Concept explainers
(a)
Interpretation: The radiotracer having shortest half-life using graph has to be identified.
Concept introduction: In a nuclear reaction, unstable nuclide disintegrates into stable nuclide with emission of various radiations spontaneously. It is called radioactivity of nucleus. Emission of rays like alpha, beta and gamma rays occur in the process which changes the identity of nucleus.
During a nuclear change, an element undergoes change in its
(a)
Answer to Problem 4RQ
Mo-99 has the shortest half life.
Explanation of Solution
Time needed for original or initial sample of radioactive element to decay to its half quantity or concentration is known as half-life time.
The formula to calculate half life ( T) is as follows:
T=tn
Where,
n is number of half lives.
t is the elapsed time.
T is the half life of radioactive nucleus.
For the graph shown for half life in days versus the radioactive element, it is clear that the half lives of elements are as follows:
I-131 has half life of 8 days. Similarly for Mo-99, P-32 and Xe-133 half lives are 3 days, 14 days and 5 days approximately; respectively.
So the shortest half life is for Mo-99.
(b)
Interpretation: The radiotracer having half-life equal to sum of half lives of Xe-133 and Mo-99 using graph has to be identified.
Concept introduction: In a nuclear reaction, unstable nuclide disintegrates into stable nuclide with emission of various radiations spontaneously. It is called radioactivity of nucleus. Emission of rays like alpha, beta and gamma rays occur in the process which changes the identity of nucleus.
During a nuclear change, an element undergoes change in its atomic mass or atomic number or both by emission of charged particles such as beta particle, alpha particle. Beta particle is one that has unit negative charge and alpha particle is one that has 2 unit positive charges and is equivalent to helium atom 42He . Emission of proton, beta particle, alpha particle causes a change in atomic number of element.
(b)
Answer to Problem 4RQ
The radiotracer having half-life equal to sum of half lives of Xe-133 and Mo-99 is I-131.
Explanation of Solution
Time needed for original or initial sample of radioactive element to decay to its half quantity or concentration is known as half-life time.
The formula to calculate half life ( T) is as follows:
T=tn
Where,
n is number of half lives.
t is the elapsed time.
T is the half life of radioactive nucleus.
For the graph shown for half life in days versus the radioactive element, it is clear that the half lives of elements are as follows:
I-131 has half life of 8 days. Similarly for Mo-99, P-32 and Xe-133 half lives are 3 days, 14 days and 5 days approximately; respectively.
On addition of half lives of Mo-99 and Xe-133 result comes out to be 8 days which is equal to half life of I-131.
(c)
Interpretation: The radiotracer will be left in least quantity after 15 days, using graph has to be identified.
Concept introduction: In a nuclear reaction, unstable nuclide disintegrates into stable nuclide with emission of various radiations spontaneously. It is called radioactivity of nucleus. Emission of rays like alpha, beta and gamma rays occur in the process which changes the identity of nucleus.
During a nuclear change, an element undergoes change in its atomic mass or atomic number or both by emission of charged particles such as beta particle, alpha particle. Beta particle is one that has unit negative charge and alpha particle is one that has 2 unit positive charges and is equivalent to helium atom 42He . Emission of proton, beta particle, alpha particle causes a change in atomic number of element.
(c)
Answer to Problem 4RQ
As per the graph, Mo-99 will be left with least quantity after 15 days.
Explanation of Solution
Time needed for original or initial sample of radioactive element to decay to its half quantity or concentration is known as half-life time.
For the graph shown for half life in days versus the radioactive element, it is clear that the half lives of elements are as follows:
I-131 has half life of 8 days. Similarly for Mo-99, P-32 and Xe-133 half lives are 3 days, 14 days and 5 days approximately; respectively.
As per the graph, Mo-99 will be left with least quantity after 15 days. This is so because its half life is 3 days, it will take 3 days to get reduced to half of its original sample. So nearly 6 days will be required to get consumed nearly full.
(d)
Interpretation: The radiotracer will be left in greatest quantity after 15 days, using graph has to be identified.
Concept introduction: In a nuclear reaction, unstable nuclide disintegrates into stable nuclide with emission of various radiations spontaneously. It is called radioactivity of nucleus. Emission of rays like alpha, beta and gamma rays occur in the process which changes the identity of nucleus.
