Consider a radioactive sample whose decay constant is λ. Initial (t=0) number of nuclei is N0. Find the total life of all the nuclei that decay in the interval of second and third half-lives (i.e.t=t1/2 to t=3t1/2)
Question
Consider a radioactive sample whose decay constant is λ. Initial (t=0) number of nuclei is N0. Find the total life of all the nuclei that decay in the interval of second and third half-lives (i.e.t=t1/2 to t=3t1/2)
Solution 1
The half-life of a radioactive substance is given by the formula t1/2 = ln(2)/λ.
The number of nuclei remaining after a certain time t is given by the formula N = N0 * e^(-λt).
- First, we need to find the number of nuclei that decay in the interval of the second and third half-lives.
At t = t1/2, the number of remaining nuclei is N0/2.
At t = 3t1/2, the number of remaining nuclei is N0/2^3 = N0/8.
So, the number of nuclei that decay in this interval is N0/2 - N0/8 = N0/4.
- The total life of these nuclei is the product of the number of nuclei and their average lifetime.
The average lifetime of a radioactive nucleus is 1/λ.
So, the total life of the nuclei that decay in the interval of the second and third half-lives is (N0/4) * (1/λ) = N0/(4λ).
Solution 2
The half-life (t1/2) of a radioactive substance is the time taken for half of the substance to decay. The decay constant (λ) is related to the half-life by the equation:
t1/2 = ln(2) / λ
The number of nuclei remaining at any time t is given by the equation:
N(t) = N0 * e^(-λt)
where N0 is the initial number of nuclei.
The total life of all the nuclei that decay in the interval of the second and third half-lives is the sum of the lives of all the nuclei that decay in that interval.
The number of nuclei that decay in the interval from t1/2 to 3t1/2 is N(t1/2) - N(3t1/2).
Substituting the expressions for N(t) and t1/2 into this equation gives:
N(t1/2) - N(3t1/2) = N0 * e^(-λt1/2) - N0 * e^(-λ3t1/2) = N0 * e^(-ln(2)) - N0 * e^(-3ln(2)) = N0/2 - N0/8 = N0/4
So, the total life of all the nuclei that decay in the interval from t1/2 to 3t1/2 is N0/4 multiplied by the length of the interval, which is 2t1/2:
Total life = N0/4 * 2t1/2 = N0/2 * t1/2 = N0 * ln(2) / (2λ)
So, the total life of all the nuclei that decay in the interval from t1/2 to 3t1/2 is N0 * ln(2) / (2λ).
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