Posted by Jim on .
The easiest fusion reaction to initiate is
11H + 11H> 42He + 10 n
The 11 indicates a small one above the other on the left side of the symbol of the element.So for 10n is a one above a zero on the left side of
n (neutron). The number on the top is the atomic mass and the one on the bottom is the atomic number.
Calculate the energy released per 42He nucleus produced and per mole of
42He produced. The atomic masses are:
21H 2.01410amu
31H 3.01605amu
42He 4.00260amu
The masses of the electron and neutron are 5.4858x10^4 and 1.00866 amu.
Maybe this might help
proton amu is 1.00782

College Chemistry (Hard) 
bobpursley,
I think you have the reaction screwed up. I suspect you mean a DT reaction
3H1 + 2H1 >> 4He2 + 1n0
add the masses on the left, you were given each. Add the masses on the right. You were given them.
Subtract the masses on the right from the masses on the left. The result is the mass deficit.
enery= massdeficit*speedlight^2
Change amu to kg. 
College Chemistry (Hard) 
DrBob222,
You have a problem with the equation.
The numbers must add up on both sides and yours don't.
The lower numbers are 2 on the left and they add to 2 on the right. Good.
The top numbers don't add up. The left side is 2 and the right side is 5. (The equation is usually see isand I'll redo the numbers so one is on the left and one is on the right1H^2 + 1H^3 ==> 2He^4 + oN^1). When you get that straightened out, add all of the electrons, protons, and neutrons on the left (their masses) and subtract the actual mass of the products. That is the change in mass that has been converted to energy. Then delta E = delta m*c^2 and calculate energy. 
College Chemistry (Hard) 
Jim,
The easiest fusion reaction to initiate is
21H + 31H> 42He + 10 n
Sorry for the mistake 
College Chemistry (Hard) 
Jim,
How would I calculate the energy released per mole of He?

College Chemistry (Hard) 
DrBob222,
If you multiply delta m (in kg) x c^2, you have the energy per molecule. A mole of He contains 6.022 x 10^23 molecules so your number x Avogadro's number gives per mole He.

College Chemistry (Hard) 
Jim,
Thank you very much DrBob222 and bobpursley

College Chemistry (Hard) 
JEns,
I have the same question. I have the answer from the book 2.820*10^12 J/nucleus and 1.698*10^12 J/mol. I have tried to what you guys have explained, but didn´t get the answer right.
delta mass= (4.00260+1.00866)(2.01410+3.01605)=1.02755 amu
convert to kg:
1.02755 amu*1.6605*10^(27)= 1.706246*10^(27) kg
E=m*c^2
E=1.706246*10^(27) kg*(3.00*10^8 m/s)^2=1.53*10^10
For the J/mol
1.706246*10^(27) kg*6.022*10^23=0.00102 kg/mol
E=m*c^2
E=0.00102 kg/mol *(3.00*10^8m/s)^2=3.08*10^5 J/mol
What´s wrong?? :D