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A radioisotope has a half-life of 5.00 min and an initial decay rate of \(6.00 \times 10^{3}\) Bq. (a) What is the decay constant? (b) What will be the decay rate at the end of (i) 5.00 min, (ii) 10.0 min, (iii) 25.0 min?

Short Answer

Expert verified
(a) The decay constant is 0.1386 min鈦宦. (b) Decay rates are 3000 Bq at 5 mins, 1500 Bq at 10 mins, and 187.5 Bq at 25 mins.

Step by step solution

01

Understand the half-life and decay constant

The half-life of a substance is the time it takes for half of the radioactive atoms to decay. The relationship between the half-life \( t_{1/2} \) and the decay constant \( \lambda \) is given by the formula \( t_{1/2} = \frac{\ln(2)}{\lambda} \). We can rearrange this to solve for the decay constant: \( \lambda = \frac{\ln(2)}{t_{1/2}} \).
02

Calculate the decay constant

Given the half-life \( t_{1/2} = 5.00 \) minutes, calculate the decay constant \( \lambda \) using the formula from Step 1: \[ \lambda = \frac{\ln(2)}{5.00} = \frac{0.693}{5.00} = 0.1386 \text{ min}^{-1}. \]
03

Understand the decay rate formula

The decay rate at any time \( t \) is given by the formula \( R = R_0 e^{-\lambda t} \), where \( R_0 \) is the initial decay rate and \( \lambda \) is the decay constant.
04

Calculate the decay rate at 5.00 min

Substitute \( t = 5.00 \) minutes and \( \lambda = 0.1386 \text{ min}^{-1} \) into the decay rate formula: \[ R = 6.00 \times 10^{3} e^{-0.1386 \times 5} = 6.00 \times 10^{3} e^{-0.693} = 6.00 \times 10^{3} \times 0.5 = 3.00 \times 10^{3} \text{ Bq}. \]
05

Calculate the decay rate at 10.0 min

Use the formula and substitute \( t = 10.0 \) minutes: \[ R = 6.00 \times 10^{3} e^{-0.1386 \times 10} = 6.00 \times 10^{3} e^{-1.386} = 6.00 \times 10^{3} \times 0.25 = 1.50 \times 10^{3} \text{ Bq}. \]
06

Calculate the decay rate at 25.0 min

Substitute \( t = 25.0 \) minutes into the formula: \[ R = 6.00 \times 10^{3} e^{-0.1386 \times 25} = 6.00 \times 10^{3} e^{-3.465} = 6.00 \times 10^{3} \times 0.03125 = 1.875 \times 10^{2} \text{ Bq}. \]

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Key Concepts

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

Half-life
The half-life of a radioactive substance is a crucial concept in understanding how it decays over time. Simply put, half-life is the time required for half of the radioactive isotopes in a sample to decay. Each half-life period sees the reduction of the remaining radioactive material by 50%.

For example, if you start with 100 grams of a radioactive isotope, after one half-life, you'll have 50 grams remaining. After the second half-life, only 25 grams would remain. This pattern continues exponentially.

The concept of half-life helps in predicting how long a radioactive substance will remain active. It is expressed in various units, such as minutes, hours, days, or even years, depending on the isotope's nature.
Decay Constant
The decay constant, symbolized by \( \lambda \), provides the probability per unit time that a given atom will decay. It directly relates to the half-life of a substance via the equation:
  • \( t_{1/2} = \frac{\ln(2)}{\lambda} \)
This shows that the decay constant and half-life are inversely proportional. A larger decay constant means a shorter half-life, indicating the substance decays quickly.

For instance, in our exercise, we calculated a decay constant \( \lambda = 0.1386 \) min\(^{-1}\). It shows the expected behavior of the radioisotope in a given time frame. By understanding the decay constant, we can accurately predict how long significant decay processes will occur.
Exponential Decay
Exponential decay describes the process by which radioactive substances lose particles. Essentially, the rate of decay of the radioactive substance decreases exponentially over time. It follows a specific mathematical formula, which in the context of radioactivity is given by:
  • \( R = R_0 e^{-\lambda t} \)
Here, \( R \) is the decay rate at any time \( t \), \( R_0 \) is the initial decay rate, and \( \lambda \) is the decay constant.

This formula implies that as time progresses, the decay rate falls off rapidly, reducing to fractions of the initial decay rate at successive multiples of the half-life. This characteristic makes radioactive decay predictable, primarily governed by natural laws rather than external factors.
Decay Rate
The decay rate or radioactivity rate indicates how many radioactive particles are emitted over time. It's a measurable quantity, often expressed in Becquerels (Bq), where one Bq equals one decay per second.

In exercises such as ours, it's essential to determine the decay rate at different points in time to understand how quickly a substance is depleting. The decay rate can be calculated using:
  • \( R = R_0 e^{-\lambda t} \)
For example, our initial decay rate was \( 6.00 \times 10^3 \) Bq. After 5 minutes, it halved to \( 3.00 \times 10^3 \) Bq. By 10 minutes, the rate was down to a quarter, and by 25 minutes, only a small fraction remained.

These calculations are based on the understanding that the decay rate decreases predictably without variations due to environmental influences.

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Most popular questions from this chapter

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 Ci. When its activity falls below 3500 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 \(6,2008 .\) It is now April \(6,2011 .\) Is the source still usable? The half-life of \(^{60} \mathrm{Co}\) is 5.271 years.

At the beginning of Section \(30.6,\) a fission process is illustrated in which \(^{235} \mathrm{U}\) is struck by a neutron and undergoes fission to produce \(^{144} \mathrm{Ba}, 89 \mathrm{Kr}\) , and three neutrons. The measured masses of these isotopes are 235.043930 u \(\left(^{235} \mathrm{U}\right)\) \(143.922953 \mathrm{u}\left(^{144} \mathrm{Ba}\right), 88.917630 \mathrm{u}\left(^{89} \mathrm{Kr}\right),\) and 1.0086649 \(\mathrm{u}\) (neutron). (a) Calculate the energy (in MeV) released by each fission reaction. (b) Calculate the energy released per gram of \(^{235} \mathrm{U},\) in \(\mathrm{MeV} / \mathrm{g} .\)

The critical density of the universe is \(5.8 \times 10^{-27} \mathrm{kg} / \mathrm{m}^{3}\) . (a) Assuming that the universe is all hydrogen, express the critical density in the number of \(\mathrm{H}\) atoms per cubic meter. (b) If the density of the universe is equal to the critical density, how many atoms, on the average, would you expect to find in a room of dimensions 4 \(\mathrm{m} \times 7 \mathrm{m} \times 3 \mathrm{m} ?\) (c) Compare your answer in part (b) with the number of atoms you would find in this room under normal conditions on the earth.

How many protons and how many neutrons are there in a nucleus of (a) neon, \(_{10}^{21} \mathrm{Ne}(\mathrm{b})\) zinc, \(_{30}^{65} \mathrm{Zn}(\mathrm{c})\) silver, \(_{47}^{108} \mathrm{Ag}\)

A sample of the radioactive nuclide \(^{199} \mathrm{Pt}\) is prepared that has an initial activity of \(7.56 \times 10^{11}\) Bq. (a) 92.4 min after the sample is prepared, the activity has fallen to \(9.45 \times 10^{10}\) Bq. What is the half-life of this nuclide? (b) How many radioactive nuclei were initially present in the sample?

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