JEE Exam > JEE Questions > The radionuclide56Mnis being produced in a cy...
Start Learning for Free
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 13.86×1010s−1. (Use ln2 = 0.693; Avogadro number = 6×1023; atomic weight of 56Mn = 56gm/mole)
Q. At what constant rate P , 56Mn nuclei are being produced in the cyclotron during the bombardment?
- a)2 x 1011 nuclei/sec
- b)6.93 x 1010 nuclei/sec
- c)9.6 x 1010 nuclei/sec
- d)13.86 x 1010 nuclei/sec
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The radionuclide56Mnis being produced in a cyclotron at a constant rat...
Given Data:
- Rate of decay of 56Mn nuclei, λ = 13.86 x 10^10 s^-1
- Half-life of 56Mn nuclei, T1/2 = 2.5 hours = 2.5 x 3600 seconds
- Atomic weight of 56Mn = 56 g/mol
Calculating Production Rate:
- Using the relation between decay constant (λ) and half-life (T1/2): λ = 0.693 / T1/2
- Therefore, λ = 0.693 / (2.5 x 3600) = 0.0000388 s^-1
- Number of 56Mn nuclei produced per second by bombardment: P = λ * N55Mn
- N55Mn = (Avogadro's number) / Atomic weight of 55Mn = 6 x 10^23 / 55 = 1.09 x 10^22
- P = 0.0000388 * 1.09 x 10^22 ≈ 4.22 x 10^17 nuclei/sec
- Converting this value to nuclei/sec: P = 4.22 x 10^17 / (56 / 56) = 7.57 x 10^16 nuclei/sec
Final Answer:
The constant rate at which 56Mn nuclei are being produced in the cyclotron during the bombardment is approximately 7.57 x 10^16 nuclei/sec, which is closest to option 'D' (13.86 x 10^10 nuclei/sec).
- Rate of decay of 56Mn nuclei, λ = 13.86 x 10^10 s^-1
- Half-life of 56Mn nuclei, T1/2 = 2.5 hours = 2.5 x 3600 seconds
- Atomic weight of 56Mn = 56 g/mol
Calculating Production Rate:
- Using the relation between decay constant (λ) and half-life (T1/2): λ = 0.693 / T1/2
- Therefore, λ = 0.693 / (2.5 x 3600) = 0.0000388 s^-1
- Number of 56Mn nuclei produced per second by bombardment: P = λ * N55Mn
- N55Mn = (Avogadro's number) / Atomic weight of 55Mn = 6 x 10^23 / 55 = 1.09 x 10^22
- P = 0.0000388 * 1.09 x 10^22 ≈ 4.22 x 10^17 nuclei/sec
- Converting this value to nuclei/sec: P = 4.22 x 10^17 / (56 / 56) = 7.57 x 10^16 nuclei/sec
Final Answer:
The constant rate at which 56Mn nuclei are being produced in the cyclotron during the bombardment is approximately 7.57 x 10^16 nuclei/sec, which is closest to option 'D' (13.86 x 10^10 nuclei/sec).
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. Can you explain this answer?
Question Description
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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. 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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. 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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. 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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. 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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. 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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. 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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. 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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. 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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. 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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. 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 of56Mnbecomes constant equal to13.86×1010s−1. (Useln2 = 0.693; Avogadro number =6×1023; atomic weight of56Mn= 56gm/mole)Q.At what constant rateP,56Mn nuclei are being produced in the cyclotron during the bombardment?a)2 x 1011nuclei/secb)6.93 x 1010nuclei/secc)9.6 x 1010nuclei/secd)13.86 x 1010nuclei/secCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup