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A radioactive isotope of mercury, \({ }^{197} \mathrm{Hg}\), decays to gold, \({ }^{197} \mathrm{Au},\) with a disintegration constant of \(0.0108 \mathrm{~h}^{-1} .\) (a) Calculate the half-life of the \({ }^{197} \mathrm{Hg}\). What fraction of a sample will remain at the end of (b) three half-lives and (c) 10.0 days?

Short Answer

Expert verified
(a) 64.17 hours; (b) 1/8 (12.5%) remains; (c) 7.4% remains.

Step by step solution

01

Calculate the Half-life

The half-life of a radioactive isotope is calculated using the formula: \[ t_{1/2} = \frac{\ln(2)}{k} \]where \( t_{1/2} \) is the half-life and \( k \) is the disintegration constant. Here, \( k = 0.0108 \, \text{h}^{-1} \). First, compute the natural logarithm of 2: \( \ln(2) \approx 0.693 \). Substitute the values into the formula:\[ t_{1/2} = \frac{0.693}{0.0108} \approx 64.17 \, \text{hours} \]
02

Determine Remaining Fraction After Three Half-lives

After each half-life, the remaining fraction of the substance is halved. After one half-life, \( \frac{1}{2} \) of the substance remains; after two half-lives, \( \left(\frac{1}{2}\right)^2 \); and after three half-lives, \( \left(\frac{1}{2}\right)^3 \). Calculate this:\[ \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]Thus, \( \frac{1}{8} \) or 12.5% of the original sample remains after three half-lives.
03

Calculate Fraction Remaining After 10 Days

First, convert 10 days to hours: \( 10 \, \text{days} \times 24 \, \frac{\text{hours}}{\text{day}} = 240 \, \text{hours} \). The decay formula is:\[ N(t) = N_0 e^{-kt} \]where \( N(t) \) is the amount remaining after time \( t \), \( N_0 \) is the initial amount, and \( k \) is the decay constant. Substitute the values:\[ N(t) = N_0 e^{-0.0108 \times 240} \]\[ N(t) = N_0 e^{-2.592} \]Compute the exponential value: \( e^{-2.592} \approx 0.074 \). Therefore, 7.4% of the original sample remains after 10 days.

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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 fundamental to understanding radioactive decay. It is the amount of time required for half of the radioactive nuclei in a sample to decay. This doesn't mean that after one half-life, the radioactive material completely disappears; rather, only half of it transforms, leaving the other half still radioactive. In the calculation of half-life, we use the formula:\[ t_{1/2} = \frac{\ln(2)}{k} \]where:
  • \( t_{1/2} \) is the half-life, typically measured in units of time.
  • \( \ln(2) \) is the natural logarithm of 2, approximately 0.693.
  • \( k \) is the disintegration constant, which indicates the rate of decay.
For example, the disintegration constant for a radioactive mercury isotope is given as \( 0.0108 \, \text{h}^{-1} \). By substituting this into the formula, the half-life of the isotope can be calculated as:\[ t_{1/2} = \frac{0.693}{0.0108} \approx 64.17 \, \text{hours} \]This result tells us that approximately every 64.17 hours, half of the isotope will have decayed.
Disintegration constant
The disintegration constant, noted as \( k \), is a crucial component in understanding how quickly a radioactive substance decays. It quantifies the probability of decay of a single nucleus per unit time and is usually expressed with units that match the time of decay, such as per hour or per second.Intuitively, a larger disintegration constant means the substance decays more quickly, while a smaller value indicates a slower rate of decay. It is directly used in calculating both the half-life and in the exponential decay formula that predicts how much of a substance remains after a certain period.Revisiting the example of radioactive mercury, the given disintegration constant \( k = 0.0108 \, \text{h}^{-1} \) conveys how rapidly this isotope decays each hour. This constant is fundamental for computing how much of a substance remains as it decays over time, using various calculations.
Exponential decay formula
Exponential decay describes how the quantity of a radioactive isotope decreases over time. This decay process follows a specific mathematical pattern, known as the exponential decay formula:\[ N(t) = N_0 e^{-kt} \]where:
  • \( N(t) \) is the quantity remaining after time \( t \).
  • \( N_0 \) is the initial quantity of the substance.
  • \( e \) is the base of the natural logarithms, approximately equal to 2.718.
  • \( k \) is the disintegration constant.
  • \( t \) is the time elapsed.
In practical terms, by knowing the disintegration constant and the time elapsed, you can use this formula to determine how much of the original substance remains. For instance, if you want to know how much of a mercury isotope remains after 10 days, convert the time to hours (as the disintegration constant is per hour) and substitute into the formula:Given that 10 days is equivalent to 240 hours, and using \( k = 0.0108 \, \text{h}^{-1} \), the formula becomes:\[ N(t) = N_0 e^{-0.0108 \times 240} \approx N_0 e^{-2.592} \approx N_0 \times 0.074 \]Thus, about 7.4% of the initial quantity remains after 10 days.

