How many whole gram of Pu238 are required to complete the mi

How many whole gram of Pu-238 are required to complete the mission? Discuss how a different generator, operating at 7& efficiency, would affect the amount of Pu-238 required to complete the mission. Use algebra to back up your claim.

Solution

(27) Answer::::

Since the half-life is 87 yrs and only has to operate for 19 yrs, that part of it does not come into the problem

let g = no. of grams required to generate 85 watts
.56 * .06 * g = 85
.0336g = 85

g = 85/.0336
g = 2530 grams or 2.53 kg required
:
Thought about this one this morning, perhaps they want the initial amt of P-238 so they would still have 2530 grams at the end of 19 yrs. Let that amt = Ao
Ao*2^(-19/87.7) = 2530
use your calc to find 2^(-19/87.7)
Ao*.86056 = 2530
Ao = 2530/.86056
Ao = 2940 grams required ..

____________________________________________________________________________

28 Answer::::

g = no. of grams required to generate 85 watts
.56 * .07 * g = 85 (where efficienccy is 7%)
.0392g = 85

g = 85/.0392

g =2168.36 grams of Pu-238 required to complete the mission.

______________________________________________________________________________