In this problem, you will verify the statement (in Section 29.4) that the \(^{14} \mathrm{C}\) activity in a living sample is \(0.25 \mathrm{Bq}\) per gram of carbon. (a) What is the decay constant \(\lambda\) for \(^{14} \mathrm{C} ?\) (b) How many \(^{14} \mathrm{C}\) atoms are in \(1.00 \mathrm{g}\) of carbon? One mole of carbon atoms has a mass of \(12.011 \mathrm{g},\) and the relative abundance of \(^{14} \mathrm{C}\) is \(1.3 \times 10^{-12} .\) (c) Using your results from parts (a) and (b), calculate the \(^{14} \mathrm{C}\) activity per gram of carbon in a living sample.

Short Answer

Expert verified
Question: Verify the statement that the activity of Carbon-14 in a living sample is 0.25 Bq per gram of carbon. Answer: The Carbon-14 activity per gram of carbon in a living sample has been calculated to be 0.25 Bq, confirming the given statement.

Step by step solution

01

Calculate the decay constant for Carbon-14

To find the decay constant, we need to use the formula for the half-life of Carbon-14: \(\lambda = \dfrac{ln(2)}{T_{1/2}}\) The half-life of Carbon-14, \(T_{1/2}\), is approximately 5730 years. We first need to convert this to seconds to use in our formula: 1 year = 365.25 days \times 24 h/day \times 3600 s/h \(T_{1/2} = 5730\; \text{years}\; \cdot 365.25\; \frac{\text{days}}{\text{year}}\; \cdot 24\; \frac{\text{h}}{\text{day}}\; \cdot 3600\; \frac{\text{s}}{\text{h}} = 1.808 \times 10^{11} \:\text{s}\) Now we can plug this half-life value into the formula to find the decay constant: \(\lambda = \dfrac{ln(2)}{1.808 \times 10^{11}\; \text{s}} = 3.842 \times 10^{-12}\; \text{s}^{-1}\) So the decay constant for Carbon-14 is \(3.842 \times 10^{-12} \;\text{s}^{-1}\).
02

Calculate the number of Carbon-14 atoms in 1 gram of carbon

We are given that one mole of carbon has a mass of 12.011 grams and the relative abundance of Carbon-14 is \(1.3 \times 10^{-12}\). To find the number of Carbon-14 atoms in 1 gram of carbon, we first need to determine how many moles are present in 1 gram of carbon: \(\text{moles} = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{1.00\; \text{g}}{12.011\; \text{g/mol}} = 0.0833\; \text{mol}\) Now, we can find the number of Carbon-12 atoms in these moles: \(\text{atoms} = \text{moles} \cdot \text{Avogadro's number} = 0.0833\; \text{mol} \cdot 6.022 \times 10^{23}\; \frac{\text{atoms}}{\text{mol}} = 5.01 \times 10^{22}\; \text{atoms}\) Finally, we can find the number of Carbon-14 atoms by multiplying the number of Carbon-12 atoms by the relative abundance: \(\text{Carbon-14 atoms} = 5.01 \times 10^{22}\; \text{atoms} \cdot 1.3 \times 10^{-12} = 6.51 \times 10^{10}\; \text{atoms}\) There are \(6.51 \times 10^{10}\) Carbon-14 atoms in 1 gram of carbon.
03

Calculate the Carbon-14 activity per gram of carbon in a living sample

Now we have the decay constant \(\lambda\) and the number of Carbon-14 atoms in 1 gram of carbon. We can use these values to calculate the activity using the formula: \(\text{Activity} = \lambda \cdot \text{number of atoms} = 3.842 \times 10^{-12} \;\text{s}^{-1} \cdot 6.51 \times 10^{10} \;\text{atoms} = 0.25\;\text{Bq}\) The Carbon-14 activity per gram of carbon in a living sample is 0.25 Bq, which confirms the statement given in the problem.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Some types of cancer can be effectively treated by bombarding the cancer cells with high energy protons. Suppose \(1.16 \times 10^{17}\) protons, each with an energy of \(950 \mathrm{keV},\) are incident on a tumor of mass $3.82 \mathrm{mg} .\( If the quality factor for these protons is \)3.0,$ what is the biologically equivalent dose?
Which decay mode would you expect for radioactive ${ }_{14}^{31} \mathrm{Si}: \alpha, \beta^{-},\( or \)\beta^{+} ?$ Explain. [Hint: Look at the neutron-to- proton ratio.]
Show that \(c^{2}=931.494\) MeV/u. [Hint: Start with the conversion factors to SI units for MeV and atomic mass units.]
The first nuclear reaction ever observed (in 1919 by Ernest Rutherford) was \(\alpha+_{7}^{14} \mathrm{N} \right arrow \mathrm{p}+\mathrm{X} .\) (a) Show that the reaction product "X" must be \(\quad\) 8O. (b) For this reaction to take place, the \(\alpha\) particle must come in contact with the nitrogen nucleus. Calculate the distance \(d\) between their centers when they just make contact. (c) If the \(\alpha\) particle and the nitrogen nucleus are initially far apart, what is the minimum value of their kinetic energy necessary to bring the two into contact? Express your answer in terms of the elementary charge \(e\), the contact distance \(d\), and whatever else you need. (d) Is the total kinetic energy of the reaction products more or less than the initial kinetic energy in part (c)? Why? Calculate this kinetic energy difference.
An isotope of sodium, \({ }_{11}^{2} \mathrm{Na}\), decays by \(\beta^{+}\) emission. Estimate the maximum possible kinetic energy of the positron by assuming that the kinetic energy of the positron by assuming that the kinetic energy of the daughter nucleus and the total energy of the neutrino emitted are both zero. [Hint: Remember to keep track of the electron masses.
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free