/*! 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 16 The radioactive nuclide \({ }^{1... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 16

The radioactive nuclide \({ }^{199} \mathrm{Pt}\) has a half-life of \(30.8 \mathrm{~min}\) utes. A sample is prepared that has an initial activity of \(7.56 \times 10^{11} \mathrm{~Bq} .\) (a) How many \({ }^{199} \mathrm{Pt}\) nuclei are initially present in the sample? (b) How many are present after \(30.8\) minutes? What is the activity at this time? (c) Repeat part (b) for a time \(92.4\) minutes after the sample is first prepared.

Short Answer

Expert verified
(a) Initial nuclei: \(2.017 \times 10^{15}\). (b) After 30.8 mins: \(1.0085 \times 10^{15}\) nuclei and activity is \(3.78 \times 10^{11}\) Bq. (c) After 92.4 mins: \(2.5213 \times 10^{14}\) nuclei and activity is \(9.45 \times 10^{10}\) Bq.

Step by step solution

01

Calculate Initial Number of Nuclei

When we have an initial activity, we can determine the initial number of nuclei using the formula for activity: \[ A = \lambda N \]where \( A \) is the activity, \( \lambda \) is the decay constant, and \( N \) is the number of nuclei. The decay constant \( \lambda \) can be found using the half-life \( T_{1/2} \):\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]Substituting the half-life given:\[ \lambda = \frac{\ln(2)}{30.8 \text{ min}} = \frac{0.693}{30.8 \times 60} \text{ s}^{-1} \]Now use this in the activity equation to find \( N \):\[ 7.56 \times 10^{11} = \left(\frac{0.693}{1848}\right) N \]\[ N = \frac{7.56 \times 10^{11} \times 1848}{0.693} \approx 2.017 \times 10^{15} \text{ nuclei} \]
02

Calculate Nuclei and Activity After One Half-life (30.8 mins)

After one half-life \((30.8 \text{ min})\), half of the initial nuclei remain:\[ N = \frac{2.017 \times 10^{15}}{2} = 1.0085 \times 10^{15} \text{ nuclei} \]The activity after one half-life is half the initial activity:\[ A = \frac{7.56 \times 10^{11}}{2} = 3.78 \times 10^{11} \text{ Bq} \]
03

Calculate Nuclei and Activity After Three Half-lives (92.4 mins)

92.4 minutes is three half-lives \((92.4/30.8 = 3)\). After each half-life, the number of nuclei is halved:\[ N = \frac{2.017 \times 10^{15}}{2^3} = \frac{2.017 \times 10^{15}}{8} = 2.5213 \times 10^{14} \text{ nuclei} \]Similarly, the activity is:\[ A = \frac{7.56 \times 10^{11}}{8} = 9.45 \times 10^{10} \text{ Bq} \]

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Key Concepts

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

Half-Life Calculations
In the world of radioactivity, the term **half-life** is essential. Half-life is the time it takes for half of the radioactive nuclei in a sample to decay. It's a constant measure and unique to each radioactive substance.
To calculate half-life mathematics, the formula used is \[\lambda = \frac{\ln(2)}{T_{1/2}}\] where \(\lambda\) represents the decay constant and \(T_{1/2}\) is the half-life duration.
Half-life calculations help you understand how quickly a radioactive sample is decreasing and are crucial for fields like archaeology in dating ancient objects.
Decay Constant
The **decay constant** \(\lambda\) expresses the probability of a single nucleus decaying per unit time. It connects directly with half-life and activity equations, making it vital in calculations.
To find the decay constant, you use the formula \[\lambda = \frac{\ln(2)}{T_{1/2}}\] This constant simplifies understanding how radioactive substances transform over time.
  • Together with activity, it gives insights into the number of nuclei in a sample.
  • It indicates the speed of radioactive decay, simplifying how we calculate remaining nuclei after any given time using exponential decay models.
Activity Measurement
**Activity** in radioactivity defines the rate at which a sample's radioactive nuclei decay. Measured in becquerels (Bq), it directly relates to the number of decays per second.
The equation for activity is given by \(A = \lambda N\), where \(A\) is the activity, \(\lambda\) is the decay constant, and \(N\) is the number of radioactive nuclei.
Knowing the initial activity of a sample helps determine how many atoms are present initially and how this number drops over time, as demonstrated by the example of \(^{199}\text{Pt}\) where activity halved each half-life.
Radioactivity Problems
**Radioactivity problems** require applying concepts like half-life, decay constant, and activity in calculations. These problems involve solving for quantities like the initial number of nuclei, remaining nuclei after certain times, and measuring changes in activity.
Such exercises allow breaks down into sequential steps:
  • First, identify known values, such as half-life and initial activity.
  • Use formulas like \(N = \frac{A}{\lambda}\) to find nuclei counts.
  • Determine how these values shift over time using half-life.
Approaching radioactivity problems systematically helps in drawing clear connections between theoretical concepts and real-world applications, enhancing one's understanding of natural radioactive decay processes.

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Most popular questions from this chapter

A laboratory stock solution is prepared with an initial activity due to \({ }^{24} \mathrm{Na}\) of \(2.5 \mathrm{mCi} / \mathrm{ml}\), and \(10 \mathrm{ml}\) of the stock solution is diluted at \(\left(t_{0}=0\right.\) ) to a working solution whose total volume is \(250 m l\). After 30 hours, a \(5 m l\) sample of the working solution is monitored with a counter. What is the measured activity? Given that half life of \({ }^{24} \mathrm{Na}\) is 15 hours

A hypothetical decay chain consists of the following elements, \(A, B_{1}, B_{2}\) and \(C\), where \(C\) is stable. The decay constants are \(\lambda, 2 \lambda\) for \(\alpha \& \beta\) decay of \(A\); and \(2 \lambda\), for \(\beta, \alpha\) decays leading to the formation of \(C\). (a) If the atomic and mass number of are \(m, n\) respectively, find the atomic and mass numbers of \(B_{1}, B_{2}\) \& respectively. (b) If at some instant, the number of atoms of \(B_{1}\) reaches a maximum value, find the ratio of the no. of atoms \(A\) to that \(B_{1}\) at this instant. (c) Do the same, when the no. of atoms of \(B_{2}\) reaches maximum, find the ratio of the no. of atoms of \(A\) to that of \(B_{2}\). (d) Find the activity of \(A\) as a function of time.

Thorium \(\frac{230}{90}\) Th decays to radium \(\frac{226}{88} \mathrm{Ra}\) by \(\alpha\) emission. The masses of the neutral atoms are \(230.033127\) u for \({ }_{90}^{230} \mathrm{Th}\) and \(226.025403 \mathrm{u}\) for \({ }_{88}^{226} \mathrm{Ra}\). If the parent thorium nucleus is at rest, what is the kinetic energy of the emitted \(\alpha\) particle? (Be sure to account for the recoil of the daughter nucleus.)

19.45 - A radioactive sample decays by \(\beta\)-emission. In first two seconds ' \(n^{\prime} \beta\)-particles are emitted and in next 2 seconds, \(0.25 n^{\prime} \beta\) -particles are emitted. The half life of radioactive nuclei is (a) \(2 \mathrm{sec}\) (b) \(4 \mathrm{sec}\) (c) \(1 \mathrm{sec}\) d) none

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 \({ }_{38}^{94} \mathrm{Sr}, 93.915360 \mathrm{u}\)
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