Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take for 43.0 % of an Am-241 sample to decay?
2007-02-17 03:50:04 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Use the formula
A=Pe^(-rt)
and r=ln2/halflife
r=ln2/432=0.0016
Let P(the starting amount) be equal to 100
If 43% will be decayed the remainig will be 57
Substituting it in the expression
57=100e^(-0.0016t)
ln (57/100)=ln e^(-0.0016t)
ln (57/100)=-0.0016t
t=ln (57/100)/-0.0016
t=350.34 years
Note: ln (or natural logarithm) is a function in the calculator, not a variable
2007-02-17 04:51:10 · answer #1 · answered by duntoktomee 2 · 1⤊ 0⤋
t(1/2) = 0.693/k, so k = 0.693/t(1/2) = 0.693/432 = 0.001604
ln (Ao/A) = kt; A = 0.57Ao
t = ln (Ao/A)/k = ln (1/0.57)/0.001604 = 0.562/0.001604 = 350.4 yr
2007-02-17 04:01:05 · answer #2 · answered by TheOnlyBeldin 7 · 2⤊ 0⤋
371.52 years.
2007-02-17 03:59:05 · answer #3 · answered by wildraft1 6 · 0⤊ 1⤋