NASA is designing a robotic spacecraft to visit Neptune. The probe will use the radioactive isotope Plutonium-238(Pu-238) as a power source. One gram of Pu-238 generates 0.56 watts of thermal power and has a half-life of 87.7 years. The generator that converts thermal power to electrical power for the probe's instruments operates at 6% efficiency (i.e. 1 watt of thermal power generates 0.06 watts of electrical power). The instruments on the spacecraft require 85 watts of electrical power and need to operate for 19 years to complete the mission.
a. How many whole grams of Pu-238 are required to complete the mission?
b. 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.
c. Suppose NASA wants the mission to last 25 years and has only 2500 grams of Pu-238 available for the probe. To what operating efficiency must the engineers design the generator? Round your answer to the nearest tenth of a percent.
I started doing 0.56*.06*x=85 for the first question. Is this right? Do I need to do anything with the 19 years thats mentioned in the problem? What about b. And c. ? This problem is so confusing.
+12531 I started doing 0.56*.06*x=85 for the first question. Is this right? Do I need to do anything with the 19 years thats mentioned in the problem? What about b. And c. ? This problem is so confusing.
+12531 c. Suppose NASA wants the mission to last 25 years and has only 2500 grams of Pu-238 available for the probe. To what operating efficiency must the engineers design the generator? Round your answer to the nearest tenth of a percent.
NASA is designing a robotic spacecraft to visit Neptune. The probe will use the radioactive isotope Plutonium-238(Pu-238) as a power source. One gram of Pu-238 generates 0.56 watts of thermal power and has a half-life of 87.7 years. The generator that converts thermal power to electrical power for the probe's instruments operates at 6% efficiency (i.e. 1 watt of thermal power generates 0.06 watts of electrical power). The instruments on the spacecraft require 85 watts of electrical power and need to operate for 19 years to complete the mission.
a. How many whole grams of Pu-238 are required to complete the mission?
b. 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.
c. Suppose NASA wants the mission to last 25 years and has only 2500 grams of Pu-238 available for the probe. To what operating efficiency must the engineers design the generator? Round your answer to the nearest tenth of a percent.
OK, young person, I'm no "rocket scientist", but will try to help you as much as I can:
a) Since the generator operates at 6% efficiency, that means in order for the generator to generate a continuous power of 85 watts, it will require:
85/.06=1,417 watts of thermal power. But, we know that one gram of Pu-238 generates 0.56 watts of thermal power and has a half-life of 87.7 years. We, there require:1,417 / .56=2,530 grams of Pu-238. But the mission will need to last for 19 years. So, the 2,530 grams will decay in that period to:
1 - 2^-(19/87.7) =86% x 2,530, which will reduce the 2,530 grams of Pu-238 to about 2,177 grams, which may compromise the 19-year mission, especially towards the end when it is transmitting all its data back to Earth. Therefore, in my humble opinion, it would be prudent for them to plan for this known problem and add an additional Pu-238 by the above percentage. So, the theoritical quantity of 2,530 grams should be increased by:2,530 / .86 =2,942 grams, just to be safe.
b) If the generator efficiency were at 7%, it would reduce the amount of Pu-238 by about 16%. They would require:85/.07/.56 =2,168 grams would be needed in that case plus allowing for the decay:2,168/.86=2,521 grams of Pu-238 would be needed.
c) in this scenario, the 2500 grams would decay to 1 - 2 ^(-25/87.7) =71.5% x 2,500 =1,787 grams.
This would only generate: 1,787 x .56 =~1,000 watts of thermal power. But to generate 85 watts of electrical power, the engine would have to have an efficiency of :85/1,000 =8.5%. I think, that is as far as I can take it. Good luck to you. Hope it helps.
Sorry, I made a math error in part c):
c) in this scenario, the 2500 grams would decay to 1 - 2 ^(-25/87.7) =82.07% x 2,500 =2,052 grams.
This would only generate: 2,052 x .56 =1,149 watts of thermal power. But to generate 85 watts of electrical power, the engine would have to have an efficiency of :85/1,149 =7.4%. Now, all 3 of us get the same figures. Any minor difference you see are due to rounding off of certain numbers.
Thank you so much for everybodys help! The fifth response is really the only one I can understand, but where on earth, for the first question, do you pull out a 1-2^-(19/87.7) ?? Where does that equation come from?
Hello young person!. That is very simple. Since Pu-238 has a half-life of 87.7 years, you have to figure out how much Pu-238 has decayed, through radioactivity, in that 19-year period. I hope you understand what half-life means. It simply means that if you had 100 grams of this Pu-238, in 87.7 years, there will be only half of it left , or just 50 grams. So, we try to find out how much has decayed and how much is left over. So, after 19 years, we have 19/87.7=0.216647 or21.66% of 1 half-life gone in 19 years. So, 1 half-life - 2^(-0.216647) (the portion that has decayed) =.86063 or 86.063% remains.
So, if you have 2500 grams, after 19 years only: 2500 x .86063=~2,151 will have remained. The formula used to do this is: A(r)=A(o) x 2^(-n/half-life), where A(r)=amount remaining after n years, A(o)=original amount of material you started with, n=number of years elapsed.
They should have taught this in school when you are dealing with radioactive elements such as Pu-238. I hope you understand it a little better now.
Is there a way to figure it out without using the A(r)=A(o) formula? A simpler way? Or you cant do anything in this problem without that formula? Because it wasnt taught to us.
No, you must use that formula. There is a much more complicated version of it, so you can't get any simpler than this. Use a calculator and practice using it. Example: Pu-239 has a half-life of 24,100 years. If you had 100 grams of it, how much of that would be left after 10,000 years from now?
Can you figure it out? Try it.
What exactly is this formula called? When I look it up, not a thing comes up about it.
It is called " Exponential Decay Formula", or sometimes "Radioactive Decay Formula". Here is another version of it but is little more involved, because you have to use Euler's constant and find "constant of decay" "k", in this formula:A=A0e^kt. But, the result of this is exactly the same as the one I gave you.
P.S. I must admit that I'm very, very surprised that you are tasked to deal with radioactive materials and yet they haven't taught you how to calculate the decay of these elements?. That is very odd.
By the way, why do you find such a simple thing "difficult"? Obviously, you don't have a "knack" for Math. I feel sorry for you. You should consider seeing a tutor in Math, or in Physics. Good luck to you.
