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.

Answer 2

1) The amount of substance that will remain after 30 years is; A(30) = 8.6 g

2) The time when the amount of the substance drop to below 20 grams is; t < 14 years

• To calculate the amount of substance that remains after t number of years of decay is;

A(t) = A_o(e^(-rt))

Where;

A_o is initial amount

r is rate of decay

t is time in years

A(t) is amount left after t years

We are given;

A_o = 40 g

r = 5% = 0.05

1) To find the amount of substance left after 30 years is;

A(30) = 40 × e^(-0.05 × 30)

A(30) = 8.925 g

Approximating to the nearest tenth gives;

A(30) = 8.9 g

2) To find the time when the amount of the substance drop to below 20 grams.

20 > 40 × e^(-0.05 × t)

20/40 > e^(-0.05 × t)

-0.05t < In 0.5

t < (In 0.5)/-0.05

t < 13.86 years

Approximating to the nearest year gives;

t < 14 years

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