How Is Energy Calculated in a Nuclear Fusion Reaction?

[Benzoate]

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

 

[dynamicsolo]

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...

 

[m2003]

.

Your solution for part a and b is correct. For part c, we can use the formula P=Energy/time, where P is power, Energy is the energy released in the reaction, and time is the time it takes for the reaction to occur. We know from part b that the energy released is -23.847 eV. Now, we need to find the time it takes for one reaction to occur. This can be found using the formula E=mc^2, where E is energy, m is mass, and c is the speed of light. Rearranging the formula, we get t=E/mc^2. Plugging in the values, we get t=(-23.847 eV)/((2.6e-8 u)(931.5 MeV/c^2)(3e8 m/s)^2)=1.9e-18 seconds. Now, to produce 1 Watt of power, we need to have 1 Joule of energy per second. Therefore, we divide 1 Watt by the energy released per reaction to get the number of reactions per second: (1 Joule/second)/(-23.847 eV/reaction)=4.2e19 reactions per second.