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A sample of radioactive material contains 1018 atoms. The half life of the material is 2 days, then the activity of the sample is
- a)3.5 x 1014 Bq
- b)3.5 x 1012 Bq
- c)7 x 1011 Bq
- d)7 x 1016 Bq
Correct answer is option 'B'. Can you explain this answer?
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A sample of radioactive material contains 1018atoms. The half life of ...
To find activity of the sample --->which is the rate of disintegration.
Since radioactivity comes under 1o kinetics.
[R]=k[A] [A]-->amount of initial sample 1018 atoms
Given,
Half-life=2days
K=0.693/2x24x60x60 sec
R=(0.693/2x24x60x60)x1018
R≈3.5x1012 Bq
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A sample of radioactive material contains 1018atoms. The half life of ...
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A sample of radioactive material contains 1018atoms. The half life of ...
Calculation of Activity of Radioactive Material
Given:
Number of atoms in the sample N = 1018
Half-life of the material t1/2 = 2 days
We know that the radioactive decay law is given by:
N = N0 x (1/2)^(t/t1/2)
where,
N0 is the initial number of atoms
t is the time elapsed since the initial number of atoms
Activity of a radioactive sample is defined as the rate of decay of the sample, which is given by:
A = λN
where,
λ is the decay constant
We know that the decay constant is related to the half-life by the equation:
λ = ln(2)/t1/2
Substituting the given values, we get:
λ = ln(2)/2 = 0.3466 day^-1
Now, substituting the values of N and λ in the equation for activity, we get:
A = λN = 0.3466 x 1018 = 3.5 x 10^12 Bq
Therefore, the activity of the radioactive material is 3.5 x 10^12 Bq (option B).
Given:
Number of atoms in the sample N = 1018
Half-life of the material t1/2 = 2 days
We know that the radioactive decay law is given by:
N = N0 x (1/2)^(t/t1/2)
where,
N0 is the initial number of atoms
t is the time elapsed since the initial number of atoms
Activity of a radioactive sample is defined as the rate of decay of the sample, which is given by:
A = λN
where,
λ is the decay constant
We know that the decay constant is related to the half-life by the equation:
λ = ln(2)/t1/2
Substituting the given values, we get:
λ = ln(2)/2 = 0.3466 day^-1
Now, substituting the values of N and λ in the equation for activity, we get:
A = λN = 0.3466 x 1018 = 3.5 x 10^12 Bq
Therefore, the activity of the radioactive material is 3.5 x 10^12 Bq (option B).
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A sample of radioactive material contains 1018atoms. The half life of the material is 2 days, then the activity of the sample isa)3.5 x 1014Bqb)3.5 x 1012Bqc)7 x 1011Bqd)7 x 1016BqCorrect answer is option 'B'. Can you explain this answer?
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A sample of radioactive material contains 1018atoms. The half life of the material is 2 days, then the activity of the sample isa)3.5 x 1014Bqb)3.5 x 1012Bqc)7 x 1011Bqd)7 x 1016BqCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A sample of radioactive material contains 1018atoms. The half life of the material is 2 days, then the activity of the sample isa)3.5 x 1014Bqb)3.5 x 1012Bqc)7 x 1011Bqd)7 x 1016BqCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sample of radioactive material contains 1018atoms. The half life of the material is 2 days, then the activity of the sample isa)3.5 x 1014Bqb)3.5 x 1012Bqc)7 x 1011Bqd)7 x 1016BqCorrect answer is option 'B'. Can you explain this answer?.
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