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Chapter 33: Problem 49

The activity of a radioactive sample goes down to about \(6 \%\) in a time of 2 hours. The half life of the sample in minutes is about: (a) 30 (b) 15 (c) 60 (d) 120

Short Answer

Expert verified
The half-life of the sample is approximately 30 minutes.

Step by step solution

01

Understand the Concept

The half-life of a substance is the time required for the activity of the substance to decrease to half its initial value. If the activity drops to 6% in 2 hours, we can find the half-life using the formula for exponential decay.
02

Define Exponential Decay Formula

Radioactive decay can be described by the equation: \[ A(t) = A_0 e^{-kt} \]Where:- \( A(t) \) is the activity at time \( t \),- \( A_0 \) is the initial activity,- \( k \) is the decay constant,- \( t \) is the time elapsed.
03

Relate Activity to Percentage

Since the activity drops to 6% of its initial value, we have:\[ A(t) = 0.06 A_0 \]Substituting into the exponential decay formula gives:\[ 0.06 A_0 = A_0 e^{-k(2 \, ext{hours})} \]
04

Solve for Decay Constant \( k \)

Dividing both sides by \( A_0 \) and solving for \( k \):\[ 0.06 = e^{-2k} \]Taking the natural logarithm of both sides gives:\[ \ ln(0.06) = -2k \ \Rightarrow k = -\frac{ln(0.06)}{2} \]
05

Determine Half-Life in Hours

The half-life \( t_{1/2} \) is related to the decay constant \( k \) by the following formula:\[ t_{1/2} = \frac{ln(2)}{k} \]Substituting the expression for \( k \) gives:\[ t_{1/2} = \frac{ln(2)}{-\frac{ln(0.06)}{2}} \]Calculating this gives \( t_{1/2} \approx 0.5 \, ext{hours} \).
06

Convert Hours to Minutes

Since one hour is 60 minutes, convert the half-life from hours to minutes:\[ 0.5 \, ext{hours} = 0.5 \times 60 = 30 \, ext{minutes} \]

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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 process where an unstable atomic nucleus loses energy by emitting radiation. Over time, this results in a decrease in radioactivity, measured in terms of decay per unit of time, such as half-life. This spontaneous process characterizes how radioactive substances transform into different elements or isotopes.
The key points of radioactive decay include:
  • It occurs naturally and cannot be stopped or reversed.
  • Common forms of decay include alpha, beta, and gamma decay.
  • The rate of decay is constant and unique to each radioactive substance.
This decay process is essential in various fields such as medicine, archaeology, and nuclear energy, where understanding the rate of decay helps in applications like cancer treatment, dating artifacts, and generating electricity.
Exponential Decay Formula
The exponential decay formula describes the decreasing behavior of quantities that reduce by a constant percentage rate over time. This mathematical representation is especially useful for modeling radioactive decay. In the formula:\[ A(t) = A_0 e^{-kt} \]
  • \(A(t)\) represents the amount of substance remaining at time \(t\).
  • \(A_0\) is the initial amount of the substance.
  • \(e\) is the base of the natural logarithm, approximately equal to 2.718.
  • \(k\) is the decay constant.
The process shows how, over time, the activity of a radioactive sample exponentially decreases. This implies each equal time period results in a consistent percentage reduction of the remaining quantity.
Decay Constant
The decay constant \(k\) is a critical factor in determining the rate of exponential decay. It directly influences how quickly a radioactive substance decreases in activity. In the context of radioactive decay, the decay constant gives us insight into how hazardous or long-lasting a substance can be.
Here鈥檚 what you need to know about decay constants:
  • A larger \(k\) value indicates a faster decay rate, meaning the substance loses its radioactivity quicker.
  • Hey, remember the relation: \[ k = -\frac{ln(0.06)}{2} \] from the solution? It demonstrates how the natural logarithm of a proportion helps uncover the decay constant.
  • The decay constant can be used to find the half-life \(t_{1/2}\) by the formula: \[ t_{1/2} = \frac{ln(2)}{k} \]
Understanding the concept of a decay constant is vital for predicting how a sample will behave over extended periods.
Natural Logarithm
The natural logarithm, denoted as \(ln\), is a mathematical function that helps solve equations involving exponential growth or decay. It is the inverse function of the exponential \(e^x\) and provides a way to deal with percentages or proportions in decay problems.
In radioactive decay, the natural logarithm is used to determine the decay constant from proportional decreases in activity, as shown:
  • The equation \( ln(0.06) = -2k \) allows us to calculate \(k\) when the remaining percentage is known.
  • Using \(ln(2)\), the half-life can be calculated as it represents the time taken for any quantity to halve itself repeatedly.
Through applications of the natural logarithm, scientists can easily translate exponential decay equations into useful information, like the half-life of a substance, by relating logarithmic values back to time contexts.

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Most popular questions from this chapter

What is the binding energy per nucleon of \({ }_{6} \mathrm{C}^{12}\) nucleus? Given: mass of \(C^{12}\left(m_{c}\right)=12.000 \mathrm{u}\) mass of proton \(\left(m_{p}\right)=1.0078 \mathrm{u}\) mass of neutron \(\left(m_{n}\right)=1.0087 \mathrm{u}\) and \(1 \mathrm{amu}=931.4 \mathrm{MeV}\) (a) \(5.26 \mathrm{MeV}\) (b) \(10.11 \mathrm{MeV}\) (c) \(15.65 \mathrm{MeV}\) (d) \(7.68 \mathrm{MeV}\)

Four vessels \(A, B, C\) and \(D\) contain respectively \(20 \mathrm{~g}\) -atom \(\left(t_{1 / 2}=5\right.\) hour \(), 2\) g-atom \(\left(t_{1 / 2}=1\right.\) hour \(), \quad 5 \quad \mathrm{~g}\) -atom \(\left(t_{1 / 2}=2\right.\) hour \()\) and \(10 \mathrm{~g}\) -atom \(\left(t_{1 / 2}=3\right.\) hour) of different radio nuclides in the beginning, the maximum activity would be exhibited by the vessel is : (a) \(A\) (b) \(B\) (c) \(\bar{C}\) (d) \(D\)

\(20 \%\) of a radioactive substance decay in 10 days. The amount of the original material left after 30 days is : (a) \(51.2 \%\) (b) \(62.6 \%\) (c) \(15 \%\) (d) \(21.27 \%\)

The radioactive decay rate of a radioactive element is found to be \(10^{3}\) disintegrations per second at a certain time. If the half life of the element is one second, the decay rate after three second is : (a) 1000 (b) 250 (c) \(\frac{1000}{3}\) (d) 125

Electric field and magnetic field do not cause deflection in : (a) \(\alpha\) -rays (b) \beta-rays (c) \(\gamma\) -rays (d) positron
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