The half-life of \({ }^{27} \mathrm{Mg}\) is \(9.50 \mathrm{~min}\). (a) Initially there were \(4.20 \times 10^{12}{ }^{27} \mathrm{Mg}\) nuclei present. How many \({ }^{27}\) Mg nuclei are left 30.0 min later? (b) Calculate the \({ }^{27} \mathrm{Mg}\) activities \((\) in \(\mathrm{Ci})\) at \(t=0\) and \(t=30.0 \mathrm{~min}\) (c) What is the probability that any one \({ }^{27} \mathrm{Mg}\) nucleus decays during a 1 -s interval? What assumption is made in this calculation?

Short Answer

Expert verified
Part (a): The number of \( ^{27}Mg \) nuclei left is \( N = 4.20 \times 10^{12} e^{\frac{-ln 2}{570} \times 1800} \) nuclei. Part (b): The activity at \( t = 0 \) is \( A_0 = \frac{\lambda N_0}{3.70 \times 10^{10}} \) Ci, and the activity at \( t = 30 \) min is \( A = \frac{\lambda N}{3.70 \times 10^{10}} \) Ci. Part (c): The decay probability in a 1 sec interval is \( P = \lambda \times 1 \) sec. The assumption is that the decay probability is constant over that time interval.

Step by step solution

01

Calculation of Remaining Nuclei.

Using the exponential decay formula \( N = N_0 e^{-\lambda t} \), where \( \lambda = \frac{ln 2}{T} \). \( N_0 \) is the initial number of nuclei, \( N \) the final number of nuclei, \( t \) the elapsed time and \( T \) the half-life. With \( N_0 = 4.20 \times 10^{12} \) nuclei, \( T = 9.5 \) min = \( 570 \) seconds, and \( t = 30 \) min = \( 1800 \) seconds, we find \( N = N_0 e^{-\lambda t} = 4.20 \times 10^{12} e^{\frac{-ln 2}{570} \times 1800} \).
02

Calculation of Activity.

The activity \( A \) is given by \( A = \lambda N \). At \( t = 0 \), \( N = N_0 \), so \( A_0 = \lambda N_0 \). At \( t = 30 \) min, \( N = N \) from Step 1, so \( A = \lambda N \). 1 Ci = \( 3.70 \times 10^{10} \) decays/s, so to convert the activity in decays/s to Ci, divide by \( 3.70 \times 10^{10} \). Thus, \( A_0 = \frac{\lambda N_0}{3.70 \times 10^{10}} \), and \( A = \frac{\lambda N}{3.70 \times 10^{10}} \).
03

Calculation of Decay Probability.

The probability \( P \) that a nucleus decays in a given time interval \( \Delta t \) is \( P = \lambda \Delta t \). We find \( P = \lambda \times 1 \) sec. The underlying assumption is that the probability does not change during that time interval.

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Key Concepts

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

Exponential Decay
Exponential decay is a process where the quantity of a particular substance decreases at a rate proportional to its current amount. In the context of nuclear physics, it's used to describe how the number of radioactive nuclei in a sample reduces over time. The key feature of exponential decay is that it happens gradually and steadily, and the rate of decay depends on a constant known as the decay constant, denoted by \( \lambda \).

The mathematical expression for exponential decay is \( N = N_0 e^{-\lambda t} \), where:\
    \
  • \\N\\ is the number of remaining nuclei after time \(t\)\
  • \
  • \\N_0\\ is the initial number of nuclei\
  • \
  • \(\lambda\) is the decay constant\
  • \
  • \(t\) is the time elapsed\
  • \
\This formula is a cornerstone in understanding how substances like \( ^{27}Mg \) diminish over time, allowing us to predict the remaining amount after any given period.
Radioactive Decay
Radioactive decay is a natural, spontaneous process by which an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. Different radioactive isotopes, or radioisotopes, have varying levels of stability and thus decay at different rates. \( ^{27}Mg \) is a radioisotope with a specific half-life, which is the amount of time it takes for half of the sample to decay.

