e^(-0.023t) = 0.5
Take natural log of both sides.
-0.023 t = -0.6932
t = 30 years
How do you set up this problem?
The decay constant for the radioactive element cesium 137 is .023 when time is measured in years. Find its half-life.
2 answers
If you are using the standard decay function equation
amount = initial amount e^-kt, where k = .023 then
.5 = e^ -.023t
-.023t = ln .5
t = ln .5/-.023 = 30.14 years
check: pick 500 grams
amount = 500 e^(-.023(30.14))
= 500 e^ -.69314..
= 250
YUP!!!
amount = initial amount e^-kt, where k = .023 then
.5 = e^ -.023t
-.023t = ln .5
t = ln .5/-.023 = 30.14 years
check: pick 500 grams
amount = 500 e^(-.023(30.14))
= 500 e^ -.69314..
= 250
YUP!!!