Hey there
I'm having a bit of difficulty with a question involving the use of differentials, and would be grateful if someone could help. The question is as follows:
In a breeder reactor, plutonium-239 is produced from uranium-238 at a constant rate, but it decays at a rate proportional to the amount of plutonium-239 present. The half-life of plutonium-239 is estimated as 24 000 years. If plutonium-239 is produced at a rate of 1 microgram per minute, solve the differential equation to find the amount present at any time after the reactor begins operation
So pretty much:
Plutonium-239 is produced at a rate of (1/1000000)g min-1
It also has a half life of 24000 years (so in 24000 years, 1g of Plutonium-239 will decay to 0.5 g)
I am quite confused as with where to start.
I tried this:
dP/dt = k - kP
dP/dt = k(1-P)
dp/(1-P) = k dt
ln(1-P) = kt + c
P = 1 - Ae^kt [Where A = e^c]
t = 0, p = 0:
0 = 1 - Ae^0
A = 1
P = 1 - e^kt
This is where I'm stuck, and thinking I did something completely wrong.
Thanks in advance for any help.
- Adam
I'm having a bit of difficulty with a question involving the use of differentials, and would be grateful if someone could help. The question is as follows:
In a breeder reactor, plutonium-239 is produced from uranium-238 at a constant rate, but it decays at a rate proportional to the amount of plutonium-239 present. The half-life of plutonium-239 is estimated as 24 000 years. If plutonium-239 is produced at a rate of 1 microgram per minute, solve the differential equation to find the amount present at any time after the reactor begins operation
So pretty much:
Plutonium-239 is produced at a rate of (1/1000000)g min-1
It also has a half life of 24000 years (so in 24000 years, 1g of Plutonium-239 will decay to 0.5 g)
I am quite confused as with where to start.
I tried this:
dP/dt = k - kP
dP/dt = k(1-P)
dp/(1-P) = k dt
ln(1-P) = kt + c
P = 1 - Ae^kt [Where A = e^c]
t = 0, p = 0:
0 = 1 - Ae^0
A = 1
P = 1 - e^kt
This is where I'm stuck, and thinking I did something completely wrong.
Thanks in advance for any help.
- Adam