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Fusion energy equation

by Kemilss
Tags: energy, equation, fusion
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Jan12-14, 01:39 PM
P: 7
I'm taking high school physics, and I've had problems in the past with flat out errors in my textbook (I've brought them up here). It's incredibly frustrating when your trying to understand something.

Anyways, I need to know which is right:

The text in my book goes through to explain,

energy released = (Eb(f) - Eb(i))C^2

or Change in E=(Mf - Mi)C^2

Fine. This gives me a negative answer, which has already given me headaches, but these turned into a eureka moment for me when I realised this was "less energy" required to be stable, or binding energy.

So now, the textbook goes on to do more then 1 example, and in all of them it's,

energy released = (Eb(i) - Eb(f))C^2

or Change in E=(Mi - Mf).

These give different answers. This website say's

and Physics forums has this example

Both contradict each other, or am I wrong?!!

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Jan12-14, 03:26 PM
Sci Advisor
P: 6,112
I won't comment on your specific examples, since I don't know what they are referring to. However in any (nuclear) reaction where there is a mass difference, the difference ends up as energy with E = mc2. If the reaction involves a loss of mass, the energy is positive. If the reaction has a gain in mass, it means energy has to be supplied to make it happen.
Jan12-14, 04:56 PM
P: 12,113
The definition of binding energy can be confusing, as binding energy is a positive value, but a larger binding energy means the nucleus has less energy.

The first set of equations is right apart from the factor of c^2 in the first line.

Jan12-14, 06:09 PM
P: 7
Fusion energy equation

hmm, I'm still a little confused. Here's an example straight out of my book.

"Calculate the energy released when three Helium-4 nuclei combine to form a carbon-12 nucleus"

Book example follows the principle (mi-mf)

or: (3*4.002603u - 12u) = 0.007809u

However, when it is explaining the process, it explains that (mf - mi) = (change in)E

or: (12u - 3*4.002603u) = -.007809u
Jan13-14, 12:51 PM
P: 12,113
Those are the actual nuclei masses, not the binding energies.

A larger mass corresponds to a smaller binding energy, so there you have to take the opposite difference.

See mathman, calculate the mass before and after the reaction, that is the easiest way to see what happens.

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