/*! 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 27 Measurements on a certain isotop... [FREE SOLUTION] | 91Ó°ÊÓ

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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 2.0 days.

Step by step solution

01

Understand the problem

We are given initial and final decay rates for an isotope over a specific time period. Our task is to find its half-life. Decay rate is related to the number of radioactive atoms present, which decreases exponentially over time.
02

Write exponential decay formula

The exponential decay formula is given by \[ N(t) = N_0 e^{-\lambda t} \] where \( N(t) \) is the decay rate at time \( t \), \( N_0 \) is the initial decay rate, and \( \lambda \) is the decay constant.
03

Calculate decay constant

Substitute the given data into the formula and solve for \( \lambda \):\[ 3091 = 8318 \, e^{-\lambda \, (4 \times 1440)} \] where 4 days are converted to minutes (4 days = 4 × 1440 minutes).Solving this equation yields \[ \lambda = \frac{1}{t} \ln\left(\frac{N_0}{N(t)}\right) = \frac{1}{5760} \ln\left(\frac{8318}{3091}\right) \approx 0.000241 \, \text{min}^{-1} \]
04

Calculate half-life

Using the decay constant \( \lambda \), the half-life \( T_{1/2} \) can be calculated using the formula \[ T_{1/2} = \frac{\ln(2)}{\lambda} \]. Substituting the value of \( \lambda \):\[ T_{1/2} = \frac{0.693}{0.000241} \approx 2876 \, \text{minutes} \]. Convert this to days:\[ T_{1/2} \approx \frac{2876}{1440} \approx 2.0 \, \text{days} \].

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Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. This can happen in the form of alpha particles, beta particles, or gamma rays. In essence, the nucleus is seeking a more stable configuration by altering its number of protons, neutrons, or both.
This change in the structure of the nucleus decreases the number of radioactive atoms over time, resulting in a lower decay rate. Understanding this process is crucial for fields like nuclear physics, archaeology (radiocarbon dating), and medicine (radiopharmaceuticals).
Here are a few key points about radioactive decay:
  • Each type of decay changes the number of protons or neutrons, transforming the isotope into a different element or another isotope.
  • This process follows a stochastic (random) pattern but can be described mathematically by exponential functions.
  • Radioactive decay is indifferent to physical or chemical changes such as temperature or pressure.
Exponential Decay Formula
The exponential decay formula describes how quantities decrease over time, where the rate of the decrease is proportional to the current value. This is particularly useful in describing radioactive decay, where the amount of remaining material decreases exponentially.
The formula is given by: \[ N(t) = N_0 e^{-\lambda t} \]In this formula:
  • \( N(t) \) refers to the quantity (or rate) left after time \( t \).
  • \( N_0 \) is the initial quantity or rate.
  • \( \lambda \) is the decay constant, which dictates how quickly the decay occurs.

The exponential nature of this function means that no matter what point in time you start observing, the substance will always take the same amount of time to reduce to half its previous value, a period known as the half-life.
Decay Constant
The decay constant, denoted as \( \lambda \), is a measure of the probability of decay of a single nucleus per unit time. Essentially, it tells us how quickly a particular isotope will decay, and it is a critical part of understanding radioactive decay dynamics.
In the exponential decay formula \( N(t) = N_0 e^{-\lambda t} \), the decay constant \( \lambda \) represents the exponential rate at which the quantity of substance decreases. The larger the value of \( \lambda \), the faster the decay process.
To calculate \( \lambda \), you can use the formula:\[ \lambda = \frac{1}{t} \ln\left( \frac{N_0}{N(t)} \right) \]Here:
  • \( t \) is the time period over which the decay is measured.
  • The natural logarithm \( \ln \) is used to linearize the exponential equation for simplifying calculations.

With the decay constant, we can also easily compute the half-life of the substance using the relation \( T_{1/2} = \frac{\ln(2)}{\lambda} \). This gives us a direct way to link \( \lambda \) to the time needed for the substance to reduce to half its initial quantity.

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Most popular questions from this chapter

What nuclide is produced in the following radioactive decays? (a) \(\alpha\) decay of \(^{239} \mathrm{Pu} ;\) (b) \(\beta^{-}\) decay of \(_{11}^{24} \mathrm{Na} ;(\mathrm{c}) \beta^{+}\) decay of \(_{8}^{15} \mathrm{O}\)

BIO A nuclear chemist receives an accidental radiation dose of 5.0 Gy from slow neutrons (RBE = 4.0). What does she receive in rad, rem, and J/kg?

BIO 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 \(^{90}\) Sr decay? (b) What percentage of the original level of \(^{90}\) 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?

What particle \((\alpha\) particle, electron, or positron) is emitted in the following radioactive decays? (a) \(_{14}^{27} \mathrm{Si} \rightarrow_{13}^{27} \mathrm{Al}\) (b) \(^{238} \mathrm{U} \rightarrow_{90}^{234} \mathrm{Th} ;\) (c) \(_{33}^{74} \mathrm{As} \rightarrow_{34}^{74} \mathrm{Se}\)

BIO A 67 -kg person accidentally ingests 0.35 Ci of tritium. (a) Assume that the tritium spreads uniformly throughout the body and that each decay leads on the average to the absorption of 5.0 \(\mathrm{keV}\) of energy from the electrons emitted in the decay. The half-life of tritium is \(12.3 \mathrm{y},\) and the RBE of the electrons is \(1.0 .\) Calculate the absorbed dose in rad and the equivalent dose in rem during one week. (b) The \(\beta^{-}\) decay of tritium releases more than 5.0 keV of energy. Why is the average energy absorbed less than the total energy released in the decay?

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