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Chapter 10: Problem 100

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) \(2.44 \mathrm{~s}\)

Short Answer

Expert verified
The average life of the isotope is \(14.4 \mathrm{~s}\), option (a).

Step by step solution

01

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.
02

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 \( T_{1/2} = \frac{0.693}{\lambda} \).
03

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 \(\lambda\): \[ \lambda = \frac{0.693}{10} = 0.0693 \mathrm{~s}^{-1} \]
04

Calculating the Average Life (\(\tau\))

Now that we have \(\lambda = 0.0693 \mathrm{~s}^{-1} \), we calculate the average life using \( \tau = \frac{1}{\lambda} \): \[ \tau = \frac{1}{0.0693} \approx 14.4 \mathrm{~s} \]
05

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.

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Key Concepts

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

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.
  • 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.
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.
  • 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.

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Most popular questions from this chapter

\(75 \%\) of a first-order reaction was completed in 32 min. When was \(50 \%\) of the reaction completed? (a) \(24 \mathrm{~min}\) (b) \(16 \mathrm{~min}\) (c) \(8 \mathrm{~min}\) (d) \(64 \mathrm{~min}\)

Under the same reactions conditions, initial concentration of \(1.386 \mathrm{~mol} \mathrm{dm}^{-3}\) of a substance becomes half in \(40 \mathrm{sec}\) and 20 sec through \(1^{\text {st }}\) order and zero order kinetics respectively. Ratio \(\left[\frac{\mathrm{K}_{1}}{\mathrm{~K}_{0}}\right]\) of the rate constants for first order \(\left(\mathrm{k}_{1}\right)\) and zero order \(\left(\mathrm{k}_{0}\right)\) of the reactions is (a) \(0.5 \mathrm{~mol}^{-1} \mathrm{dm}^{-3}\) (b) \(1 \mathrm{~mol} \mathrm{dm}^{-3}\) (c) \(1.5 \mathrm{~mol} \mathrm{dm}^{-3}\) (d) \(2 \mathrm{~mol}^{-1} \mathrm{dm}^{3}\)

The following data pertains to the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) $$ \begin{array}{llll} \hline \text { S. No. } & {[\mathrm{A}] \mathrm{mol} \mathrm{L}^{-1}} & {[\mathrm{~B}] \mathrm{mol} \mathrm{L}^{-1}} & \begin{array}{l} \text { Rate mol } \\ \mathrm{L}-\mathbf{S}^{-1} \end{array} \\ \hline \text { 1. } & 1 \times 10^{-2} & 2 \times 10^{-2} & 2 \times 10^{-4} \\\ 2 . & 2 \times 10^{-2} & 2 \times 10^{-2} & 4 \times 10^{-4} \\ 3 . & 2 \times 10^{-2} & 4 \times 10^{-2} & 8 \times 10^{-8} \\ \hline \end{array} $$ Which of the following inferences are drawn from the above data? (1) rate constant of the reaction is \(10^{-4}\) (2) rate law of the reaction is \([\mathrm{A}][\mathrm{B}]\) (3) rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: (a) 1 and 3 (b) 2 and 3 (c) land 2 (d) 1,2 and 3

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 becomes \(100 \mathrm{~min}\). Find the order. (a) zero (b) first (c) second (d) third

For a chemical reaction \(\mathrm{A} \longrightarrow \mathrm{B}\), the rate of reaction doubles when the concentration of \(\mathrm{A}\) is in creased four times. The order of reaction for \(\mathrm{A}\) is (a) zero (b) one (c) two (d) half
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