Anonymous
Mar 2
The half-life of a radioactive nucleus radium 226 is 1.6×10^3 seconds. If a sample initially contains 3.0×10^16 such nuclei. Determine the initial activities in curries?
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ChatGPT
Mar 2
The initial activity of a radioactive sample is given by the formula:
A0 = λN0
where:
A0 = initial activity
λ = decay constant
N0 = initial number of radioactive nuclei
The decay constant (λ) can be calculated using the formula:
λ = ln(2) / half-life
Given that the half-life of radium 226 is 1.6×10^3 seconds, we can calculate the decay constant:
λ = ln(2) / 1.6×10^3
λ ≈ 4.34×10^-4 s^-1
Now, we can calculate the initial activity:
A0 = 4.34×10^-4 * 3.0×10^16
A0 ≈ 1.30×10^13 curies
Therefore, the initial activity of the sample containing 3.0×10^16 radium 226 nuclei is approximately 1.30×10^13 curies.
A0 = λN0
where:
A0 = initial activity
λ = decay constant
N0 = initial number of radioactive nuclei
The decay constant (λ) can be calculated using the formula:
λ = ln(2) / half-life
Given that the half-life of radium 226 is 1.6×10^3 seconds, we can calculate the decay constant:
λ = ln(2) / 1.6×10^3
λ ≈ 4.34×10^-4 s^-1
Now, we can calculate the initial activity:
A0 = 4.34×10^-4 * 3.0×10^16
A0 ≈ 1.30×10^13 curies
Therefore, the initial activity of the sample containing 3.0×10^16 radium 226 nuclei is approximately 1.30×10^13 curies.
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Shaiphan Aila
Mar 2
What of our first answer being written as 1.3×10^13 Bq per year
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