During a nuclear change, an element undergoes change in its atomic mass or atomic number or both by emission of charged particles such as beta particle, alpha particle. Beta particle is one that has unit negative charge and alpha particle is one that has 2 unit positive charges and is equivalent to helium atom 42He . Emission of proton, beta particle, alpha particle causes a change in atomic number of element.
(d)
Answer to Problem 4RQ
As per the graph, P-32 will be left with greatest quantity after 15 days.
Explanation of Solution
Time needed for original or initial sample of radioactive element to decay to its half quantity or concentration is known as half-life time.
For the graph shown for half life in days versus the radioactive element, it is clear that the half lives of elements are as follows:
I-131 has half life of 8 days. Similarly for Mo-99, P-32 and Xe-133 half lives are 3 days, 14 days and 5 days approximately; respectively.
As per the graph, P-32 will be left in greatest quantity after 15 days. This is so because its half life is 14 days approximately; it will take 14 days to get reduced to half of its original sample. So nearly 28 days more will be required to get consumed nearly full. Whereas I-131, Xe-133 and Mo-99 will require 16 days, 10 days and 6 days to consumed nearly completely respectively.
Chapter 19 Solutions
World of Chemistry
- Draw the mechanism for the following Friedel-Craft reaction. AlBr3 Brarrow_forward(a) Draw the structures of A and B in the following reaction. (i) NaNH2, NH3(1) A + B (ii) H3O+arrow_forwardFor the reaction 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 →> NO₂+ NO3_(K1) NO2 + NO3 → N2O5 (k-1) NO2 + NO3 → → NO2 + O2 + NO (K2) NO + N2O5- NO2 + NO2 + NO2 (K3) d[N₂O5] __2k‚k₂[N2O5] Indicate whether the following rate expression is acceptable: dt k₁₁+ k₂arrow_forward
- Consider the following decomposition reaction of N2O5(g): For the reaction 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 → NO2 + NO3 (K1) NO2 + NO3 → N2O5 (k-1) NO2 + NO3 → NO2 + O2 + NO (K2) NO + N2O5 → NO2 + NO2 + NO2 (K3) Indicate whether the following rate expression is acceptable: d[N2O5] = -k₁[N₂O₂] + K¸₁[NO₂][NO3] - K¸[NO₂]³ dtarrow_forwardIn a reaction of A + B to give C, another compound other than A, B or C may appear in the kinetic equation.arrow_forwardFor the reaction 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 →> NO₂+ NO3_(K1) NO2 + NO3 → N2O5 (k-1) NO2 + NO3 → → NO2 + O2 + NO (K2) NO + N2O5- NO2 + NO2 + NO2 (K3) d[N₂O5] __2k‚k₂[N2O5] Indicate whether the following rate expression is acceptable: dt k₁₁+ k₂arrow_forward
- Given the reaction R + Q → P, indicate the rate law with respect to R, with respect to P and with respect to P.arrow_forwardSteps and explanations. Also provide, if possible, ways to adress this kind of problems in general.arrow_forwardk₁ Given the reaction A B, indicate k-1 d[A] (A). the rate law with respect to A: (B). the rate law with respect to B: d[B] dt dtarrow_forward
- k₁ Given the reaction R₂ R + R, indicate k-1 (A). the rate law with respect to R2: (B). the rate law with respect to R: d[R₂] dt d[R] dtarrow_forwardGiven the reaction R+ Q → P, indicate (A). the rate law with respect to P: (B). the rate law with respect to R: (C). the rate law with respect to Q: d[P] dt d[R] dt d[Q] dtarrow_forwardThe reaction for obtaining NO2 from NO and O2 has the rate equation: v = k[NO]2[O2]. Indicate which of the following options is correct.(A). This rate equation is inconsistent with the reaction consisting of a single trimolecular step.(B). Since the overall order is 3, the reaction must necessarily have some trimolecular step in its mechanism.(C). A two-step mechanism: 1) NO + NO ⇄ N2O2 (fast); 2) N2O2 + O2 → NO2 + NO2 (slow).(D). The mechanism must necessarily consist of three unimolecular elementary steps with very similar rate constants.arrow_forward
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY