The exponential decay function A = A0(1/2)^t/P can be used to determine the amount A, of a radioactive substance present at time t, if A0 re

Question

The exponential decay function A = A0(1/2)^t/P can be used to determine the amount A, of a radioactive substance present at time t, if A0 represents the initial amount and P represents the half-life of the substance.
If a substance loses 70% of its radioactivity in 500 days, determine the period of the half-life.
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Trúc Chi 4 years 2021-08-03T20:51:43+00:00 1 Answers 108 views 0

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  1. Delwyn
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    2021-08-03T20:53:27+00:00
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    Answer:

    The half-life of the substance is about 288 days.

    Step-by-step explanation:

    The exponential decay function:

    \displaystyle A=A_0\left(\frac{1}{2}\right)^{t/P}

    Can determine the amount A of a radioactive substance present at time t. A₀ represents the initial amount and P is the half-life of the substance.

    We are given that a substance loses 70% of its radioactivity in 500 days, and we want to determine the period of the half-life.

    In other words, we want to determine P.

    Since the substance has lost 70% of its radioactivity, it will have only 30% of its original amount. This occured in 500 days. Therefore, A = 0.3A₀ when t = 500 (days). Substitute:

    \displaystyle 0.3A_0=A_0\left(\frac{1}{2}\right)^{500/P}

    Divide both sides by A₀:

    \displaystyle 0.3=\left(\frac{1}{2}\right)^{500/P}

    We can take the natural log of both sides:

    \displaystyle \ln(0.3)=\ln\left(\left(\frac{1}{2}\right)^{500/P}\right)

    Using logarithmic properties:

    \displaystyle \ln(0.3)=\frac{500}{P}\left(\ln\left(\frac{1}{2}\right)\right)

    So:

    \displaystyle \frac{500}{P}=\frac{\ln(0.3)}{\ln(0.5)}

    Take the reciprocal of both sides:

    \displaystyle \frac{P}{500}=\displaystyle \frac{\ln(0.5)}{\ln(0.3)}

    Use a calculator:

    \displaystyle P=\frac{500\ln(0.5)}{\ln(0.3)}\approx287.86

    The half-life of the substance is about 288 days.

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