If the activity of a sample of cesium-137 decreases from 7.48E-3 Ci to 1.87E-3 Ci in 60.4 years, what is the half-life of cesium-137?

30.2 years

Explanation:

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

Original amount (Nₒ) = 7.48E-3

Amount remaining (N) = 1.87E-3

Time (t) = 60.4 years

Half-life (t½) =?

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

Original amount (Nₒ) = 7.48E-3 = 7.48×10¯³

Amount remaining (N) = 1.87E-3 = 1.87×10¯³

Number of half-lives (n) =?

N = 1/2ⁿ × Nₒ

1.87×10¯³ = 1/2ⁿ × 7.48×10¯³

Cross multiply

1.87×10¯³ × 2ⁿ = 7.48×10¯³

Divide both side by 1.87×10¯³

2ⁿ = 7.48×10¯³ / 1.87×10¯³

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 cesium-137. This can be obtained as follow:

Time (t) = 60.4 years

Number of half-lives (n) = 2

Half-life (t½) =?

n = t / t½

2 =60.4 / t½

Cross multiply

2 × t½ = 60.4

Divide both side by 2

t½ = 60.4 / 2

t½ = 30.2 years

Thus, the half-life of the cesium-137 is 30.2 years

1.solid

2.carbon dioxide

3.helium

it would be a gas formation