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Measurements on a certain isotope tell you that the decay rate decreases from 8318 decays/min to 3091 decays/min in 4.00 days. What is the half-life of this isotope?

Short Answer

Expert verified
The half-life of the isotope is approximately 2.47 days.

Step by step solution

01

Understand the Problem

We are given that the decay rate of an isotope decreases from 8318 decays/min to 3091 decays/min in 4.00 days, and we need to find the half-life of the isotope. The half-life is the time required for the decay rate to reduce to half of its initial value.
02

Use the Exponential Decay Formula

The decay of radioactive isotopes follows the exponential decay model, represented by the equation \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the quantity at time \( t \), \( N_0 \) is the initial quantity, \( k \) is the decay constant, and \( t \) is time.
03

Calculate Decay Constant \( k \)

Given \( N_0 = 8318 \) and \( N(t) = 3091 \) at \( t = 4.00 \) days, substitute these into the equation: \( 3091 = 8318 e^{-4k} \). Solve for \( k \) by first dividing both sides by 8318 to get \( \frac{3091}{8318} = e^{-4k} \). Take the natural logarithm of both sides to find \( k \): \( -4k = \ln\left(\frac{3091}{8318}\right) \), which gives \( k = -\frac{1}{4} \ln\left(\frac{3091}{8318}\right) \).
04

Solve for Half-Life \( T_{1/2} \)

The half-life \( T_{1/2} \) is related to the decay constant \( k \) by the formula \( T_{1/2} = \frac{\ln(2)}{k} \). Using the value of \( k \) calculated previously, substitute into the half-life formula to find the half-life.
05

Calculate the Numerical Answer

First calculate \( k \) from Step 3 using \( k = -\frac{1}{4} \ln\left(\frac{3091}{8318}\right) \). After calculating, use \( T_{1/2} = \frac{\ln(2)}{k} \) to find the half-life. Ensure that all calculations maintain unit consistency, with \( t \) in 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 crucial in understanding radioactive decay. It refers to the time it takes for half of a given amount of a radioactive substance to decay. In this context, the decay rate of a radioactive isotope is the number of disintegrations per minute, and finding the half-life helps in predicting how long it will take for the isotope to reduce to half its initial activity.
The half-life is expressed using the formula:\[ T_{1/2} = \frac{\ln(2)}{k} \]where:
  • \( T_{1/2} \): Half-life
  • \( k \): Decay constant
  • \(\ln(2)\): Natural logarithm of 2, approximately equal to 0.693
By knowing the half-life, you can determine the longevity and decay behavior of the isotope over time, making it an essential metric for understanding the stability and risks associated with radioactive materials.
Radioactive Decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This spontaneous transformation results in the conversion of an atom of one element into an atom of a different element, often with the emission of smaller particles like alpha, beta particles, and sometimes gamma rays.
The decay is described by an exponential function, emphasizing that the rate of decay is proportional to the number of atoms present. As time progresses, the radioactive atoms undergo exponential decay, gradually reducing their numbers.
  • The decay rate is often measured in units of decays per minute or per second.
  • It's a random process on the microscopic level, which means you cannot predict when a single atom will decay.
  • Overall, the statistical nature of decay processes allows for precise predictions of behavior in large samples.
Understanding radioactive decay is essential in fields such as nuclear medicine, environmental science, and archaeological dating (using Carbon-14 isotope).
Decay Constant
The decay constant, denoted by \( k \), is a critical factor in determining the rate of radioactive decay. It represents the probability of decay of a radioactive isotope per unit time and is a unique value for each type of isotope.
This constant is used in exponential decay equations of the form:\[ N(t) = N_0 e^{-kt}\]where:
  • \( N(t) \): Number of radioactive atoms remaining at time \( t \)
  • \( N_0 \): Initial number of radioactive atoms
  • \( e \): Base of the natural logarithm, approximately equal to 2.718
  • \( k \): Decay constant
To determine \( k \), you can rearrange the decay equation and use logarithms, as shown in the example problem.
By knowing \( k \), you can calculate how fast or slow an isotope decays and thus predict the reduction of the isotope over time.
Isotopes
Isotopes are different forms of the same element, which have the same number of protons but different numbers of neutrons in their nuclei. This gives rise to variations in atomic mass, making some isotopes stable and others unstable.
Unstable isotopes are radioactive and undergo decay, while stable isotopes generally do not change over time.
  • Each isotope of an element has its unique decay properties, including its decay constant.
  • Some isotopes are naturally occurring, while others are artificially created in laboratories.
  • Radioactive isotopes can be used for dating archaeological finds, medical diagnostics and treatments, and energy generation in nuclear reactors.
Understanding isotopes is fundamental in numerous scientific fields, as they provide insight into atomic structure and serve practical applications across various industries.

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Most popular questions from this chapter

\(^2$$^3$$^8\)U decays spontaneously by \(\alpha\) emission to \(^2$$^3$$^4\)Th. Calculate (a) the total energy released by this process and (b) the recoil velocity of the \(^2$$^3$$^4\)Th nucleus. The atomic masses are 238.050788 u for \(^2$$^3$$^8\)U and 234.043601 u for \(^2$$^3$$^4\)Th.

Consider the fusion reaction \(^{2}_{1}H\) + \(^{2}_{1}H \rightarrow ^{3}_{2}He + ^{1}_{0}n\). (a) Estimate the barrier energy by calculating the repulsive electrostatic potential energy of the two \(^{2}_{1}H\) nuclei when they touch. (b) Compute the energy liberated in this reaction in MeV and in joules. (c) Compute the energy liberated \(per\) \(mole\) of deuterium, remembering that the gas is diatomic, and compare with the heat of combustion of hydrogen, about \(2.9 \times 10^{5} J/mol\).

\(\textbf{We Are Stardust.}\) In 1952 spectral lines of the element technetium-99 (\(^9$$^9\)Tc) were discovered in a red giant star. Red giants are very old stars, often around 10 billion years old, and near the end of their lives. Technetium has no stable isotopes, and the half-life of \(^9$$^9\)Tc is 200,000 years. (a) For how many halflives has the \(^9$$^9\)Tc been in the red giant star if its age is 10 billion years? (b) What fraction of the original \(^9$$^9\)Tc would be left at the end of that time? This discovery was extremely important because it provided convincing evidence for the theory (now essentially known to be true) that most of the atoms heavier than hydrogen and helium were made inside of stars by thermonuclear fusion and other nuclear processes. If the \(^9$$^9\)Tc had been part of the star since it was born, the amount remaining after 10 billion years would have been so minute that it would not have been detectable. This knowledge is what led the late astronomer Carl Sagan to proclaim that "we are stardust".

Which reaction produces \(^{131}\)Te in the nuclear reactor? (a) \(^{130}Te + n \rightarrow ^{131}Te\); (b) \(^{130}I + n \rightarrow ^{131}Te\); (c) \(^{132}Te + n \rightarrow ^{131}Te\); (d) \(^{132}I + n \rightarrow ^{131}Te\).

Radioisotopes are used in a variety of manufacturing and testing techniques. Wear measurements can be made using the following method. An automobile engine is produced using piston rings with a total mass of 100 g, which includes \(9.4 \mu Ci\) of \(^{59}Fe\) whose half-life is 45 days. The engine is test-run for 1000 hours, after which the oil is drained and its activity is measured. If the activity of the engine oil is 84 decays/s, how much mass was worn from the piston rings per hour of operation?

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