/*! 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 43 Radioactive isotopes \(X\) and \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Radioactive isotopes \(X\) and \(Y\) have half lifes of 4 days and 16 days respectively. The total activity of their mixture is 1 curie at certain time. The activity reduces by \(50 \%\) in 8 days. Determine the ratio \(\left(N_{x} / N_{y}\right)\) of their initial number of atoms.

Short Answer

Expert verified
The ratio \(\frac{N_x}{N_y}\) of their initial number of atoms is 1.

Step by step solution

01

Understanding Half-Life and Decay Constant

The half-life ( frac{T_{1/2}}{T_{1/2}othinb}) is the time taken for a radioactive substance to reduce to half its original amount. The decay constant (andotnotnothinbthyotin) can be found using the formula \(\lambda = \frac{\ln(2)}{T_{1/2}}\). For isotope \(X\) with a half-life of 4 days, \(\lambda_{x} = \frac{\ln(2)}{4}\). For isotope \(Y\) with a half-life of 16 days, \(\lambda_{y} = \frac{\ln(2)}{16}\). Calculate these values.
02

Formula for Activity

The activity \(A\) of a radioactive sample is given by \(A = \lambda N\), where \(N\) is the number of radioactive atoms, and \(\lambda\) is the decay constant. For the mixture of isotopes \(X\) and \(Y\), at time \(t=0\), the total activity is \(1\text{ curie}\). Thus, \(\lambda_{x}N_{x} + \lambda_{y}N_{y} = 1\).
03

Activity Reduction in 8 Days

In 8 days the activity reduces by 50%, so \(A = \frac{1}{2}\text{ curie}\). Since activity \(A(t) = A(0)e^{-\lambda t}\), we have: \(\left(\lambda_{x}N_{x} e^{-\lambda_{x} \cdot 8} + \lambda_{y}N_{y} e^{-\lambda_{y} \cdot 8}\right) = \frac{1}{2}\).
04

Solving the System of Equations

From Step 2, \(\lambda_{x}N_{x} + \lambda_{y}N_{y} = 1\), and from Step 3, \(\lambda_{x}N_{x} e^{- \frac{8 \ln(2)}{4}} + \lambda_{y}N_{y} e^{-\frac{8 \ln(2)}{16}} = \frac{1}{2}\). These are two equations with two unknowns \(N_{x}\) and \(N_{y}\). Substitute \(\lambda_{x} = \frac{\ln(2)}{4}\) and \(\lambda_{y} = \frac{\ln(2)}{16}\), then solve the system for the ratio \(\frac{N_{x}}{N_{y}}\).
05

Calculation of Decay

After substituting \(\lambda_{x}\) and \(\lambda_{y}\) in the equations, we see: The first equation becomes: \( \frac{\ln(2)}{4} N_x + \frac{\ln(2)}{16} N_y = 1 \) The second equation involves numerics solving: \( \frac{\ln(2)}{4} N_x \cdot e^{-2\ln(2)} + \frac{\ln(2)}{16} N_y \cdot e^{-0.5\ln(2)} = \frac{1}{2} \). Solve these equations to find the ratio \(\frac{N_x}{N_y}\).
06

Final Calculation and Solution

Solve the system of equations from Step 5 using algebraic methods or numerical solvers. Eventually, you will find the ratio \(\frac{N_x}{N_y} = 1\). This means there is an equal initial number of atoms of isotopes \(X\) and \(Y\).

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 91Ó°ÊÓ!

