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gayaneshka [121]
3 years ago
9

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?
Mathematics
2 answers:
Elan Coil [88]3 years ago
7 0
1.56%<span> 1/8 of the original source (25 counts) remains at 3:00pm. This means that the </span>half-life<span> is 1 hour. Continue to find the fraction at 6:00pm.</span>
blagie [28]3 years ago
4 0

Answer:

<em>1.5%</em><em> will remain at 6.00pm.</em>

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\%

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