Question

at the beginning of an experiment, a scientist has 368 grams of radioactive goo. after 105 minutes, her sample has decayed to 11.5 grams.
what is the half - life of the goo in minutes? (round to one decimal place)
the half - life of the goo is
find a formula for ( g(t) ), the amount of goo remaining at time ( t ).
( g(t)=)
how many grams of goo will remain after 80 minutes? (round to one decimal place)
there will be grams of goo after 80 minutes.

Explanation:

Step1: Find the decay constant \( k \)

The radioactive decay formula is \( G(t)=G_0e^{-kt} \), where \( G_0 = 368 \) grams (initial amount), \( t = 105 \) minutes, and \( G(105)=11.5 \) grams.
Substitute into the formula: \( 11.5 = 368e^{-105k} \).
Divide both sides by 368: \( \frac{11.5}{368}=e^{-105k} \).
Simplify \( \frac{11.5}{368}=\frac{1}{32} \), so \( \frac{1}{32}=e^{-105k} \).
Take the natural logarithm of both sides: \( \ln(\frac{1}{32})=- 105k \).
Since \( \ln(\frac{1}{32})=\ln(1)-\ln(32)=- \ln(2^{5})=-5\ln(2)\approx - 3.4657 \), then \( -3.4657=-105k \).
Solve for \( k \): \( k=\frac{3.4657}{105}\approx0.033 \).

Step2: Find the half - life \( T \)

The formula for half - life is \( T=\frac{\ln(2)}{k} \).
Substitute \( k = 0.033 \): \( T=\frac{\ln(2)}{0.033}\approx\frac{0.693}{0.033}\approx21.0 \) minutes.

Step3: Find the formula for \( G(t) \)

Since \( G_0 = 368 \) and \( k\approx0.033 \), the formula is \( G(t)=368e^{-0.033t} \).

Step4: Find the amount after \( t = 80 \) minutes

Substitute \( t = 80 \) into \( G(t)=368e^{-0.033t} \).
\( G(80)=368e^{-0.033\times80}=368e^{-2.64} \).
Since \( e^{-2.64}\approx0.071 \), then \( G(80)=368\times0.071\approx26.1 \) grams.

Answer:

The half - life of the goo is \( 21.0 \) minutes.
\( G(t)=368e^{-0.033t} \)
There will be \( 26.1 \) grams of goo after 80 minutes.