Is this nuclear physics question's answer correct?

In summary, the correct power output of a reactor fueled by uranium-235, if it takes 30 days to use up 10 Kg of fuel and each fission gives 200 MeV of energy, is approximately 316 MW.
  • #1
Toyona10
31
0

Homework Statement



What is the power output of a reactor fueled by uranium-235 if it takes 30 days to use up 10 Kg of fuel and if each fission gives 200 MeV of energy.

Homework Equations





The Attempt at a Solution


(I was absent that day so this answer is from someone else's notebook)
No of U-235 nuclei in 10 kg:
(6.023*10^23*10000)/235 = 2.56*10^25

Fission rate= (2.56*10^25)/(30*24*3600)= 9.76*10^18 J/s (?? how is it joules per second, isn't that number above just no. of nuclei?)

Therefore, energy produced through fission per second:
= 200* 9.76*10^18= 1.95*10^15 MW

So...is this correct?
Thank you
 
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  • #2
Toyona10 said:

Homework Statement



What is the power output of a reactor fueled by uranium-235 if it takes 30 days to use up 10 Kg of fuel and if each fission gives 200 MeV of energy.

Homework Equations





The Attempt at a Solution


(I was absent that day so this answer is from someone else's notebook)
No of U-235 nuclei in 10 kg:
(6.023*10^23*10000)/235 = 2.56*10^25

Fission rate= (2.56*10^25)/(30*24*3600)= 9.76*10^18 J/s (?? how is it joules per second, isn't that number above just no. of nuclei?)

Therefore, energy produced through fission per second:
= 200* 9.76*10^18= 1.95*10^15 MW

So...is this correct?
Thank you

As you notice above, because those calculations don't carry the units through from left to right, it's hard to figure out if things are correct by dimensional analysis alone. So you will want to start out by defining each factor introduced along with their correct units. When you do this, you quickly realize that the calculations don't make sense because the units don't work out.

For example,

(6.023*10^23*10000)/235 = 2.56*10^25

is missing a conversion from atomic mass units to grams. The fission rate is, as you noticed, not in J/s. Finally

= 200* 9.76*10^18= 1.95*10^15 MW

doesn't account for the fact that when we list an energy in MeV, we really mean MeV/c^2, where c is the speed of light. In order to obtain an expression in SI units, we need to use the value of c to convert units.

There could be other problems, but these were the most obvious ones. Hopefully it gives you a good idea of how to do the correct calculation.
 
  • #3
fzero said:
As you notice above, because those calculations don't carry the units through from left to right, it's hard to figure out if things are correct by dimensional analysis alone. So you will want to start out by defining each factor introduced along with their correct units. When you do this, you quickly realize that the calculations don't make sense because the units don't work out.

For example,

(6.023*10^23*10000)/235 = 2.56*10^25

is missing a conversion from atomic mass units to grams. The fission rate is, as you noticed, not in J/s. Finally

= 200* 9.76*10^18= 1.95*10^15 MW

doesn't account for the fact that when we list an energy in MeV, we really mean MeV/c^2, where c is the speed of light. In order to obtain an expression in SI units, we need to use the value of c to convert units.

There could be other problems, but these were the most obvious ones. Hopefully it gives you a good idea of how to do the correct calculation.

Hmmm...lol

well, this is how I tried it:
total energy= energy per fission*no of atoms
= 200 MeV * 2.56*10^25
=8.19*10^14 J
so Pout=total E / time =(8.19*10^14)/(30*24*3600)
=315.97MW
 
  • #4
Toyona10 said:
Hmmm...lol

well, this is how I tried it:
total energy= energy per fission*no of atoms
= 200 MeV * 2.56*10^25
=8.19*10^14 J
so Pout=total E / time =(8.19*10^14)/(30*24*3600)
=315.97MW

That looks better, but I haven't looked up the precise conversion factors, so I'm just assuming you have them correct. I'd round it off to 2 or 3 significant digits, since the way the number of atoms was computed was rough.
 
  • #5
htrytr said:
agree

to which one? the 1st answer or the second one? :D
 
Last edited by a moderator:

1. Is there a definitive way to determine if a nuclear physics question's answer is correct?

No, there is no one definitive way to determine if a nuclear physics question's answer is correct. It often requires multiple approaches and cross-checking with other sources or experiments.

2. What are some common mistakes that can lead to incorrect answers in nuclear physics questions?

Some common mistakes in nuclear physics questions include incorrect assumptions or simplifications, incorrect calculations or use of formulas, and misinterpretation of data or experimental results.

3. How can I improve my chances of getting the correct answer in a nuclear physics question?

To improve your chances of getting the correct answer in a nuclear physics question, make sure to carefully read and understand the question, double-check your calculations and assumptions, and consult with reliable sources or experts if needed.

4. Are there any techniques or strategies for solving nuclear physics questions more effectively?

Yes, there are various techniques and strategies that can help with solving nuclear physics questions more effectively. These include breaking down the problem into smaller parts, using diagrams or visual aids, and practicing with similar problems to build problem-solving skills.

5. Can the answer to a nuclear physics question change over time?

Yes, the answer to a nuclear physics question can change over time as new research and advancements are made in the field. It is important to stay updated and re-evaluate answers if new information becomes available.

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