# The initial mass of a radioactive goo is 180 grams. After 195 minutes, the sample has decayed to 2.8125 grams. Find the half-life of the substance in minutes? Find a formula for the amount, G(t), remaining at time t.? Find grams remain after 35 min?

Mar 14, 2018

We know that a form of the decay formula is:

$G \left(t\right) = G \left(0\right) {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\frac{1}{2}\right)} \text{ }$

Where $G \left(0\right)$ is the initial amount and ${t}_{\frac{1}{2}}$ is the half-life.

Given $G \left(0\right) = 180 \text{ g" and G(195" min") = 2.8125" g}$

Substitute the above into equation :

2.8125" g" = (180" g")(1/2)^((195" min")/t_(1/2))

Solve for half-life, ${t}_{\frac{1}{2}}$.

Divide both sides by $180 \text{ g}$:

(2.8125" g")/(180" g") = (1/2)^((195" min")/t_(1/2))

Take the natural logarithm of both sides:

ln((2.8125" g")/(180" g")) = ln((1/2)^((195" min")/t_(1/2)))

Use the property of logarithms that allow one to move the exponent to the outside as a coefficient:

$\ln \left(\frac{2.8125 \text{ g")/(180" g")) = ((195" min}}{t} _ \left(\frac{1}{2}\right)\right) \ln \left(\frac{1}{2}\right)$

Multiply both sides by ${t}_{\frac{1}{2}}$

t_(1/2)ln((2.8125" g")/(180" g")) = (195" min")ln(1/2)

Divide both sides by $\ln \left(\left(2.8125 \text{ g")/(180" g}\right)\right)$:

t_(1/2) = (195" min")ln(1/2)/ln((2.8125" g")/(180" g"))

${t}_{\frac{1}{2}} = 32.5 \text{ min}$

The formula for $G \left(t\right)$ is:

G(t) = G(0)(1/2)^(t/(32.5" min")

The amount of the 180-gram sample remaining after 35 min:

G(35" min") = (180" g")(1/2)^((35" min")/(32.5" min"))

G(35" min") = 85.327" g"