(a)
Interpretation:
Half-life of the radionuclide has to be determined if after 3.2 days, 1/8 fraction of undecayed nuclide is present.
Concept Introduction:
Radioactive nuclides undergo disintegration by emission of radiation. All the radioactive nuclide do not undergo the decay at a same rate. Some decay rapidly and others decay very slowly. The nuclear stability can be quantitatively expressed by using the half-life.
The time required for half quantity of the radioactive substance to undergo decay is known as half-life. It is represented as
The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,
(a)
Answer to Problem 11.28EP
Half-life of the radionuclide is 1.1 days.
Explanation of Solution
Number of half-lives can be determined as shown below,
As the bases are equal, the power can be equated. This gives the number of half-lives that have elapsed as 3 half-lives.
In the problem statement it is given that the time is 3.2 days. From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,
Therefore, the half-life of the given sample is determined as 1.1 days.
Half-life of the given sample is determined.
(b)
Interpretation:
Half-life of the radionuclide has to be determined if after 3.2 days, 1/128 fraction of undecayed nuclide is present.
Concept Introduction:
Radioactive nuclides undergo disintegration by emission of radiation. All the radioactive nuclide do not undergo the decay at a same rate. Some decay rapidly and others decay very slowly. The nuclear stability can be quantitatively expressed by using the half-life.
The time required for half quantity of the radioactive substance to undergo decay is known as half-life. It is represented as
The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,
(b)
Answer to Problem 11.28EP
Half-life of the radionuclide is 0.46 day.
Explanation of Solution
Number of half-lives can be determined as shown below,
As the bases are equal, the power can be equated. This gives the number of half-lives that have elapsed as 7 half-lives.
In the problem statement it is given that the time is 3.2 days. From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,
Therefore, the half-life of the given sample is determined as 0.46 day.
Half-life of the given sample is determined.
(c)
Interpretation:
Half-life of the radionuclide has to be determined if after 3.2 days, 1/32 fraction of undecayed nuclide is present.
Concept Introduction:
Radioactive nuclides undergo disintegration by emission of radiation. All the radioactive nuclide do not undergo the decay at a same rate. Some decay rapidly and others decay very slowly. The nuclear stability can be quantitatively expressed by using the half-life.
The time required for half quantity of the radioactive substance to undergo decay is known as half-life. It is represented as
The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,
(c)
Answer to Problem 11.28EP
Half-life of the radionuclide is 0.64 day.
Explanation of Solution
Number of half-lives can be determined as shown below,
As the bases are equal, the power can be equated. This gives the number of half-lives that have elapsed as 5 half-lives.
In the problem statement it is given that the time is 3.2 days. From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,
Therefore, the half-life of the given sample is determined as 0.64 day.
Half-life of the given sample is determined.
(d)
Interpretation:
Half-life of the radionuclide has to be determined if after 3.2 days, 1/512 fraction of undecayed nuclide is present.
Concept Introduction:
Radioactive nuclides undergo disintegration by emission of radiation. All the radioactive nuclide do not undergo the decay at a same rate. Some decay rapidly and others decay very slowly. The nuclear stability can be quantitatively expressed by using the half-life.
The time required for half quantity of the radioactive substance to undergo decay is known as half-life. It is represented as
The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,
(d)
Answer to Problem 11.28EP
Half-life of the radionuclide is 0.36 day.
Explanation of Solution
Number of half-lives can be determined as shown below,
As the bases are equal, the power can be equated. This gives the number of half-lives that have elapsed as 9 half-lives.
In the problem statement it is given that the time is 3.2 days. From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,
Therefore, the half-life of the given sample is determined as 0.36 day.
Half-life of the given sample is determined.
Want to see more full solutions like this?
Chapter 11 Solutions
General, Organic, and Biological Chemistry
- The density of sulfuric acid is 0.875 g/cm3. If a procedure needed 4.00 mL of sulfuricacid, how many grams would you measure out?arrow_forwardQuestion Suggest a mechanism for the following reactions. Each will require multiple types of concerted pericyclic reactions (cycloaddition, electrocyclic, and sigmatropic. Classify each reaction type. CN a. NC 180 °Carrow_forwardDon't used Ai solution and don't used hand raitingarrow_forward
- Q2: Ranking Acidity a) Rank the labeled protons in the following molecule in order of increasing pKa. Briefly explain the ranking. Use Table 2.2 as reference. Ha Нь HC H-N Ha OHe b) Atenolol is a drug used to treat high blood pressure. Which of the indicated N-H bonds is more acidic? Explain. (Hint: use resonance structures to help) Name the functional groups on atenolol. H H-N atenolol Ν H-N OH Нarrow_forwardAnswer d, e, and farrow_forwardIf the rotational constant of a molecule is B = 120 cm-1, it can be stated that the transition from 2←1:a) gives rise to a line at 120 cm-1b) is a forbidden transitionc) gives rise to a line at 240 cm-1d) gives rise to a line at 480 cm-1arrow_forward
- Briefly indicate the coordination forms of B and Si in borates and silicates, respectively.arrow_forwardCan you please draw out the Lewis structure for these two formulasarrow_forwardIn a rotational Raman spectrum of a diatomic molecule it is correct to say that:a) anti-Stokes lines occur at frequencies higher than the excitatory oneb) Stokes lines occur at frequencies higher than the excitatory onec) Rayleigh scattering is not observedd) Rayleigh scattering corresponds to delta J = 0arrow_forward
- Of the molecules: H2, N2, HCl, CO2, indicate which ones can give Raman vibration-rotation spectra:a) H2, N2 and HClb) H2, N2, HCl and CO2c) H2 and N2d) all of themarrow_forwardCan you please help me with drawing the Lewis structure of each molecular formula?I truly appreciate you!arrow_forwardCan you please help me with drawing the Lewis structure of each molecular formula?I truly appreciate you!arrow_forward
- General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage LearningPrinciples of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage Learning
- World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningGeneral Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage Learning