Problem 100
Question
If the half life period of a radioactive isotope is \(10 \mathrm{~s}\), then its average life will be (a) \(14.4 \mathrm{~s}\) (b) \(1.44 \mathrm{~s}\) (c) \(0.144 \mathrm{~s}\) (d) \(2.44 \mathrm{~s}\)
Step-by-Step Solution
Verified Answer
The average life of the isotope is \(14.4 \mathrm{~s}\), option (a).
1Step 1: Understanding Half-Life and Average Life
The half-life of a radioactive substance represents the time it takes for half of the material to decay. The average life, also known as mean life, is defined as the reciprocal of the decay constant (\(\lambda\)), and is related to the half-life.
2Step 2: Formula for Average Life
The average life \( \tau \) of a radioactive isotope can be calculated using the formula \( \tau = \frac{1}{\lambda} \). The relationship between half-life \( T_{1/2} \) and the decay constant \( \lambda \) is given by \( T_{1/2} = \frac{0.693}{\lambda} \).
3Step 3: Calculating the Decay Constant (\(\lambda\))
We know the half-life \( T_{1/2} = 10 \mathrm{~s} \). Using the formula \( T_{1/2} = \frac{0.693}{\lambda} \), we can rearrange to solve for \(\lambda\): \[ \lambda = \frac{0.693}{10} = 0.0693 \mathrm{~s}^{-1} \]
4Step 4: Calculating the Average Life (\(\tau\))
Now that we have \(\lambda = 0.0693 \mathrm{~s}^{-1} \), we calculate the average life using \( \tau = \frac{1}{\lambda} \): \[ \tau = \frac{1}{0.0693} \approx 14.4 \mathrm{~s} \]
5Step 5: Selection of Correct Answer
Comparing the calculated average life with the given options, we find that option (a) \(14.4 \mathrm{~s}\) matches our calculated result.
Key Concepts
Understanding Half-LifeCalculating Average LifeRole of the Decay Constant
Understanding Half-Life
Half-life is one of the most important concepts in radioactive decay. It refers to the time it takes for half of the radioactive isotopes in a sample to decay. During this time, the radioactive material transforms, emitting particles or energy, leaving behind only half the original quantity.
Understanding half-life is crucial because it helps us predict how long a radioactive element will remain active. Different elements have different half-lives, which can range from fractions of a second to billions of years. This means that some substances decay quickly, while others persist over long periods.
If we consider a radioactive isotope with a half-life of 10 seconds, as in our exercise, after 10 seconds, only half of the original substance would remain active.
Understanding half-life is crucial because it helps us predict how long a radioactive element will remain active. Different elements have different half-lives, which can range from fractions of a second to billions of years. This means that some substances decay quickly, while others persist over long periods.
- The shorter the half-life, the faster the decay.
- Longer half-lives indicate slower decay rates.
If we consider a radioactive isotope with a half-life of 10 seconds, as in our exercise, after 10 seconds, only half of the original substance would remain active.
Calculating Average Life
Average life, sometimes referred to as mean life, of a radioactive isotope provides a complementary perspective to half-life. While half-life gives us the time for half the substance to decay, the average life gives the average time each isotope atom remains before decaying.
The average life (\(\tau\)) can be calculated using the decay constant (\(\lambda\)) of the isotope: \[ \tau = \frac{1}{\lambda} \]
It's important to note the relationship between average life and half-life:
- The more complex relation is given by \(\tau = \frac{T_{1/2}}{0.693}\).- This equation shows that average life is approximately 1.44 times the half-life.
Using the example provided, where the half-life is 10 seconds, we use the formula for average life to understand that it results in roughly 14.4 seconds.
The average life (\(\tau\)) can be calculated using the decay constant (\(\lambda\)) of the isotope: \[ \tau = \frac{1}{\lambda} \]
It's important to note the relationship between average life and half-life:
- The more complex relation is given by \(\tau = \frac{T_{1/2}}{0.693}\).- This equation shows that average life is approximately 1.44 times the half-life.
Using the example provided, where the half-life is 10 seconds, we use the formula for average life to understand that it results in roughly 14.4 seconds.
Role of the Decay Constant
The decay constant (\(\lambda\)) is a key player in understanding radioactive decay. It indicates the probability per unit time that a given particle will decay.
The decay constant helps connect both half-life and average life with the actual decay process. The relationship between half-life and the decay constant is expressed as: \[ T_{1/2} = \frac{0.693}{\lambda} \] This formula helps us calculate the decay constant if we know the half-life, or vice versa.
In essence, the decay constant is fundamental for performing these types of calculations and for understanding the speed at which the decay process happens.
The decay constant helps connect both half-life and average life with the actual decay process. The relationship between half-life and the decay constant is expressed as: \[ T_{1/2} = \frac{0.693}{\lambda} \] This formula helps us calculate the decay constant if we know the half-life, or vice versa.
- With our example, knowing the half-life of 10 seconds, we can calculate the decay constant: \(\lambda = \frac{0.693}{10} = 0.0693 \mathrm{s}^{-1}\).
- This decay constant value then helps us directly determine the average life.
In essence, the decay constant is fundamental for performing these types of calculations and for understanding the speed at which the decay process happens.
Other exercises in this chapter
Problem 97
The half-life of a chemical reaction at a particular concentration is \(50 \mathrm{~min}\), when the concentration of reactants is doubled, the half-life become
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The half-life of a chemical reaction at a particular concentration is \(50 \mathrm{~min}\), when the concentration of reactants is doubled, the half-life become
View solution Problem 101
In the first-order reaction, half of the reaction is com pleted in 100 seconds. The time for \(99 \%\) reaction to occur will be (a) \(664.64 \mathrm{~s}\) (b)
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For a certain reaction, the activation energy is zero. What is the value of rate constant at \(300 \mathrm{~K}\), if \(\mathrm{K}=1.6\) \(\times 10^{6} \mathrm{
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