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Chapter 45: Problem 26

A certain isotope has a half-life of \(7.0 \mathrm{~h}\). How many seconds does it take for 10 percent of the sample to decay?

Short Answer

Expert verified
It takes approximately 3667 seconds for 10% of the isotope to decay.

Step by step solution

01

Understand the Half-Life Concept

The half-life, denoted by \( t_{1/2} \), is the time required for half of the radioactive substance to decay. In this problem, \( t_{1/2} = 7 \) hours. You want to find the time taken for 10% of the substance to decay.
02

Convert Half-Life to Seconds

First, convert the half-life from hours to seconds because the final answer needs to be in seconds. Since there are 3600 seconds in an hour, calculate:\[t_{1/2} = 7 imes 3600 = 25200 \text{ seconds}\]
03

Use the Decay Law Formula

The decay law formula is \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the quantity of the substance remained after time \( t \), \( N_0 \) is the initial quantity, and \( k \) is the decay constant. First, calculate \( k \) using the half-life formula:\[k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{25200}\]
04

Set Up the Equation for 10% Decay

Since you want 10% to decay, 90% of the substance remains. Set up the equation as follows:\[0.9N_0 = N_0 e^{-kt}\]Cancel \( N_0 \) from both sides to get:\[0.9 = e^{-kt}\]
05

Solve for Time \( t \)

Take the natural logarithm of both sides to solve for \( t \):\[\ln(0.9) = -kt\]Substitute \( k \) from Step 3:\[\ln(0.9) = -\left(\frac{0.693}{25200}\right)t\]Solve for \( t \):\[t = \frac{\ln(0.9)}{-\frac{0.693}{25200}} \approx 3667 \text{ seconds}\]
06

Verify the Solution

Verify that the calculated time results in a decay that equates to 10% of the initial amount decayed. Recheck calculations by plugging \( t \) back into the decay equation to confirm it aligns with the expected 10% decay.

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Key Concepts

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

Half-Life Calculation
The concept of half-life is crucial when understanding radioactive decay. The half-life of a substance is defined as the time it takes for half of the radioactive atoms in a sample to decay. This measure is a constant for each radioactive isotope and does not depend on the initial quantity of the substance.
  • For example, in our exercise, the half-life of the isotope is 7 hours (or 25200 seconds).
  • During each successive half-life period, another half of the remaining atoms will decay.
Knowing the half-life helps us predict how quickly a sample will reduce over time, which is vital in calculations like the one we are addressing.
Decay Constant
The decay constant, denoted by the symbol \( k \), is a critical component in the analysis of radioactive decay rates. This constant reflects the probability of decay per unit time and ties directly into the half-life.
The relationship between the decay constant \( k \) and the half-life \( t_{1/2} \) is given by:\[k = \frac{\ln 2}{t_{1/2}}\]This formula means that the decay constant is derived from the natural logarithm of 2, divided by the half-life.
  • The natural logarithm of 2 is approximately 0.693.
  • In our problem, substituting the half-life of 25200 seconds yields a specific decay constant used to determine the time it takes for a specific fraction of the material to decay.
Understanding this constant allows us to express how quickly a radioactive substance undergoes decay.
Exponential Decay Formula
Radioactive decay is best described by the exponential decay formula, which helps us determine the quantity of material left after a given time. The formula is expressed as:\[N(t) = N_0 e^{-kt}\]where:
  • \(N(t)\) is the quantity remaining after time \(t\).
  • \(N_0\) is the original quantity.
  • \(k\) is the decay constant.
  • \(t\) is time.
This equation demonstrates how the quantity of a radioactive substance decreases over time.
In our exercise, it's used to set up the equation \(0.9N_0 = N_0 e^{-kt}\) when 10% of the isotope has decayed. This means that 90% or 0.9 of the initial quantity \(N_0\) remains. The constant decay factor \(e^{-kt}\) is crucial here in implicitly showing how decay happens over continuous time frames.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a fundamental concept in the calculation of exponential decay. It's defined as logarithm to the base of the mathematical constant \( e \), which is approximately equal to 2.718.
When solving equations concerning decay, the natural logarithm helps us isolate variables like time \( t \).
In our solution, taking the natural logarithm on both sides of the decay equation \( 0.9 = e^{-kt} \) allows us to simplify the expression to solve for \( t \):\[\ln(0.9) = -kt\]This step is essential as it transforms the equation from an exponential form to a linear form that can be rearranged to solve for unknowns.
Natural logarithms are particularly useful in decay problems as they provide a straightforward way to deal with exponential equations.

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Most popular questions from this chapter

What mass of \(\frac{60}{27}\) Co has an activity of \(1.0 \mathrm{Ci}\) ? The half-life of cobalt-60 is \(5.25\) years.