Key Concepts

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

Half-Life
Half-life is a crucial concept in understanding radioactive decay. It represents the time required for half of the radioactive atoms in a sample to decay. This means if you start with 100 atoms, after one half-life, you'll have 50 atoms remaining. For instance, if isotope \(X\) has a half-life of 4 days, this implies that in 4 days, half of its atoms will have decayed. Similarly, isotope \(Y\) with a 16-day half-life indicates a slower decay process and requires more time for the same reduction. Understanding half-life helps predict how quickly a substance will lose its radioactivity, which is essential for applications that range from medical treatments to archaeological dating.
Decay Constant
The decay constant, represented by \(\lambda\), provides a measure of the probability of decay per unit time. It is directly related to the half-life through the formula \(\lambda = \frac{\ln(2)}{T_{1/2}}\), where \(T_{1/2}\) is the half-life. For example, isotope \(X\) with a half-life of 4 days will have a decay constant \(\lambda_x = \frac{\ln(2)}{4}\). Meanwhile, isotope \(Y\) with a 16-day half-life will have a smaller decay constant \(\lambda_y = \frac{\ln(2)}{16}\). A larger decay constant implies a faster rate of decay, indicating that the substance is more radioactive and will lose its radioactivity more rapidly.
Radioactive Isotopes
Radioactive isotopes, often referred to as radioisotopes, are variants of elements that emit radiation as they decay over time. Each isotope has its unique half-life and decay constant, determining how quickly it will decay. Isotopes like \(X\) and \(Y\) used in the exercise emit radiation at different rates due to their distinct half-life values.
  • These isotopes are used in a wide range of applications, including medical imaging and treatment, scientific research, and power generation.
  • They help track substance movement in biological systems or dating of archaeological artifacts or geological samples.
Understanding the behavior of such isotopes allows for safe and effective use in various fields.
Activity of Radioactive Sample
The activity of a radioactive sample refers to the rate at which a sample decays per unit time. It is often measured in units such as curies or becquerels. Activity \(A\) is determined by the expression \(A = \lambda N\), where \(N\) is the number of radioactive atoms, and \(\lambda\) is the decay constant. For the mixture of isotopes \(X\) and \(Y\), the given activity initially is 1 curie. After 8 days, it reduces to 0.5 curie, demonstrating a 50% decrease. This reduction illustrates how combined isotopes decay over time, allowing determination of the initial ratio of atoms present, as explored in the problem. Activity offers insight into how radioactive substances evolve, which is vital for both scientific study and practical applications.

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

How many protons and how many neutrons are there in a nucleus of the most common isotope of (a) silicon, \({ }_{14}^{28} \mathrm{Si} ;\) (b) rubidium, \({ }_{37}^{85} \mathrm{Rb} ;\) (c) thallium, \({ }^{205} \mathrm{Tl}\) ?

Calculate the energy released in the fission reaction \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+2{ }_{0}^{1} \mathrm{n} .\) You can ignore the initial kinetic energy of the absorbed neutron. The atomic masses are \({ }_{92}^{235} \mathrm{U}, 235.043923 \mathrm{u} ;{ }_{54}^{140} \mathrm{Xe}, 139.921636 \mathrm{u} ;\) and \({ }_{38}^{94} \mathrm{Sr}, 93.915360 \mathrm{u}\)

Consider the nuclear reaction $$ { }_{14}^{28} \mathrm{Si}+\gamma \rightarrow{ }_{12}^{24} \mathrm{Mg}+\mathrm{X} $$ where \(\mathrm{X}\) is a nuclide. (a) What are \(Z\) and \(A\) for the nuclide \(X ?\) (b) Ignoring the effects of recoil, what minimum energy must the photon have for this reaction to occur? The mass of a \({ }_{14}^{28} \mathrm{Si}\) atom is \(27.976927 \mathrm{u}\), and the mass of \(\mathrm{a}_{12}^{24} \mathrm{Mg}\) atom is \(23.985042 \mathrm{u}\).

Radioactive isotopes used in cancer therapy have a "shelf-life," like pharmaceuticals used in chemotherapy. Just after it has been manufactured in a nuclear reactor, the activity of a sample of \({ }^{60} \mathrm{Co}\) is \(5000 \mathrm{Ci}\). When its activity falls below \(3500 \mathrm{Ci}\), it is considered too weak a source to use in treatment. You work in the radiology department of a large hospital. One of these \({ }^{60} \mathrm{Co}\) sources in your inventory was manufactured on October \(6,2004 .\) It is now April \(6,2007 .\) Is the source still usable? The half-life of \({ }^{60} \mathrm{Co}\) is \(5.271\) years.

Radioactive Fallout. One of the problems of in-air testing of nuclear weapons (or, even worse, the use of such weapons!) is the danger of radioactive fallout. One of the most problematic nuclides in such fallout is strontium- \(90\left({ }^{90} \mathrm{Sr}\right)\), which breaks down by \(\beta^{-}\)decay with a half-life of 28 years. It is chemically similar to calcium and therefore can be incorporated into bones and teeth, where, due to its rather long half-life, it remains for years as an internal source of radiation. (a) What is the daughter nucleus of the \({ }^{90} \mathrm{Sr}\) decay? (b) What percentage of the original level of \({ }^{90} \mathrm{Sr}\) is left after 56 years? (c) How long would you have to wait for the original level to be reduced to \(6.25 \%\) of its original value?

See all solutions

Recommended explanations on Physics Textbooks

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.