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?

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Answer 1 · 2018-03-14T02:16:02

We know that a form of the decay formula is:

$G (t) = G (0) {(\frac{1}{2})}^{\frac{t}{t_{\frac{1}{2}}}} [1]$ $G(t) = G(0)(1/2)^(t/t_(1/2))" [1]"$

Where $G (0)$ $G(0)$ is the initial amount and $t_{\frac{1}{2}}$ $t_(1/2)$ is the half-life.

Given $G (0) = 180 g and G (195 min) = 2.8125 g$ $G(0) = 180" g" and G(195" min") = 2.8125" g"$

Substitute the above into equation [1]:

$2.8125 g = (180 g) {(\frac{1}{2})}^{\frac{195 min}{t_{\frac{1}{2}}}}$ $2.8125" g" = (180" g")(1/2)^((195" min")/t_(1/2))$

Solve for half-life, $t_{\frac{1}{2}}$ $t_(1/2)$ .

Divide both sides by $180 g$ $180" g"$ :

$\frac{2.8125 g}{180 g} = {(\frac{1}{2})}^{\frac{195 min}{t_{\frac{1}{2}}}}$ $(2.8125" g")/(180" g") = (1/2)^((195" min")/t_(1/2))$

Take the natural logarithm of both sides:

$ln (\frac{2.8125 g}{180 g}) = ln ({(\frac{1}{2})}^{\frac{195 min}{t_{\frac{1}{2}}}})$ $ln((2.8125" g")/(180" g")) = ln((1/2)^((195" min")/t_(1/2)))$

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

$ln (\frac{2.8125 g}{180 g}) = (\frac{195 min}{t_{\frac{1}{2}}}) ln (\frac{1}{2})$ $ln((2.8125" g")/(180" g")) = ((195" min")/t_(1/2))ln(1/2)$

Multiply both sides by $t_{\frac{1}{2}}$ $t_(1/2)$

$t_{\frac{1}{2}} ln (\frac{2.8125 g}{180 g}) = (195 min) ln (\frac{1}{2})$ $t_(1/2)ln((2.8125" g")/(180" g")) = (195" min")ln(1/2)$

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

$t_{\frac{1}{2}} = (195 min) \frac{ln (\frac{1}{2})}{ln (\frac{2.8125 g}{180 g})}$ $t_(1/2) = (195" min")ln(1/2)/ln((2.8125" g")/(180" g"))$

$t_{\frac{1}{2}} = 32.5 min$ $t_(1/2) = 32.5" min"$

The formula for $G (t)$ $G(t)$ is:

$G (t) = G (0) {(\frac{1}{2})}^{\frac{t}{32.5 min}}$ $G(t) = G(0)(1/2)^(t/(32.5" min")$

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

$G (35 min) = (180 g) {(\frac{1}{2})}^{\frac{35 min}{32.5 min}}$ $G(35" min") = (180" g")(1/2)^((35" min")/(32.5" min"))$

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