Question 8

Question

Americium-241 is a radioactive substance used in smoke detectors. The half life of americium is 432 years. If a smoke detector initially contains 1 gram of Americium 241, how much will remain in 1,296 years?

a
0.125 g
b
0.5 g
c
0.25 g
d
1.0 g

Answer 1

To determine how much Americium-241 will remain after 1,296 years, we can use the concept of half-life.

The half-life of Americium-241 is 432 years. This means that every 432 years, half of the Americium will decay.

First, we’ll calculate how many half-lives fit into 1,296 years:

\[ \text{Number of half-lives} = \frac{\text{Total time}}{\text{Half-life}} = \frac{1296 \text{ years}}{432 \text{ years}} = 3 \]

This means that 1,296 years corresponds to 3 half-lives.

Now, we can determine how much remains after each half-life:

1. After the first half-life (432 years), the amount of Americium-241 will be: \[ 1 \text{ g} \times \frac{1}{2} = 0.5 \text{ g} \]

2. After the second half-life (864 years), the amount will be: \[ 0.5 \text{ g} \times \frac{1}{2} = 0.25 \text{ g} \]

3. After the third half-life (1296 years), the amount will be: \[ 0.25 \text{ g} \times \frac{1}{2} = 0.125 \text{ g} \]

Therefore, after 1,296 years, 0.125 grams of Americium-241 will remain.

So the answer is:

a) 0.125 g