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Most popular questions from this chapter

How much energy is released when a \({ }^{238} \mathrm{U}\) nucleus decays by emitting (a) an alpha particle and (b) a sequence of neutron, proton, neutron, proton? (c) Convince yourself both by reasoned argument and by direct calculation that the difference between these two numbers is just the total binding energy of the alpha particle. (d) Find that binding energy. Some needed atomic and particle masses are $$ \begin{array}{llll} { }^{238} \mathrm{U} & 238.05079 \mathrm{u} & { }^{234} \mathrm{Th} & 234.04363 \mathrm{u} \\ { }^{237} \mathrm{U} & 237.04873 \mathrm{u} & { }^{4} \mathrm{He} & 4.00260 \mathrm{u} \\ { }^{236} \mathrm{~Pa} & 236.04891 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ { }^{235} \mathrm{~Pa} & 235.04544 \mathrm{u} & \mathrm{n} & 1.00866 \mathrm{u} \end{array} $$

Generally, more massive nuclides tend to be more unstable to alpha decay. For example, the most stable isotope of uranium, \({ }^{238} \mathrm{U},\) has an alpha decay half-life of \(4.5 \times 10^{9} \mathrm{y} .\) The most stable isotope of plutonium is \({ }^{244} \mathrm{Pu}\) with an \(8.0 \times 10^{7} \mathrm{y}\) half-life, and for curium we have \({ }^{248} \mathrm{Cm}\) and \(3.4 \times 10^{5} \mathrm{y}\). When half of an original sample of \({ }^{238} \mathrm{U}\) has decayed, what fraction of the original sample of (a) plutonium and (b) curium is left?

In a Rutherford scattering experiment, assume that an incident alpha particle (radius \(1.80 \mathrm{fm}\) ) is headed directly toward a target gold nucleus (radius \(6.23 \mathrm{fm}\) ). What energy must the alpha particle have to just barely "touch" the gold nucleus?

What is the binding energy per nucleon of the rutherfordium isotope \({ }_{104}^{259} \mathrm{Rf}\) ? Here are some atomic masses and the neutron mass. $$\begin{array}{lr}\frac{259}{104} \mathrm{Rf} & 259.10563 \mathrm{u} \\\\\mathrm{n} & 1.008665 \mathrm{u}\end{array} { }^{1} \mathrm{H} \quad 1.007825 \mathrm{u}$$

Radioactive element \(A A\) can decay to either element \(B B\) or element \(C C\). The decay depends on chance, but the ratio of the resulting number of \(B B\) atoms to the resulting number of \(C C\) atoms is always \(2 / 1 .\) The decay has a half-life of 8.00 days. We start with a sample of pure \(A A .\) How long must we wait until the number of \(C C\) atoms is 1.50 times the number of \(A A\) atoms?

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