The half-life is crucial in radioactive decay because it's a constant value unique to each isotope and can be used to calculate decay activity, remaining radioactive material, and potential risks associated with the radioisotope. In nuclear chemistry and physics, understanding the half-life provides insight into the stability of the nucleus and the duration for which it will remain radioactive.
Decay Activity Calculation
Decay activity calculation involves determining the rate at which a radioactive substance undergoes decay. The activity, \(A\), quantifies the amount of decay occurring in a specified time and is measured in units like becquerels (Bq) or curies (Ci). Activity is directly proportional to the number of radioactive nuclei present; as the substance decays and the number of nuclei decreases, so does the activity.

To calculate the activity of a radioactive substance, you can use the formula \( A = \lambda N \), where:\
    \
  • \(\lambda\) is the decay constant\
  • \
  • \(N\) is the number of undecayed nuclei at the time of measurement\
  • \
\For instance, for \( ^{27}Mg \) at the starting point \(t = 0\), the activity \(A_0\) can be found using the initial number of nuclei \(N_0\). Whereas at a later time (such as 30 minutes) the remaining nuclei \(N\) would be used in the activity calculation, reflecting the decreased rate of decay as the number of undecayed nuclei lessens.

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Most popular questions from this chapter

Write complete nuclear equations for these processes: (a) tritium, \({ }^{3} \mathrm{H},\) undergoes \(\beta\) decay; \((\mathrm{b}){ }^{242} \mathrm{Pu}\) undergoes \(\alpha\) -particle emission; \((\mathrm{c})^{131} \mathrm{I}\) undergoes \(\beta\) decay; (d) \(^{251} \mathrm{Cf}\) emits an \(\alpha\) particle.

Calculate the nuclear binding energy (in J) and the nuclear binding energy per nucleon of the following isotopes: (a) \({ }_{3}^{7} \mathrm{Li}(7.01600 \mathrm{amu})\) and (b) 35 17 Cl \((34.95952 \mathrm{amu})\)

Nuclear waste disposal is one of the major concerns of the nuclear industry. In choosing a safe and stable environment to store nuclear wastes, consideration must be given to the heat released during nuclear decay. As an example, consider the \(\beta\) decay of \({ }^{90} \mathrm{Sr}\) \((89.907738 \mathrm{amu})\) $$ { }_{38}^{90} \mathrm{Sr} \longrightarrow{ }_{39}^{90} \mathrm{Y}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=28.1 \mathrm{yr} $$ The \({ }^{90} \mathrm{Y}(89.907152 \mathrm{amu})\) further decays as follows: $$ { }_{39}^{90} \mathrm{Y} \longrightarrow{ }_{40}^{90} \mathrm{Zr}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=64 \mathrm{~h} $$ Zirconium-90 (89.904703 amu) is a stable isotope. (a) Use the mass defect to calculate the energy released (in joules) in each of the preceding two decays. (The mass of the electron is \(5.4857 \times\) \(10^{-4}\) amu. ( b) Starting with 1 mole of \({ }^{90}\) Sr, calculate the number of moles of \(9^{9}\) Sr that will decay in a year. (c) Calculate the amount of heat released (in kilojoules) corresponding to the number of moles of \({ }^{90} \mathrm{Sr}\) decayed to \({ }^{90} \mathrm{Zr}\) in \((\mathrm{b})\)

For each pair of isotopes listed, predict which one is less stable: (a) \({ }_{3}^{6} \mathrm{Li}\) or \({ }_{3}^{9} \mathrm{Li},\) (b) \({ }_{11}^{23} \mathrm{Na}\) or \({ }_{11}^{25} \mathrm{Na},\) (c) \({ }_{20}^{48} \mathrm{Ca}\) or \({ }_{21}^{48} \mathrm{Sc}\).

Cobalt- 60 is an isotope used in diagnostic medicine and cancer treatment. It decays with \(\gamma\) ray emission. Calculate the wavelength of the radiation in nanometers if the energy of the \(\gamma\) ray is \(2.4 \times 10^{-13} \mathrm{~J} / \mathrm{photon}\)

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