Question

A researcher measures 200 counts per minute coming from a radioactive source at noon. At 3:00pm, she finds that this has dropped to 25 counts per minute. What percentage of the original source will remain at 6:00pm?

Answer 1

1.56% 1/8 of the original source (25 counts) remains at 3:00pm. This means that the half-life is 1 hour. Continue to find the fraction at 6:00pm.

Answer 2

Answer:

1.5% will remain at 6.00pm.

Step-by-step explanation:

The function for exponential decay is,

$y(t)=Ae^{rt}$

where,

y(t) = the future amount,

A = initial amount,

r = rate of growth,

t = time.

A researcher measures 200 counts per minute coming from a radioactive source at noon.

At 3:00pm, i.e after 3 hours,she finds that this has dropped to 25 counts per minute.

Putting the values,

$\Rightarrow 25=200e^{r\times 3}$

$\Rightarrow \ln \dfrac{25}{200}=\ln e^{r\times 3}$

$\Rightarrow \ln \dfrac{1}{8}={r\times 3}\times \ln e$

$\Rightarrow \ln \dfrac{1}{8}={r\times 3}\times 1$

$\Rightarrow \ln \dfrac{1}{8}={r\times 3}$

$\Rightarrow r=\dfrac{\ln \dfrac{1}{8}}{3}=-0.6931$

-ve sign means the amount is decreasing.

As we have to find the amount at 6.00pm i.e after 6 hours, so

$y(t)=200e^{-0.6931\times 6}=3.13\approx 3$

As number of bacteria can't be in decimal.

So the percentage will be,

$=\dfrac{3}{200}=0.015=1.5\%$