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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?

Mar 2

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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.

Shaiphan Aila

Mar 2

What of our first answer being written as 1.3×10^13 Bq per year