JEE Exam > JEE Questions > A sample of radioactive material A, that has ...
Start Learning for Free
A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :
- a)20 days and 5 days
- b)20 days and 10 days
- c)5 days and 10 days
- d)10 days and 40 days
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A sample of radioactive material A, that has an activity of 10 mCi(1 C...
Most Upvoted Answer
A sample of radioactive material A, that has an activity of 10 mCi(1 C...
Given information:
- Activity of radioactive material A = 10 mCi
- Activity of radioactive material B = 20 mCi
- Number of nuclei in material A is twice the number of nuclei in material B
Conversion:
1 Ci = 3.7 x 10^10 decays/s
Therefore, 1 mCi = 3.7 x 10^7 decays/s
Calculating the number of decays per second:
- Activity of material A = 10 mCi = 10 x 3.7 x 10^7 decays/s = 37 x 10^7 decays/s
- Activity of material B = 20 mCi = 20 x 3.7 x 10^7 decays/s = 74 x 10^7 decays/s
Calculating the number of nuclei:
Since the number of nuclei in material A is twice the number of nuclei in material B:
Number of nuclei in material A = 2 x Number of nuclei in material B
Calculating the half-life:
The decay of radioactive material follows an exponential decay equation, given by:
A(t) = A(0) * (1/2)^(t/T)
Where:
A(t) = Activity at time t
A(0) = Initial activity
t = Time
T = Half-life
For material A:
A(t) = A(0) * (1/2)^(t/T_A)
Given that the initial activity of material A is 37 x 10^7 decays/s, and the half-life is T_A, we can substitute these values into the equation:
37 x 10^7 = 37 x 10^7 * (1/2)^(t/T_A)
Since the initial activity cancels out, we are left with:
1 = (1/2)^(t/T_A)
Taking the logarithm base 2 of both sides:
log_2(1) = log_2((1/2)^(t/T_A))
0 = (t/T_A) * log_2(1/2)
Since log_2(1/2) = -1, we have:
0 = -(t/T_A)
Therefore, t = 0, which means there is no decay. This implies that the half-life of material A is infinity.
For material B:
A(t) = A(0) * (1/2)^(t/T_B)
Given that the initial activity of material B is 74 x 10^7 decays/s, and the half-life is T_B, we can substitute these values into the equation:
74 x 10^7 = 74 x 10^7 * (1/2)^(t/T_B)
Since the initial activity cancels out, we are left with:
1 = (1/2)^(t/T_B)
Taking the logarithm base 2 of both sides:
log_2(1) = log_2((1/2)^(t/T_B))
0 = (t/T_B) * log_2(1/2)
Since log_2(1/2) = -1, we have:
0 = -(t/T_B)
Therefore, t = 0, which means there
- Activity of radioactive material A = 10 mCi
- Activity of radioactive material B = 20 mCi
- Number of nuclei in material A is twice the number of nuclei in material B
Conversion:
1 Ci = 3.7 x 10^10 decays/s
Therefore, 1 mCi = 3.7 x 10^7 decays/s
Calculating the number of decays per second:
- Activity of material A = 10 mCi = 10 x 3.7 x 10^7 decays/s = 37 x 10^7 decays/s
- Activity of material B = 20 mCi = 20 x 3.7 x 10^7 decays/s = 74 x 10^7 decays/s
Calculating the number of nuclei:
Since the number of nuclei in material A is twice the number of nuclei in material B:
Number of nuclei in material A = 2 x Number of nuclei in material B
Calculating the half-life:
The decay of radioactive material follows an exponential decay equation, given by:
A(t) = A(0) * (1/2)^(t/T)
Where:
A(t) = Activity at time t
A(0) = Initial activity
t = Time
T = Half-life
For material A:
A(t) = A(0) * (1/2)^(t/T_A)
Given that the initial activity of material A is 37 x 10^7 decays/s, and the half-life is T_A, we can substitute these values into the equation:
37 x 10^7 = 37 x 10^7 * (1/2)^(t/T_A)
Since the initial activity cancels out, we are left with:
1 = (1/2)^(t/T_A)
Taking the logarithm base 2 of both sides:
log_2(1) = log_2((1/2)^(t/T_A))
0 = (t/T_A) * log_2(1/2)
Since log_2(1/2) = -1, we have:
0 = -(t/T_A)
Therefore, t = 0, which means there is no decay. This implies that the half-life of material A is infinity.
For material B:
A(t) = A(0) * (1/2)^(t/T_B)
Given that the initial activity of material B is 74 x 10^7 decays/s, and the half-life is T_B, we can substitute these values into the equation:
74 x 10^7 = 74 x 10^7 * (1/2)^(t/T_B)
Since the initial activity cancels out, we are left with:
1 = (1/2)^(t/T_B)
Taking the logarithm base 2 of both sides:
log_2(1) = log_2((1/2)^(t/T_B))
0 = (t/T_B) * log_2(1/2)
Since log_2(1/2) = -1, we have:
0 = -(t/T_B)
Therefore, t = 0, which means there
Free Test
FREE
| Start Free Test |
Community Answer
A sample of radioactive material A, that has an activity of 10 mCi(1 C...
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer?
Question Description
A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer?.
A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. 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 A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :a)20 days and 5 daysb)20 days and 10 daysc)5 days and 10 daysd)10 days and 40 daysCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
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