# Binding Energy problem

1. Sep 26, 2007

### Benzoate

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
2^H = 1875.613 MeV ; 4^He=3727.379 MeV
1 u= 931.5 MeV/c^2
P=Energy/time
3. 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

2. Sep 26, 2007

### dynamicsolo

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

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

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...