The half-life of cesium-137 is 30 years. Suppose we have a 40 mg sample. Exercise (a) Find the mass that remains after t years. Step 1 Let y (t) be the mass (in mg) remaining after t years. Then we know the following. y (t) = y (0) ekt = · ekt Exercise (b) How much of the sample remains after 20 years? Step 1 After 20 years we have the following. y (20) = 40 · 2 = mg (Round your answer to two decimal places.)

1) y (t) = (40.0 mg) (e∧ ( - (0.0231 year⁻¹) t).

2) 25.2 mg.

Explanation:

(a) Find the mass that remains after t years. Step 1 Let y (t) be the mass (in mg) remaining after t years. Then we know the following. y (t) = y (0) ekt

For first order reactions: y (t) = y (0) (e∧-kt)

where, y (t) is the mass of the substance at any time (t).

y (0) is the initial concentration of the substance at (t = 0) (y (0) = 40.0 mg).

k is the rate constant of the reaction.

t is the time of the reaction.

For first order reactions: k = ln2 / (t1/2) = 0.693 / (30 years) = 0.0231 year⁻¹.

∴ y (t) = y (0) (e∧-kt)

y (t) = (40.0 mg) (e∧ ( - (0.0231 year⁻¹) t).

Exercise (b) How much of the sample remains after 20 years?

∵ y (t) = y (0) (e∧kt)

k = 0.03465 year⁻¹, t = 20.0 years, y (0) = 40.0 mg.

∴ y (t) = y (0) (e∧-kt) = (40.0 mg) e∧ - (0.0231 year⁻¹) (20.0 years) = 25.2 mg.