Asked by Taylor Candice
Plutonium-239 has a half-life of 24 000 years. What percent of plutonium-239 remains after 10 000 years? (Answer=74.9%)
AND
After 30 hours, a sample of Plutonium 243 (PU^ 243) has decayed to 4 ^ -1/3 of its original mass. What is the half life of PU^ 243. (Answer=45 hours)
AND
After 30 hours, a sample of Plutonium 243 (PU^ 243) has decayed to 4 ^ -1/3 of its original mass. What is the half life of PU^ 243. (Answer=45 hours)
Answers
Answered by
Anonymous
A is the Amount
A = Ai e^-kt
when A/Ai = .5, t = 24,000
so
.5 = e^-24,000 k
ln .5 = -24,000 k
-.693 = -24,000 k
so
k = 2.888 * 10^-5
now we can do any old time t
at t = 10,000 = 10^4
A/Ai = e^-2.888*10^-5 *10^4= e^-.2888
= .749
as a percent that is 74.9%
A = Ai e^-kt
when A/Ai = .5, t = 24,000
so
.5 = e^-24,000 k
ln .5 = -24,000 k
-.693 = -24,000 k
so
k = 2.888 * 10^-5
now we can do any old time t
at t = 10,000 = 10^4
A/Ai = e^-2.888*10^-5 *10^4= e^-.2888
= .749
as a percent that is 74.9%
Answered by
MathMate
L=half-life in years,
t=time lapse in years
Residue(t)=Initial*(1/2)^(-t/L)
Residue(10,000)
=100%(1/2)^(10000/24000)
=74.92%
t=time lapse in years
Residue(t)=Initial*(1/2)^(-t/L)
Residue(10,000)
=100%(1/2)^(10000/24000)
=74.92%
Answered by
Steve
(1/2)^(10000/24000) = 0.74915
(1/2)^(30/n) = 4^(-1/3)
n = 45
(1/2)^(30/n) = 4^(-1/3)
n = 45
Answered by
MathMate
Residue(t)=Initial*(1/2)^(t/L)
[the negative sign removed]
[the negative sign removed]
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