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The count rate of a radioactive nuclei falls from 992 counts per minute to 62 counts per minute in 10 hour. The half-life of the element is : (a) 1 hour (b) \(2.5\) hour (c) 5 hour (d) 6 hour

Short Answer

Expert verified
The half-life is \(2.5\) hours.

Step by step solution

01

Understanding the Decay Formula

In radioactive decay, the number of undecayed nuclei decreases over time. This is expressed by the formula \( N = N_0 \times e^{-\lambda t} \), where \( N_0 \) is the initial count rate, \( N \) is the final count rate, \( \lambda \) is the decay constant, and \( t \) is the time period.
02

Convert Count Rate to Decay Constant

We know that the count rate changes from 992 to 62 in 10 hours. We can substitute these values into the decay formula:\[ 62 = 992 \times e^{-\lambda \times 10} \]Rearrange the formula to solve for the decay constant (\(\lambda\)).
03

Solve for Lambda (\(\lambda\))

Divide both sides of the equation by 992:\[ \frac{62}{992} = e^{-\lambda \times 10} \]Take the natural logarithm of both sides:\[ \ln\left(\frac{62}{992}\right) = -\lambda \times 10 \]Solve for \( \lambda \):\[ \lambda = -\frac{1}{10} \times \ln\left(\frac{62}{992}\right) \]
04

Calculate the Half-Life

The half-life \( T_{1/2} \) is related to the decay constant by the formula \( T_{1/2} = \frac{\ln(2)}{\lambda} \). Use the previously calculated \( \lambda \):\[ T_{1/2} = \frac{\ln(2)}{-\frac{1}{10} \times \ln\left(\frac{62}{992}\right)} \]Calculate \( T_{1/2} \) using this expression.
05

Calculate and Compare

After calculating, you find that \( T_{1/2} \approx 2.490 \) hours. Round to the nearest provided option, which is 2.5 hours.

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Key Concepts

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

Half-life calculation
Half-life is a crucial concept in understanding radioactive decay. Half-life (symbolized as \(T_{1/2}\)) is the time required for half of the radioactive nuclei in a sample to decay. Essentially, during this period, the count rate of radioactive nuclei decreases to half of its original value.

To calculate the half-life, we use the relationship between the decay constant \(\lambda\) and the half-life, expressed through the formula: \[ T_{1/2} = \frac{\ln(2)}{\lambda} \] This equation tells us that the half-life is the natural logarithm of 2 divided by the decay constant. By analyzing how quickly or slowly a substance decays, this calculation helps us determine how much time it takes for the quantity of a radioactive substance to reduce significantly.

Understanding half-life aids in various practical scenarios, such as carbon dating and determining the longevity of radioactive sources in medicine and industry.
Decay constant
The decay constant, denoted by \(\lambda\), is another fundamental parameter in the study of radioactive decay. This constant represents the probability per unit time that a single nucleus will decay. It is directly linked to the stability of a radioactive substance – a higher decay constant suggests a less stable nucleus that decays quickly.

In the context of mathematical equations, the decay constant features prominently in the exponential decay model: \[ N = N_0 \times e^{-\lambda t} \] Here, \(N_0\) is the initial quantity of nuclei, \(N\) is the quantity remaining after time \(t\), and \(e\) is the base of the natural logarithm system.

This formula captures how a given substance experiences a steady, predictable decline over time. To find \(\lambda\), we often rearrange the equation to solve it given known parameters, using logarithmic methods as illustrated in the solved exercise. Grasping \(\lambda\) is vital for interpreting anything from measured count rates to predicting future behaviors of decaying samples.
Radioactive nuclei count rate
The radioactive nuclei count rate is a practical way of measuring radioactive decay. Think of it as the pulse of a decaying substance, typically measured in counts per minute or counts per second.

At the core, the count rate provides a snapshot of the number of disintegrations occurring in a radioactive sample within a specific period. This rate decreases as the substance decays, serving as a tangible indicator of the substance's declining radioactivity.

In scientific investigations and practical applications like the exercise, changes in count rate over time allow us to apply decay models to determine half-life and decay constants. For instance, if an initial count rate reduces significantly over time, it signifies an exponential decay pattern. Using the decay formula, this data is essential for solving problems related to nuclear physics, health physics (radiation protection), and other fields where understanding the rate of decay is necessary.

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Most popular questions from this chapter

Half-life of radioactive \({ }_{6} \mathrm{C}^{14}\) is 8000 years. What will be the age of wooden article if its \({ }_{6} \mathrm{C}^{14}\) activity is \(1 / 3\) of that of newly cut wood? (Take \(\log _{10} 3=0.477\) ) (a) 6788 (b) 8748 (c) 8788 (d) None of these

The energy released per nucleon of the reactant, in the thermonuclear reaction is \(3{ }_{1} \mathrm{H}^{2} \longrightarrow{ }_{2} \mathrm{He}^{4}+{ }_{1} \mathrm{H}^{1}+{ }_{0} n^{1}+21.6 \mathrm{MeV}\) (a) \(21.6 \mathrm{MeV}\) (b) \(7.2 \mathrm{MeV}\) (c) \(3.6 \mathrm{MeV}\) (d) \(1.8 \mathrm{MeV}\)

When the nucleus of \({ }^{\text {' }} \mathrm{U}_{92}\) disintegrates to give one nucleus of \({ }^{206} \mathrm{~Pb}_{82}\), the number of \(\alpha\) -particles emitted and the number of \(\beta\) -particles emitted are : (a) 4 and 8 respectively (b) 6 and 8 respectively (c) 8 and 6 respectively (d) 8 and 10 respectively

A particular nucleus in a large population of identical radioactive nuclei survives 10 half lives of that isotopes. The probability that this surviving nucleus will survive the next half-life is : (a) \(1 / 10\) (b) \(2 / 5\) (c) \(1 / 2\) (d) \(1 / 2^{10}\)

The binding energies per nucleon are \(5.3 \mathrm{MeV}, 6.2 \mathrm{MeV}\) and \(7.4 \mathrm{MeV}\) for the nuclei with mass numbers 3,4 and 5 respectively. If one nucleus of mass number 3 combines with one nucleus of mass number 5 to give two nuclei of mass number 4, then : (a) \(0.3 \mathrm{MeV}\) energy is absorbed (b) \(0.3 \mathrm{MeV}\) energy is released (c) \(28.1 \mathrm{MeV}\) energy is absorbed (d) \(3.3 \mathrm{MeV}\) energy is absorbed

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