Cesium-124 has a half-life of \(31 \mathrm{~s}\). What fraction of a cesium-124 sample will remain after \(0.10 \mathrm{~h}\) ?

An experiment is done to determine the half-life of a radioactive substance that emits one beta particle for each decay process. Measurements show that an average of \(8.4\) beta particles are emitted each second by \(2.5 \mathrm{mg}\) of the substance. The atomic mass of the substance is 230 . Find the half-life of the substance.

Technetium-99 \(\left({ }_{43}^{99} \mathrm{Tc}\right)\) has an excited state that decays by emission of a gamma ray. The half-life of the excited state is \(360 \mathrm{~min}\). What is the activity, in curies, of \(1.00 \mathrm{mg}\) of this excited isotope? Because we have the half-life \(\left(t_{1 / 2}\right)\) we can determine the decay constant since \(\lambda t_{1 / 2}=0.693 .\) The activity of a sample is \(\lambda N\). In this case, $$ \lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{21600 \mathrm{~s}}=3.21 \times 10^{-5} \mathrm{~s}^{-1} $$ We also know that \(99.0 \mathrm{~kg}\) of Tc contains \(6.02 \times 10^{26}\) atoms. A mass \(m\) will therefore contain \([m /(99.0 \mathrm{~kg})]\left(6.02 \times 10^{26}\right)\) atoms. In our case, \(m=1.00 \times 10^{-6} \mathrm{~kg}\), and so $$ \begin{aligned} \text { Activity } &=\lambda N=\left(3.21 \times 10^{-5} \mathrm{~s}^{-1}\right)\left(\frac{1.00 \times 10^{-6} \mathrm{~kg}}{99.0 \mathrm{~kg}}\right)\left(6.02 \times 10^{26}\right) \\ &=1.95 \times 10^{14} \mathrm{~s}^{-1}=1.95 \times 10^{14} \mathrm{~Bq} \end{aligned} $$

Complete the following nuclear equations: (a) \({ }^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{8}^{17} \mathrm{O}+\) ? (d) \({ }_{15}^{30} \mathrm{P} \rightarrow{ }_{14}^{30} \mathrm{Si}+\) ? (b) \({ }_{4}^{9} \mathrm{Be}+{ }_{2}^{3} \mathrm{He} \rightarrow{ }_{6}^{12} \mathrm{C}+?\) (e) \({ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+?\) (c) \({ }_{4}^{9} \operatorname{Be}(p, \alpha)\) ? (f) \({ }_{20}^{43} \mathrm{Ca}(\alpha, ?) \frac{46}{21} \mathrm{Sc}\) (a) The sum of the subscripts on the left is \(7+2=9 .\) The subscript of the first product on the right is 8 . Hence, the second product on the right must have a subscript (net charge) of \(1 .\) Also, the sum of the superscripts on the left is \(14+4=18 .\) The superscript of the first product is \(17 .\) Hence, the second product on the right must have a superscript (mass number) of 1 . The particle with nuclear charge 1 and mass number 1 is the proton, \({ }_{1}^{1} \underline{H}\). (b) The nuclear charge of the second product particle (its subscript) is \((4+2)-6=0 .\) The mass number of the particle (its superscript) is \((9+4)-12=1 .\) Hence, the particle must be the neutron, \(\downarrow{n}\). (c) The reactants \({ }_{4}^{9} \mathrm{Be}\) and \({ }_{1}^{1} \mathrm{H}\) have a combined nuclear charge of 5 and a mass number of \(10 .\) In addition to the alpha particle, a product will be formed of charge \(5-2=3\) and mass number \(10-4=6\). This is \(\frac{6}{3} \mathrm{Li}\). (d) The nuclear charge of the second product particle is \(15-14=+1\). Its mass number is \(30-30=0\). Hence, the particle must be a positron, \({ }_{+1} e\). (e) The nuclear charge of the second product particle is \(1-2=-1 .\) Its mass number is \(3-3=0 .\) Hence. the particle must be a beta particle (an electron), \(_{-1}^{0} e\). ( \(f\) ) The reactants, \(_{20}^{43} \mathrm{Ca}\) and \({ }_{2}^{4} \mathrm{He}\), have a combined nuclear charge of 22 and mass number of \(47 .\) The ejected product will have charge \(22-21=1\), and mass number \(47-46=1 .\) This is a proton and should be represented in the parentheses by \(p\). In some of these reactions a neutrino and/or a photon are emitted. We ignore them for this discussion since the charge for both are zero. Moreover the mass of the photon is zero and the mass of each of the several neutrinos, although not zero, is negligibly small.
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