The isotope \({ }^{212}_{83} \mathrm{Bi}\) has a half-life of \(1.01 \mathrm{yr}\). What mass (in mg) of a 2.00-mg sample will remain after \(3.75 \times 10^{3} \mathrm{~h}\) ?

Short Answer

Expert verified
1.496 mg

Step by step solution

01

- Convert Hours to Years

First, convert the given time from hours to years. Use the conversion factor: 1 year = 8760 hours.equation:\t\[3.75 \times 10^3\,\mathrm{h} \times \frac{1\,\mathrm{yr}}{8760\,\mathrm{h}} = 0.428\,\mathrm{yr}\]
02

- Calculate the Number of Half-Lives

Next, determine the number of half-lives that have elapsed. Divide the time in years by the half-life.equation:\[\frac{{0.428\,\mathrm{yr}}}{{1.01\,\mathrm{yr/half\text{-}life}}} = 0.423\text{-}half\text{-}lives\]
03

- Calculate the Remaining Mass

Use the formula for exponential decay to find the remaining mass. The formula is \[m(t) =m_0 \times (\frac{1}{2} )^{n}\] where: - m(t) is the remaining mass after time t - m_0 is the initial mass - n is the number of half-livesequation:\[m(0.428\,\text{yr}) = 2.00\,\text{mg} \times \left(\frac{1}{2}\right) ^ {0.423} = 2.00\,\text{mg} \times 0.748 = 1.496\,\text{mg}\]

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Key Concepts

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

isotopes
Isotopes are different forms of the same element that have the same number of protons but different numbers of neutrons. This means they occupy the same place in the periodic table but have different atomic masses. For example, the isotope \({ }^{212}_{83} \mathrm{Bi}\) has 83 protons (which defines it as bismuth) but has a different mass number, 212, due to the number of neutrons it contains.

The properties of isotopes can include differences in stability, leading to radioactive isotopes, which can undergo decay over time. Stable isotopes, on the other hand, do not undergo radioactive decay. Naturally occurring isotopes can have a range of applications, such as in medical imaging, cancer treatment, and dating ancient artifacts.
half-life
The half-life of a radioactive substance is the time it takes for half of the material to decay. This concept is crucial in understanding the rate of decay of isotopes. For instance, in the given exercise, the half-life of \({ }^{212}_{83} \mathrm{Bi}\) is 1.01 years.

Knowing the half-life allows us to calculate how much of the substance will remain after a certain amount of time. For example, if you start with 2 mg of a substance with a half-life of 1.01 years, after 1.01 years, you will have 1 mg remaining. The half-life is constant and unaffected by the initial amount of substance.

This principle is used in various fields like archaeology for dating artifacts (carbon dating), medicine for determining appropriate dosages of radioisotopes, and even in environmental science for studying the effects of pollutants.
exponential decay
Exponential decay describes the process by which the quantity of a substance decreases at a rate proportional to its current value. The formula used to express this is \[ m(t) = m_0 \times (\frac{1}{2})^{n} \] where:
  • \( m(t) \) is the remaining mass after time t.
  • \( m_0 \) is the initial mass.
  • \( n \) is the number of half-lives that have passed.
In the exercise, we used this formula to calculate how much of a 2.00 mg sample of \( { }^{212}_{83} \mathrm{Bi} \) would remain after 3750 hours (about 0.428 years). As we found that 0.423 half-lives had passed, we plugged this into the formula to find:

\[ m(0.428 \text{yr}) = 2.00 \text{mg} \times \left(\frac{1}{2}\right)^{0.423} = 2.00 \text{mg} \times 0.748 = 1.496 \text{mg} \]

Exponential decay is a fundamental concept in various scientific fields and can model many real-world phenomena, such as population decline, cooling of objects, and charge reduction in capacitors.

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