Question

Element X decays radioactively with a half-life of 14 minutes if there are 680 grams of element X how long to the nearest 10th of a minute would it take the element to decay 17 grams

Answer 1

Answer:

It would take 74.5 minutes for the element to decay 17 grams.

Step-by-step explanation:

The amount of element X after t minutes is given by the follwoing equation:

$X(t) = X(0)e^{rt}$

In which X(0) is the initial amount of the substance and r is the decay rate.

Half life of 14 minutes.

This means that $X(14) = 0.5X(0)$

So

$X(t) = X(0)e^{rt}$

$0.5X(0) = X(0)e^{14r}$

$e^{14r} = 0.5$

$\ln{e^{14r}} = \ln{0.5}$

$14r = \ln{0.5}$

$r = \frac{\ln{0.5}}{14}$

$r = -0.0495$

So

$X(t) = X(0)e^{-0.0495t}$

There are 680 grams of element X

This means that $X(0) = 680$

$X(t) = X(0)e^{-0.0495t}$

$X(t) = 680e^{-0.0495t}$

How long would it take the element to decay 17 grams

This is t for which X(t) = 17. So

$X(t) = 680e^{-0.0495t}$

$17 = 680e^{-0.0495t}$

$e^{-0.0495t} = \frac{17}{680}$

$e^{-0.0495t} = 0.025$

$\ln{e^{-0.0495t}} = \ln{0.025}$

$-0.0495t = \ln{0.025}$

$0.0495t = -\ln{0.025}$

$t = -\frac{\ln{0.025}}{0.0495}$

$t = 74.5$

It would take 74.5 minutes for the element to decay 17 grams.