\). Can you suggest a simple reason for this difference?

As discussed in this chapter, the heaviest nuclides are probably created in supernova explosions and become distributed in the galactic matter from which later stars (and their planets) form. Under the assumption that equal amounts of the \({ }^{235} \mathrm{U}\) and \({ }^{236} \mathrm{U}\) now in the earth were created in this way in the same supernova, calculate how long ago this occurred from their respective observed relative abundances of \(0.71\) and \(99.3\) percent and respective half-lives of \(7.0 \times 10^{8} \mathrm{y}\) and \(4.5 \times 10^{9} \mathrm{y}\).

The activity of a sample of an unknown radionuclide is measured at daily intervals. The results, in MBq, are 32.1, \(27.2,23.0,19.5\), and \(16.5\). Find the half-life of the radionuclide.

Derive Eq. \((12,11), \mathrm{KE}_{a}=(A-4) Q / A\), for the kinetic energy of the alpha particle released in the decay of a mucleus of mass number \(A\). Assume that the ratio \(M_{a} / M_{d}\) between the mass of an alpha particle and the mass of the daughter is \(=4 /(A-4)\).

A slab of absorber is exactly one mean free path thick for a beam of certain incident particles. What percentage of the particles will cmerge from the slab?
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Electromagnetism

Read Explanation

Radiation

Read Explanation

Force

Read Explanation

Modern Physics

Read Explanation

Torque and Rotational Motion

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.