153k views
3 votes
10 g of a radioactive substance was placed in an air tight container 50 years ago. When the container was opened, only 2.5 g remains. What was the half life of this material?

1 Answer

7 votes

Answer:

25 years.

Step-by-step explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 10 g

Time (t) = 50 years

Amount remaining (N) = 2.5 g

Half-life (t½) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Original amount (N₀) = 10 g

Amount remaining (N) = 2.5 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

2.5 = 1/2ⁿ × 10

Cross multiply

2.5 × 2ⁿ = 10

Divide both side by 2.5

2ⁿ = 10 / 2.5

2ⁿ = 4

Express 4 in index form with 2 as the base.

2ⁿ = 2²

n =2

Thus, 2 half-lives has elapsed.

Finally, we shall determine the half-life of the radioactive substance. This can be obtained as follow:

Time (t) = 50 years

Number of half-lives (n) = 2

Half-life (t½) =?

n = t / t½

2 = 50 / t½

Cross multiply

2 × t½ = 50

Divide both side by 2

t½ = 50 / 2

t½ = 25 years.

Thus, the half-life of the radioactive substance is 25 years.

User WJM
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.