How Is Energy Calculated in a Nuclear Fusion Reaction?

Click For Summary
In a nuclear fusion reaction involving two deuterium (2^H) atoms combining to form helium (4^He), the binding energy is calculated to be 23.85 MeV, resulting in a mass decrease of approximately 2.6e-8 unified mass units. The energy released in the reaction is equivalent to the binding energy, which is -23.847 eV depending on the sign convention used. To determine how many reactions are needed to produce 1 watt of power, the energy per reaction must be converted from MeV to joules, as 1 watt equals 1 joule per second. The discussion emphasizes the importance of understanding these conversions and the relationship between energy release and power output. Overall, the calculations illustrate the energy dynamics in nuclear fusion processes.
Benzoate
Messages
418
Reaction score
0

Homework Statement


In a nuclear fusion reaction two 2^H atoms are combined to produced 4^He (2 is not raised to the H power; 4 is not raised to the He power) . a) calculate the decrease in rest mass in unified mass units b) How much energy is released in this reaction? c) How many such reactions must take place per second to produce 1 W of power?


Homework Equations


2^H = 1875.613 MeV ; 4^He=3727.379 MeV
1 u= 931.5 MeV/c^2
P=Energy/time

The Attempt at a Solution


In part a, I first have to find the Binding Energy and in order to find the Binding energy I first have to write out the reaction of hydrogen and helium: 2^H +2^H => 4^He . therefore BE=(1875.613 MeV + 1875.613 MeV)-(3727.379 MeV)= 23.85 MeV ; therefore the decrease in mass is: 23.85e6eV/931.5MeV=2.6e-8 u

In part b, the energy released would just be equal to the magnitude to of the binding energy: The reaction is in reverse: 4^He => 2^H + 2^H => (3727.379)-(1875.613 + 1875.613)= -23.847 eV

In part c, I not sure how to relate the number of reactions to 1 Watt per second
 
Physics news on Phys.org
Benzoate said:

The Attempt at a Solution


In part a, I first have to find the Binding Energy and in order to find the Binding energy I first have to write out the reaction of hydrogen and helium: 2^H +2^H => 4^He . therefore BE=(1875.613 MeV + 1875.613 MeV)-(3727.379 MeV)= 23.85 MeV ; therefore the decrease in mass is: 23.85e6eV/931.5MeV=2.6e-8 u

First off, 23.85e6 eV = 23.85 MeV. So that conversion should be...?

In part b, the energy released would just be equal to the magnitude to of the binding energy.

Correct. As to whether the sign is positive or negative depends on which convention you are using in your course.


In part c, I not sure how to relate the number of reactions to 1 Watt per second

I think this should say "the number of reactions per second to 1 Watt". I guess they're asking how many such fusion reactions per second yield a power output of 1 W.
You now know how many MeV per reaction are released; what is that in joules? Since a watt is 1 joule/second, you should be able to find your way from here...
 
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 10 ·
Replies
10
Views
3K
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
13
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 54 ·
2
Replies
54
Views
10K
  • · Replies 2 ·
Replies
2
Views
2K