Energy released in the fission of one uranium-235 nucleus? (extra neutron)

Click For Summary
The discussion centers on the energy released during the fission of one uranium-235 nucleus, with a calculated value of 4.73×10^-11 joules. Participants reference Einstein's equation, E=mc², to explain the energy-mass relationship. There is a focus on the importance of significant figures in calculations, emphasizing that the final answer should reflect the precision of the input values. The conversation also highlights the need to find the mass difference before calculating energy, adhering to proper scientific methodology. Overall, the thread underscores the significance of accurate calculations in nuclear physics.
mccarthyp64
Messages
5
Reaction score
0
1. How much energy is released during the
fission of one uranium-235 nucleus?




2. e=mc2 lol :smile:



3. I thought of putting them together but then how to separate them?
1.67495×10-27+3.903 05×10–25×c2=1.7539207313048996244767495×10-7



By the way, the answer is 4.73×10-11
 

Attachments

  • physics question.png
    physics question.png
    23.4 KB · Views: 1,513
Physics news on Phys.org
mccarthyp64 said:
1. How much energy is released during the
fission of one uranium-235 nucleus?

2. e=mc2 lol :smile:
3. I thought of putting them together but then how to separate them?
What are "them"?

1.67495×10-27+3.903 05×10–25×c2=1.7539207313048996244767495×10-7
If you start using six significant figures, it is silly to have more than six in your answer. And since you are adding two quantities of different orders of magnitude, you should have even fewer. Your answer also needs to have units.

Solve the problem by doing exactly what the problem says to do: find the mass difference, then find the energy.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 1 ·
Replies
1
Views
3K