/*! 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 26 The half-life of a radioactive i... [FREE SOLUTION] | 91Ó°ÊÓ

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The half-life of a radioactive isotope is 140 d. How many days would it take for the decay rate of a sample of this isotope to fall to one-fourth of its initial value?

Short Answer

Expert verified
The decay rate falls to one-fourth of its initial value in 280 days.

Step by step solution

01

Understand the Concept of Half-life

The half-life of a substance is the time it takes for half of the material or its activity to decay. In this problem, the substance has a half-life of 140 days.
02

Set Up the Decay Formula

The decay formula relating the decay constant \( k \), half-life \( t_{1/2} \), and decay is: \( N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \). We want to find \( t \) for which \( N(t) = \frac{1}{4}N_0 \).
03

Set up the Equation for the Problem

Using the decay formula, \( \left(\frac{1}{2}\right)^{\frac{t}{140}} = \frac{1}{4} \). Since \( \frac{1}{4} = \left(\frac{1}{2}\right)^2 \), the equation simplifies to \( \left(\frac{1}{2}\right)^{\frac{t}{140}} = \left(\frac{1}{2}\right)^2 \).
04

Solve the Equation

Because the bases are equal, equate the exponents: \( \frac{t}{140} = 2 \). Solve for \( t \) by multiplying both sides by 140, which gives \( t = 280 \) days.

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Key Concepts

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

Half-life
The concept of half-life is fundamental in understanding radioactive decay. It refers to the amount of time it takes for half of a given quantity of a radioactive isotope to decay into a more stable form. In other words, if you start with a sample of 100 atoms, after one half-life, only 50 atoms of the original isotope remain, assuming no other processes affect this atom count. The remaining atoms would have decayed into another element or isotope.

Half-life is a constant value unique to each radioactive substance, and it helps scientists predict the rate at which a sample will decay.
For example:
  • A shorter half-life means the substance decays rapidly.
  • A longer half-life indicates a slower decay process.
The half-life of an element can be used in various fields such as archaeology for dating artifacts, medicine for cancer treatment, and nuclear energy production. By knowing the half-life, you can determine how long it will take for a substance to reach a certain level of radioactivity or to stabilize.
Decay Rate
Decay rate refers to the speed at which a radioactive isotope undergoes decay. This term is often used interchangeably with activity, which is a measure of how many decay events occur per unit of time.
Decay rate decreases over time as the radioactive material diminishes, but initially, it can be quite high. This rate is directly related to the amount of a substance and its half-life.

When you are considering a sample's
  • Initial decay rate: This is when the sample has just been measured.
  • Future decay rate: This is calculated by understanding how much of the substance remains after a certain period of time, usually measured in half-lives.
In our problem, the decay rate falls to one-fourth of its initial value over 280 days. Understanding decay rate is crucial to predicting how radioactive materials behave over time, which has significant implications for safety measures and environmental assessments.
Decay Constant
A decay constant is a number that gives more precise information about how rapidly a radioactive isotope decays. It represents the probability per unit time of an atom decaying. This constant is symbolized by the letter \(k\) and can be calculated using the half-life with the formula:
\[ k = \frac{\ln(2)}{t_{1/2}} \]where \(t_{1/2}\) is the half-life.

The value of the decay constant \(k\) is crucial for calculating decay processes because:
  • It quantifies the likelihood of decay per unit time for a single remaining atom or unit of substance.
  • It helps describe the exponential nature of radioactive decay mathematically.
The decay constant is vital for understanding how quickly a given radioactive isotope will diminish. In practical terms, knowing both the decay constant and the amount of substance can help scientists and engineers design systems that manage the decay products effectively.

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Most popular questions from this chapter

The isotope \({ }^{40} \mathrm{~K}\) can decay to either \({ }^{40} \mathrm{Ca}\) or \({ }^{40} \mathrm{Ar} ;\) assume both decays have a half-life of \(1.26 \times 10^{9} \mathrm{y}\). The ratio of the Ca produced to the Ar produced is \(8.54 / 1=8.54 .\) A sample originally had only \({ }^{40} \mathrm{~K}\). It now has equal amounts of \({ }^{40} \mathrm{~K}\) and \({ }^{40} \mathrm{Ar}\); that is, the ratio of \(\mathrm{K}\) to \(\mathrm{Ar}\) is \(1 / 1=1 .\) How old is the sample? (Hint: Work this like other radioactive-dating problems, except that this decay has two products.)

What is the binding energy per nucleon of the europium isotope \(\frac{152}{63} \mathrm{Eu}\) ? Here are some atomic masses and the neutron mass. $$\begin{array}{lr}\frac{152}{63} \mathrm{Eu} & 151.921742 \mathrm{u} \\\\\mathrm{n} & 1.008665 \mathrm{u}\end{array} { }^{1} \mathrm{H} \quad 1.007825 \mathrm{u}$$

(a) Show that the total binding energy \(E_{\mathrm{bc}}\) of a given nuclide is $$E_{\mathrm{be}}=Z \Delta_{\mathrm{H}}+N \Delta_{\mathrm{n}}-\Delta$$ where \(\Delta_{\mathrm{H}}\) is the mass excess of \({ }^{1} \mathrm{H}, \Delta_{\mathrm{n}}\) is the mass excess of a neutron, and \(\Delta\) is the mass excess of the given nuclide. (b) Using this method, calculate the binding energy per nucleon for \({ }^{197}\) Au. Compare your result with the value listed in Table \(42-1 .\) The needed mass excesses, rounded to three significant figures, are \(\Delta_{\mathrm{H}}=+7.29 \mathrm{MeV}\) \(\Delta_{n}=+8.07 \mathrm{MeV},\) and \(\Delta_{197}=-31.2 \mathrm{MeV} .\) Note the economy of calculation that results when mass excesses are used in place of the actual masses.

A \({ }^{7} \mathrm{Li}\) nucleus with a kinetic energy of \(3.00 \mathrm{MeV}\) is sent toward a \({ }^{232} \mathrm{Th}\) nucleus. What is the least center-to- center separation between the two nuclei, assuming that the (more massive) \({ }^{232} \mathrm{Th}\) nucleus does not move?

A \(5.00 \mathrm{~g}\) charcoal sample from an ancient fire pit has a \({ }^{14} \mathrm{C}\) activity of 63.0 disintegrations/min. A living tree has a \({ }^{14} \mathrm{C}\) activity of 15.3 disintegrations/min per \(1.00 \mathrm{~g}\). The half-life of \({ }^{14} \mathrm{C}\) is \(5730 \mathrm{y}\). How old is the charcoal sample?

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