Question

A radioactive element decays at a rate of 5% annually. There are 40 grams of the substance present. • How much of the substance remains after 30 years (to the nearest tenth)? • When will the amount of the substance drop to below 20 grams (to the nearest year)?

Answer 1

Answer:

8.6 grams of the substance remains after 30 years.

The amount of the substance will drop to below 20 grams in 14 years.

Step-by-step explanation:

The amount of the substance after t years is given by the following equation:

$A(t) = A(0)(1-r)^{t}$

In which A(0) is the initial amount and r is the yearly decay rate.

A radioactive element decays at a rate of 5% annually. There are 40 grams of the substance present.

This means, respectively, that $r = 0.05, A(0) = 40$

So

$A(t) = A(0)(1-r)^{t}$

$A(t) = 40(0.95)^{t}$

How much of the substance remains after 30 years (to the nearest tenth)?

This is A(30).

$A(30) = 40(0.95)^{30} = 8.6$

8.6 grams of the substance remains after 30 years

When will the amount of the substance drop to below 20 grams (to the nearest year)?

This is t when A(t) = 20. So

$A(t) = 40(0.95)^{t}$

$20 = 40(0.95)^{t}$

$(0.95)^{t} = \frac{20}{40}$

$(0.95)^{t} = 0.5$

$\log{(0.95)^{t}} = \log{0.5}$

$t\log{0.95} = \log{0.5}$

$t = \frac{\log{0.5}}{\log{0.95}}$

$t = 13.51$

Rounding to the nearest year

The amount of the substance will drop to below 20 grams in 14 years.