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Chapter 42: Problem 49

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) 12.5%; (c) 8.2%.

Step by step solution

01

Identify Half-Life Formula

The half-life \( t_{1/2} \) of a substance is related to its decay constant \( k \) by the formula: \[ t_{1/2} = \frac{\ln(2)}{k} \]. The natural logarithm \( \ln(2) \approx 0.693 \).
02

Calculate Half-Life

Substitute the given disintegration constant \( k = 0.0108 \, \text{h}^{-1} \) into the half-life formula: \[t_{1/2} = \frac{0.693}{0.0108} \approx 64.17 \, \text{hours} \].
03

Calculate Fraction Remaining After Three Half-Lives

After three half-lives, the fraction of the original sample remaining is given by \( \left(\frac{1}{2}\right)^3 \). Calculate this: \[\left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125 \]. This means 12.5% of the sample remains.
04

Convert 10 Days to Hours for Decay Calculation

Firstly, convert 10 days into hours: \( 10 \times 24 = 240 \) hours.
05

Calculate Number of Half-Lives in 10 Days

Use the half-life calculated in Step 2 to find how many half-lives occur in 240 hours using: \[ n = \frac{240}{64.17} \approx 3.74 \].
06

Calculate Fraction Remaining After 10 Days

The fraction remaining is given by \( \left(\frac{1}{2}\right)^n \). So, \[\left(\frac{1}{2}\right)^{3.74} \approx 0.082 \]. This means approximately 8.2% of the sample remains.

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Key Concepts

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

half-life
The half-life of a radioactive substance is a crucial concept in understanding radioactive decay. It represents the time it takes for half of the radioactive nuclei in a sample to decay. This time period is constant for a given substance, irrespective of how much substance exists or its concentration. In the context of the exercise, we calculate the half-life of the isotope
  • The formula used is: \[ t_{1/2} = \frac{\ln(2)}{k} \], where \( \ln(2) \approx 0.693 \).
  • For the disintegration constant \( k = 0.0108 \, \text{h}^{-1} \), the half-life calculates to approximately 64.17 hours.
This half-life signifies that every 64.17 hours, the quantity of ^{197}Hg becomes half of its previous amount until it decays further. Knowing the half-life helps in predicting how much of the radioisotope will remain after a given time period.
disintegration constant
The disintegration constant, often denoted by \( k \), defines the rate at which a radioactive substance decays. It's a probability rate constant that signifies how quickly or slowly the disintegration process occurs over time. The unit of \( k \) indicates the time frame; for instance, \( \, \text{h}^{-1} \) suggests the time constant is based in hours.
  • A higher disintegration constant implies a faster decay rate.
  • For our isotope ^{197}Hg, \( k = 0.0108 \, \text{h}^{-1} \), which guides us in calculating the half-life and helps in understanding the decay progression.
A precise grasp of the disintegration constant assists scientists in estimating how rapidly a sample will lose its radioactivity over time.
radioisotope decay
Radioisotope decay is a natural process through which an unstable atomic nucleus loses energy by emitting radiation. This process leads to the transmutation of elements, where the original isotope changes into a different element or isotope. For example, in the exercise, the mercury isotope decays into gold.
  • The decay process can be accurately described through the exponential decay formula, establishing a predictable pattern for the rate and amount of decay.
  • Understanding how much of the substance remains after certain time periods, such as after three half-lives or 10 days, is essential for practical applications, like determining the age of fossils or managing nuclear waste.
Studying radioisotope decay enriches our comprehension of elements' life-cycles and the energy released during transformation.
exponential decay formula
The exponential decay formula is a mathematical description of the decay process. It is crucial in modeling the rate at which a radioactive substance decreases over time and predicting the amount left after a certain period. The formula is expressed as:\[ N(t) = N_0 e^{-kt} \],where:
  • \( N(t) \) is the remaining quantity at time \( t \).
  • \( N_0 \) is the initial quantity.
  • \( e \) is the base of the natural logarithm.
  • \( k \) is the disintegration constant.
This equation helps calculate the fraction of a sample remaining after any specified duration. For instance, to determine how much of a sample remains after 10 days, we first convert days to hours and then use the formula to find the fraction remaining. This allows clear and consistent modeling of decay processes, which is fundamental in both theoretical studies and practical scenarios.

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Most popular questions from this chapter

A periodic table might list the average atomic mass of magnesium as being \(24.312 \mathrm{u}\), which is the result of weighting the atomic masses of the magnesium isotopes according to their natural abundances on Earth. The three isotopes and their masses are \({ }^{24} \mathrm{Mg}(23.98504 \mathrm{u}),{ }^{25} \mathrm{Mg}(24.98584 \mathrm{u})\), and \({ }^{26} \mathrm{Mg}(25.98259 \mathrm{u})\). The natural abundance of \({ }^{24} \mathrm{Mg}\) is \(78.99 \%\) by mass (that is, \(78.99 \%\) of the mass of a naturally occurring sample of magnesium is due to the presence of \({ }^{24} \mathrm{Mg}\) ). What is the abundance of (a) \({ }^{25} \mathrm{Mg}\) and (b) \({ }^{26} \mathrm{Mg}\) ?

When an alpha particle collides elastically with a nucleus, the nucleus recoils. Suppose a \(5.00 \mathrm{MeV}\) alpha particle has a head-on elastic collision with a gold nucleus that is initially at rest. What is the kinetic energy of (a) the recoiling nucleus and (b) the rebounding alpha particle?

In 1992, Swiss police arrested two men who were attempting to smuggle osmium out of Eastern Europe for a clandestine sale. However, by error, the smugglers had picked up \({ }^{137} \mathrm{Cs}\). Reportedly, each smuggler was carrying a \(1.0 \mathrm{~g}\) sample of \({ }^{137} \mathrm{Cs}\) in a pocket! In (a) bequerels and (b) curies, what was the activity of each sample? The isotope \({ }^{137} \mathrm{Cs}\) has a half-life of \(30.2 \mathrm{y}\). (The activities of radioisotopes commonly used in hospitals range up to a few millicuries.)

A source contains two phosphorus radionuclides, \({ }^{32} \mathrm{P}\left(T_{1 / 2}=14.3\right.\) d) and \({ }^{33} \mathrm{P}\left(T_{1 / 2}=25.3 \mathrm{~d}\right)\). Initially, \(10.0 \%\) of the decays come from \({ }^{33} \mathrm{P}\). How long must one wait until \(95.0 \%\) do so?

A \(6.00 \mathrm{~g}\) charcoal sample from an ancient fire pit has a \({ }^{14} \mathrm{C}\) activity of \(63.0\) disintegrations/min. A living tree has a \({ }^{14} \mathrm{C}\) activity of \(15.3\) disintegrations/min per \(1.00 \mathrm{~g}\). The half-life of \({ }^{14} \mathrm{C}\) is \(5730 \mathrm{y}\) How old is the charcoal sample?
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