Question

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.)

Answer 1

Answer:

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.