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)
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%
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%
(1/2)^(10000/24000) = 0.74915
(1/2)^(30/n) = 4^(-1/3)
n = 45
Residue(t)=Initial*(1/2)^(t/L)
[the negative sign removed]
To calculate the percent of plutonium-239 remaining after 10,000 years, we can use the formula:
Percent remaining = (1/2)^(time elapsed / half-life) * 100
Given that the half-life of plutonium-239 is 24,000 years, we can plug in the values:
Percent remaining = (1/2)^(10,000 / 24,000) * 100
Calculating this expression gives us approximately 74.9%. Therefore, 74.9% of plutonium-239 remains after 10,000 years.
Now let's move on to the second question about Plutonium-243 (PU^243). We are given that after 30 hours, a sample of PU^243 has decayed to 4^(-1/3) of its original mass.
To find the half-life of PU^243, we can use the formula:
Fraction remaining = (1/2)^(time elapsed / half-life)
Given that the fraction remaining is 4^(-1/3), we can rewrite it as:
(1/2)^(time elapsed / half-life) = 4^(-1/3)
To simplify the equation, we express 4^(-1/3) as 2^(-2/3):
(1/2)^(time elapsed / half-life) = 2^(-2/3)
Now, we can equate the exponents:
time elapsed / half-life = -2/3
To find the value of the half-life, we can cross-multiply:
3 * time elapsed = -2 * half-life
Solving for the half-life:
half-life = (-3 * time elapsed) / 2
Given that the time elapsed is 30 hours, we can substitute it into the equation:
half-life = (-3 * 30) / 2 = -90 / 2 = -45 hours.
Since half-life cannot be negative, we take the absolute value and get the half-life of PU^243 as 45 hours.