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.
Show your work.

Answer 1

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.