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Chapter 11: Problem 86

A sample of radioactive substance gave 630 counts per minute and 610 counts per minute at times differing by 1 hour. The decay constant \((\lambda)\) in \(\min ^{-1}\) is given by a. \(\lambda=\frac{630}{610} \times 60\) b. \(\mathrm{e}^{60 \lambda}=\frac{630}{610}\) c. \(\lambda=\frac{2.303}{60} \log \frac{610}{630}\) d. \(\lambda=\frac{2.303}{60} \times \frac{630}{610}\)

Short Answer

Expert verified
Option b is correct.

Step by step solution

01

Understand the decay process

Radioactive decay follows an exponential decay model, given by \( N(t) = N_0 e^{-\lambda t} \), where \( N(t) \) is the number of counts per minute at time \( t \), \( N_0 \) is the initial count rate, and \( \lambda \) is the decay constant.
02

Use the data given

From the problem statement, we know that \( N_0 = 630 \) and \( N(t) = 610 \) at a time \( t = 60 \) minutes (1 hour). We need to find \( \lambda \) such that \( 610 = 630 \cdot e^{-60\lambda} \).
03

Rearrange the decay formula

To find \( \lambda \), rearrange the decay formula to \( e^{60\lambda} = \frac{630}{610} \). This requires taking the natural logarithm next.
04

Identify the correct answer

Using the equation from Step 3, \( \mathrm{e}^{60 \lambda} = \frac{630}{610} \). Comparing this with the given options, option b, \( \mathrm{e}^{60 \lambda} = \frac{630}{610} \), matches our derived equation.

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Key Concepts

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

Decay Constant
The decay constant, often symbolized by \( \lambda \), is a crucial parameter in the study of radioactive decay. It quantifies the rate at which a radioactive substance undergoes disintegration. Essentially, it provides us with a measure of how quickly the substance is losing its radioactivity over time.
The mathematical representation of the decay constant can be found in the equation \( N(t) = N_0 e^{-\lambda t} \), where:
  • \( N(t) \) is the radioactive count at time \( t \)
  • \( N_0 \) is the initial count rate
  • \( \lambda \) is the decay constant

In the context of radioactive problems, finding the decay constant is vital for predicting how the radioactive sample's activity changes over time. By knowing \( \lambda \), one can calculate the time it will take for the substance to reduce to half its original count or activity, commonly known as the half-life. This makes \( \lambda \) incredibly useful in various fields such as nuclear physics and radiometric dating.
Exponential Model
Radioactive decay is best understood through the exponential model. This model helps explain how substances decay at rates proportional to their current amount. This relationship is exponential because each unit of time sees the substance reduce by a constant percentage rather than a constant amount.
The function reflects this:
  • \( N(t) = N_0 e^{-\lambda t} \)

The formula shows that the decrease in substance is not linear. Instead, it resembles a slope that gets less steep over time, implying that while initially the decay seems rapid, it slows down as time progresses. The beauty of the exponential model lies in its ability to mathematically describe how a substance never fully reaches zero but approaches it asymptotically as time goes on.
This exponential model's characteristics are particularly relevant when analyzing how quickly radioactive substances decay over certain periods, which is critical for both scientific applications and safety assessments.
Radioactive Substance Counts
Radioactive substance counts are a way to express the activity level of a radioactive sample. The count rate, usually measured in counts per minute (cpm), involves detecting and counting the number of disintegrations occurring in the sample over a given time period.
In problems involving radioactive decay, initial and subsequent counts help us determine how the substance is decaying. For example, in the exercise provided, the counts at the beginning \( (N_0 = 630) \) and after some time \( (N(t) = 610) \) provide data needed to compute the decay constant \( \lambda \).
The change in count from 630 to 610 after 60 minutes represents the decay process over that period. Counts are integral to applying the decay formula, allowing us to link experimental data to theoretical models like the exponential model of decay. Consequently, understanding and interpreting these counts are essential for accurate predictions and calculations related to radioactive materials.

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Most popular questions from this chapter

The first trans-uranium element was synthesis by bombarding \({ }_{92} \mathrm{U}^{238}\) with neutrons. After capturing one neutron, the resulting nuclide was unstable and decayed by beta emission. What was the product of these two nuclear reactions? a. \({ }_{90} \mathrm{Th}^{235}\) b. \({ }_{92} \mathrm{U}^{239}\) c. \({ }_{93} \mathrm{~Np}^{239}\) d. \({ }_{91} \mathrm{~Pa}^{239}\)

Match the lists I and II and pick the correct matching from the codes given below. List I List II A. Isotope (p) \(_{38} R a^{228}\) and \({ }_{39} A c^{228}\) B. Isobar(q) (q) \({ }_{18} \mathrm{Ar}^{39}\) and \({ }_{19} \mathrm{~K}^{40}\) C. Isotone (r) \({ }_{1} \mathrm{H}^{2}\) and \({ }_{1} \mathrm{H}^{3}\) D. Isosters (s) \({ }_{92} \mathrm{U}^{235}\) and \({ }_{92} \mathrm{U}^{2 \text { ss }}\) (t) \(\mathrm{CO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}\)

Loss of a \(\beta\)-particle is equivalent to a. increase of one proton only b. decrease of one neutron only c. both a and \(\mathbf{b}\) d. none of these

Select the correct statements: a. In the reaction \({ }_{11} \mathrm{Na}^{23}+\mathrm{Q} \rightarrow{ }_{12} \mathrm{Mg}^{23}+{ }_{0} \mathrm{n}^{1}\), the bombarding par- ticle \(\mathrm{q}\) is deutron b. In the reaction \({ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow 56 \mathrm{Ba}^{140}+2\) \({ }_{0} \mathrm{n}^{1}+\mathrm{p}\), produced \(\mathrm{p}\) is \({ }_{36} \mathrm{Kr}^{94}\) c. In a fission reaction, a loss in mass occurs releasing a huge amount of energy d A huge amount of energy is produced during nuclear fission and nuclear fussion reaction

The radionucliede \({ }_{90} \mathrm{Th}^{234}\) undergoes two successive \(\beta\)-decays followed by one \(\alpha\)-decay. The atomic number and the mass number respectively of the resulting radio nucliede are a. 92 and 234 b. 94 and 230 c. 90 and 230 d. 92 and 230
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