Problem 37

  If the activity of a radioactive substance is initially 398 disintegrations/min and two days later it is 285 disintegrations/min, what is the activity four days later still, or six days after the start? Give your answer in disintegrations/min.

SOLUTION: The activity of a radioactive substance is defined as the number of disintegrations per unit time. Since radioactive decay is random, the number of disintegrations per unit time is proportional to the number of radioactive atoms; that is,

$$\frac{\Delta N}{\Delta t} = -\lambda N ,$$

so the activity is

$$A = \lambda N .$$

The solution for N(t) is

$$N(t) = N_o e^{-\lambda t}$$

so if we multiply this by $\lambda$, we get

$$\lambda N(t) = \lambda N_o e^{-\lambda t}$$

$$A(t) = A_o e^{-\lambda t}$$

where A(t) is the activity at time t and A_o is the initial activity. We need to determine $\lambda$

$$ln \left ( \frac{A(t)}{A_o} \right ) = -\lambda t$$

$$\lambda = -\frac{1}{t} ln \left ( \frac{A(t)}{A_o} \right )$$

$$\lambda = -\frac{1}{2 days} ln \left ( \frac{285 dis/min}{398 dis/min} \right ) = 0.167 days^{-1} .$$

Then we can plug four days into our equation to get the activity at that time.

A(4 days) = (285 dis/min) e^{(0.167 days<<55>>-1)(4.00 days)}

$$A = 146 \:disintegrations/min$$