Rates of Decay Reactions

Question

The half-life of cesium-129 is 32.0 hours. How much time is required for the activity of a sample of cesium-129 to fall to 10.4 percent of its original value?

Answer 1

Radioactive decay is a 1st order event, and so we can use 1st order kinetics to determine the time required for the activity to fall to 104 percent of the original value. There are a couple of methods to address this.

(1). FR = 0.5ⁿ

FR = fraction remaining = 0.104 (10.4% as a fraction)

n = number of half lives that have elapsed = ?

Solving for n ...

0.104 = 0.5ⁿ

log 0.104 = n log 0.5

-0.983 = -0.301 n

n = 3.266 half lives have elapsed

Time required = 3.266 half lives x 32.0 hours / half life = 105 hours (3 sig. figs.)

(2). ln [A] = -kt + ln [A]_o

[A]_o = initial concentration = 1

[A] = concentration at time t = 0.104

k = rate constant = ln 2 / t_1/2 = 0.693 / 32 hrs = 0.0217 hrs^-1

t = time = ?

Solving for t ..

ln 0.104 = - 0.0217 t + ln 1

-2.26 = - 0.0217 t + 0

t = 104 hrs