Power produced from alpha decay

AI Thread Summary
To calculate the power produced from a 14 kg sample of 240Pu through alpha decay, first determine the number of plutonium atoms using Avogadro's number. The energy released per decay is approximately 5.25572 MeV, which can be converted to Joules. Given the half-life of Pu-240 is about 6560 years, calculate the decay rate to find how many atoms decay per second. Multiply the energy per decay by the decay rate and then apply the 60% efficiency factor to find the total power output in watts. The expected result is around 59.4 watts.
MattHorbacz
Messages
18
Reaction score
0

Homework Statement


Assume that a 14 kg sample of 240Pu is used to produce electrical power from its α decay. If your device is 60% efficient in producing electrical power, how much power can be produced? (in watts)

Homework Equations


Q=mass of plutonium-mass alpha-mass daughter; 240 plutonium= alpha +uranium236

The Attempt at a Solution


I found Q to be 5.25572 Mev. The number of Plutonium atoms is 14*10^3 grams * 6.022*10^23/240. Multiply this by Q to get total energy in ev. Then multiply by 1.602176565e-19 to get in terms of Joules. Then multiply by .6, and that's where I get stuck. No idea how to get to watts. the answer should be 59.4 watts.
 
Physics news on Phys.org
MattHorbacz said:

Homework Statement


Assume that a 14 kg sample of 240Pu is used to produce electrical power from its α decay. If your device is 60% efficient in producing electrical power, how much power can be produced? (in watts)

Homework Equations


Q=mass of plutonium-mass alpha-mass daughter; 240 plutonium= alpha +uranium236

The Attempt at a Solution


I found Q to be 5.25572 Mev. The number of Plutonium atoms is 14*10^3 grams * 6.022*10^23/240. Multiply this by Q to get total energy in ev. Then multiply by 1.602176565e-19 to get in terms of Joules. Then multiply by .6, and that's where I get stuck. No idea how to get to watts. the answer should be 59.4 watts.

You need to know how quickly Pu-240 decays. Do you know the half-life of Pu-240?
 
I looked it up at it seems to be about 6560 years
 
MattHorbacz said:
I looked it up at it seems to be about 6560 years
OK, from that, you need to figure out how many atoms of Pu-240 are decaying each second in your original 14 kg of this substance.

Knowing that, you have already computed how much energy is released with each decay, so finding the power should be trivial.
 
  • Like
Likes MattHorbacz
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top