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The radionuclide 56Mn is being produced in a cyclotron at a constant rate P by bombarding a manganese target with deutrons. 56Mn has a half life of 2.5 hours and the target contains large number of only the stable manganese isotope 55Mn. The reaction that produces 56Mn is:
55Mn+d→56Mn+P
After being bombarded for a long time, the activity of 56Mn becomes constant equal to 13.86×10s−1. (Use ln2 = 0.693; Avogadro number = 6×1023 atomic weight of 56Mn = 56gm/mole)
Q. After a long time bombardment, number of 56Mn nuclei present in the target depends upon
(a) The number of 56Mn nuclei present at the start of the process
(b) Half life of the 56Mn
(c) The constant rate of production P
(a) The number of 56Mn nuclei present at the start of the process
(b) Half life of the 56Mn
(c) The constant rate of production P
- a)All (a), (b) and (c) are correct
- b)Only (a) and (b) are correct
- c)Only (b) and (c) are correct
- d)Only (a) and (c) are correct
Correct answer is option 'A'. Can you explain this answer?
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The radionuclide56Mnis being produced in a cyclotron at a constant rat...
+ d -> 56Mn + p
Since the radionuclide 56Mn is being produced at a constant rate P, we can use the concept of radioactive decay to determine the rate at which 56Mn is decaying.
The decay of 56Mn follows first-order kinetics, which means that the rate of decay is proportional to the amount of 56Mn present. The decay constant (λ) can be calculated using the half-life (t1/2) as follows:
λ = ln(2) / t1/2
Substituting the given half-life of 2.5 hours, we can calculate the decay constant:
λ = ln(2) / 2.5 hours
Now, let's consider the rate of production and decay of 56Mn. Since the rate of production is constant (P), the rate of decay (R) must balance the rate of production to maintain a constant amount of 56Mn.
R = P
The rate of decay can be calculated using the decay constant and the amount of 56Mn present (N):
R = λN
Since N is not given, we can express it in terms of the amount of 55Mn present (N55Mn) and the total amount of manganese present (NTotal):
N = NTotal - N55Mn
Now, let's substitute the expressions for R and N into the equation:
λN = P
λ(NTotal - N55Mn) = P
Rearranging the equation, we can solve for the amount of 55Mn present:
N55Mn = NTotal - P / λ
In summary, the amount of stable manganese isotope 55Mn present in the target can be calculated using the total amount of manganese present (NTotal), the rate of production of 56Mn (P), and the decay constant (λ):
N55Mn = NTotal - P / λ
Since the radionuclide 56Mn is being produced at a constant rate P, we can use the concept of radioactive decay to determine the rate at which 56Mn is decaying.
The decay of 56Mn follows first-order kinetics, which means that the rate of decay is proportional to the amount of 56Mn present. The decay constant (λ) can be calculated using the half-life (t1/2) as follows:
λ = ln(2) / t1/2
Substituting the given half-life of 2.5 hours, we can calculate the decay constant:
λ = ln(2) / 2.5 hours
Now, let's consider the rate of production and decay of 56Mn. Since the rate of production is constant (P), the rate of decay (R) must balance the rate of production to maintain a constant amount of 56Mn.
R = P
The rate of decay can be calculated using the decay constant and the amount of 56Mn present (N):
R = λN
Since N is not given, we can express it in terms of the amount of 55Mn present (N55Mn) and the total amount of manganese present (NTotal):
N = NTotal - N55Mn
Now, let's substitute the expressions for R and N into the equation:
λN = P
λ(NTotal - N55Mn) = P
Rearranging the equation, we can solve for the amount of 55Mn present:
N55Mn = NTotal - P / λ
In summary, the amount of stable manganese isotope 55Mn present in the target can be calculated using the total amount of manganese present (NTotal), the rate of production of 56Mn (P), and the decay constant (λ):
N55Mn = NTotal - P / λ
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The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer?
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The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer?.
The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer?.
Solutions for The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The radionuclide56Mnis being produced in a cyclotron at a constant ratePby bombarding a manganese target with deutrons.56Mnhas a half life of 2.5 hours and the target contains large number of only the stable manganese isotope55Mn. The reaction that produces56Mnis:55Mn+d→56Mn+PAfter being bombarded for a long time, the activity of56Mn becomes constant equal to13.86×10s−1. (Useln2 = 0.693; Avogadro number =6×1023atomic weight of56Mn = 56gm/mole)Q. After a long time bombardment, number of56Mn nuclei present in the target depends upon(a) The number of56Mnnuclei present at the start of the process(b) Half life of the56Mn(c) The constant rate of productionPa)All (a), (b) and (c) are correctb)Only (a) and (b) are correctc)Only (b) and (c) are correctd)Only (a) and (c) are correctCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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