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A freshly prepared sample of a certain radioactive isotope has an activity of 10.0 mCi. After 4.00 hours, its activity is 8.00 mCi.

Find the sample's activity 30.0 hours after it is prepared.

Answer :

The radioactive decay is first order kinetics whose rate of decay is directly proportional to the number of radioactive nuclei present in the sample. From the given values, the decay constant of radioactive sample is calculated using k = 2.303 / t [log (A1 / A2)]. By using this constant, the activity of sample at time t = 30.0 hours is calculated using A = A0 e^(-kt) which is 4.03 mCi.

Given data Initial activity of radioactive sample = 10.0 mCi After 4 hours, activity of radioactive sample = 8.00 mCi

Now we have to find the activity of the sample after 30 hours from the time it is prepared.

Method usedRadioactive decay follows first-order kinetics. According to this, the rate of decay of a radioactive substance is directly proportional to the number of radioactive nuclei present in the sample.

So, the rate of decay can be expressed as : k = 2.303 / t [log (A1 / A2)]Where k = decay constant = time A1 = initial activity of radioactive sampleA2 = final activity of radioactive sample

Now, the time t can be calculated as : t = 0.693 / k

The activity of radioactive substance at any time t can be expressed as : A = A0 e^(-kt)Where A = activity of sample at time tA0 = initial activity of sample

Now we can apply the above formulas to find the required values.

Determination of decay constant

The decay constant of radioactive sample can be calculated as follows :k = 2.303 / t [log (A1 / A2)]Putting the given values, we get :k = 2.303 / 4 [log (10.0 / 8.00)]k = 0.0190 h^-1Determination of activity of radioactive sample after 30 hours

Now we can calculate the activity of sample at time t = 30.0 hours :A = A0 e^(-kt)A = 10.0 mCi * e^(-0.0190 h^-1 * 30.0 h)A = 4.03 mCi (rounded off to 3 decimal places)

Therefore, the activity of the radioactive sample 30 hours after it is prepared is 4.03 mCi.

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