Question
The half of 14C isotope is 5730 year .if a sample of 14C contains 10^22 nuclei , what is the activity of the sample
Answers
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The activity of a radioactive sample is given by:
Activity = decay constant x number of nuclei
The decay constant (λ) for 14C isotope can be calculated using the half-life formula:
t½ = ln(2) / λ
where t½ is the half-life and ln is the natural logarithm.
Substituting the given values, we get:
5730 years = ln(2) / λ
Solving for λ, we get:
λ = ln(2) / 5730 years
λ = 1.21 x 10^-4 year^-1
Now, we can calculate the activity of the sample:
Activity = (1.21 x 10^-4) x (10^22)
Activity = 1.21 x 10^18 decays per second (Bq)
Therefore, the activity of the sample is 1.21 x 10^18 Bq.
Activity = decay constant x number of nuclei
The decay constant (λ) for 14C isotope can be calculated using the half-life formula:
t½ = ln(2) / λ
where t½ is the half-life and ln is the natural logarithm.
Substituting the given values, we get:
5730 years = ln(2) / λ
Solving for λ, we get:
λ = ln(2) / 5730 years
λ = 1.21 x 10^-4 year^-1
Now, we can calculate the activity of the sample:
Activity = (1.21 x 10^-4) x (10^22)
Activity = 1.21 x 10^18 decays per second (Bq)
Therefore, the activity of the sample is 1.21 x 10^18 Bq.