Question

Determine the algebraic modeling a. One type of Iodine disintegrates continuously at a constant rate of displaystyle{8.6}% per day. Suppose the origin

Modeling
ANSWERED
asked 2020-10-28
Determine the algebraic modeling a.
One type of Iodine disintegrates continuously at a constant rate of \(\displaystyle{8.6}\%\) per day.
Suppose the original amount, \(\displaystyle{P}_{{0}}\), is 10 grams, and let t be measured in days.
Because the Iodine is decaying continuously at a constant rate, we use the model \(\displaystyle{P}={P}_{{0}}{e}^{k}{t}\) for the decay equation, where k is the rate of continuous decay.
Using the given information, write the decay equation for this type of Iodine.
b.
Use your equation to determine the half-life ofthis type of Fodine, That is, find ‘out how many days it will take for half of the original amount to be left. Show an algebraic solution using logs.

Answers (1)

2020-10-29
a. The model of the decay equation is given by
\(\displaystyle{P}={P}_{{0}}{e}^{k}{t}\)
Here \(\displaystyle{P}_{{0}}={10}\) grams of iodine
\(\displaystyle{k}=\text{rate of continuous rate}=-{8.6}\%\) { negative sign implies the decay}
Which implies \(\displaystyle{k}=-{0.086}\)
t is measured in days
Therefore, the decay equation for this type of Iodine is
\(\displaystyle{P}={10}{e}^{{-{0.086}{t}}}\)
b. To find the half life of iodine
\(\displaystyle{\left({i}.{e}\right)}{t}=?\ \text{then}\ {P}=\frac{{P}_{{0}}}{{2}}=\frac{10}{{2}}={5}\) grams of iodine
\(\displaystyle{P}={10}{e}^{{-{0.086}{t}}}\)
substitute \(\displaystyle{P}={5}\) in the above equation
\(\displaystyle{5}={10}{e}^{{-{0.086}{t}}}\)
Dividing both sides by 10 we get,
\(\displaystyle\frac{5}{{10}}={e}^{{-{0.086}{t}}}\)
\(\displaystyle\Rightarrow{e}^{{-{0.086}{t}}}=\frac{1}{{2}}\)
\(\displaystyle{e}^{{{0.086}{t}}}={2}\)
Taking log on both sides we get,
\(\displaystyle{0.086}{t}={{\log}_{{e}}{2}}\)
\(\displaystyle\Rightarrow{t}=\frac{{{{\log}_{{e}}{2}}}}{{{0.086}}}=\frac{{{0.6931}}}{{{0.086}}}={8.0598}\approx{8}\)
Therefore it took 8 days for the iodine reduces to 5 grams
0
 
Best answer

expert advice

Need a better answer?
...