/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 The half-life of \({ }^{131} \ma... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 22

The half-life of \({ }^{131} \mathrm{I}\) is \(8.04\) days. (a) Calculate the decay constant for this isotope. (b) Find the number of \({ }^{131} \mathrm{I}\) nuclei necessary to produce a sample with an activity of \(0.5 \mu \mathrm{Ci}\).

Short Answer

Expert verified
The decay constant for this isotope \({ }^{131} \mathrm{I}\) is approximately \(0.0864\) per day, and the number of \({ }^{131} \mathrm{I}\) nuclei necessary to produce a sample with an activity of \(0.5 \mu \mathrm{Ci}\) is approximately \(1.85 \times 10^{16}\).

Step by step solution

01

Calculate the decay constant

In this step, the decay constant \(\lambda\) is calculated using the half-life formula. \(\lambda = ln(2) / T_{1/2}\). Substituting the given half life (\(8.04\) days) into this equation results in \(\lambda \approx 0.0864\) per day.
02

Calculate the number of nuclei

Next, calculate the number of nuclei using the formula \(A = \lambda*N\). The given activity is \(0.5 \mu \mathrm{Ci}\), but this needs to be converted to decays per day. \(1 \mu \mathrm{Ci} = 3.7 \times 10^{10}\) decays per second, so \(0.5 \mu \mathrm{Ci} = 0.5 \times 3.7 \times 10^{10} = 1.85 \times 10^{10}\) decays per second. Convert this to decays per day by multiplying by the number of seconds in a day (\(86400\)): \(1.85 \times 10^{10} \times 86400 \approx 1.6 \times 10^{15}\) decays per day. Solving the equation \(1.6 \times 10^{15} = 0.0864 \times N\) gives \(N \approx 1.85 \times 10^{16}\).
03

Final calculation and unit conversion

So, the number of \({ }^{131} \mathrm{I}\) nuclei necessary to produce a sample with an activity of \(0.5 \mu \mathrm{Ci}\) is approximately \(1.85 \times 10^{16}\).

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Key Concepts

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

Half-life Calculation
Half-life is an important concept in radioactive decay. It refers to the time it takes for half of a radioactive substance to transform into a different element. This helps us understand how long a radioactive substance remains active.
The formula used to calculate the decay constant (\(\lambda\)) from half-life is given by:
  • \(\lambda = \frac{\ln(2)}{T_{1/2}}\)
where \(T_{1/2}\) is the half-life of the isotope. In this exercise, we're working with \(^{131}\mathrm{I}\) which has a half-life of 8.04 days. Plugging this value into our formula provides the decay constant \(\lambda \approx 0.0864\, \text{per day}\).
This step is crucial as it links the physical property of the substance directly to the mathematical models we use for calculations in radioactive decay.
Decay Constant
The decay constant (\(\lambda\)) is pivotal in understanding how quickly a radioactive isotope decays. It represents the probability per unit time that a nucleus will decay. Decay is a random process, but the decay constant provides a statistical look at this process.
With the decay constant \(\lambda\) obtained, we use it in many calculations, such as determining the activity or the number of nuclei needed for a specific activity level. Here, it allows us to connect the half-life of a nucleus to its activity. Remember, the decay constant tells us how many nuclei will decay in a given time frame and is linked directly to our activity calculations.
Activity Conversion
Activity describes the rate at which a sample of radioactive material undergoes nuclear decay. It's essential to understand it in specific units, typically disintegrations per second (\(Bq\)) or curies (\(\mu \mathrm{Ci}\)).
In our example, the activity was initially given as \(0.5\, \mu \mathrm{Ci}\). To convert this to disintegrations per day, we use:
  • \(1 \mu \mathrm{Ci} = 3.7 \times 10^{10}\) decays per second
  • Multiply by seconds in a day: \(86400 s\)
  • This leads to \(0.5 \times 3.7 \times 10^{10} \times 86400 \approx 1.6 \times 10^{15}\) decays per day
This conversion is vital for comparing activities in different units or calculating the necessary number of radioactive nuclei, allowing us to find practically useful results from mathematical models.

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Most popular questions from this chapter

A freshly prepared sample of a certain radioactive isotope has an activity of \(10 \mathrm{mCi}\). After an elapsed time of \(4 \mathrm{~h}\), its activity is \(8 \mathrm{mCi}\). (a) Find the decay constant and half-life of the isotope. (b) How many atoms of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity \(30 \mathrm{~h}\) after it is prepared?

A certain African artifact is found to have a carbon-14 activity of \((0.12 \pm 0.01)\) Bq per gram of carbon. Assume the uncertainty is negligible in the half-life of \({ }^{14} \mathrm{C}\) ( \(\left.5730 \mathrm{yr}\right)\) and in the activity of atmospheric carbon \((0.25 \mathrm{~Bq}\) per gram \() .\) The age of the object lies within what range?

(a) Can \({ }^{57}\) Co decay by \(\mathrm{e}^{+}\) emission? Explain. (b) Can \({ }^{14} \mathrm{C}\) decay by \(\mathrm{e}^{-}\) emission? Explain. (c) If either answer is yes, what is the range of kinetic energies available for the beta particle?

(a) In the liquid-drop model of nuclear structure, why does the surface-effect term \(-C_{2} A^{2 / 3}\) have a minus sign? (b) The binding energy of the nucleus increases as the volume-to-surface ratio increases. Calculate this ratio for both spherical and cubical shapes, and explain which is more plausible for nuclei.

A sample of radioactive material is said to be carrier-free when no stable isotopes of the radioactive element are present. Calculate the mass of strontium in a carrierfree \(5-\mathrm{mCi}\) sample of \({ }^{90} \mathrm{Sr}\) whose half-life is \(28.8 \mathrm{yr}\).
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