Calculating the energy that is released in nuclear fusion

AI Thread Summary
The discussion revolves around calculating the energy released in nuclear fusion, specifically addressing a question from an AQA exam paper. Participants clarify that the binding energy corresponds to the mass defect, and the energy released can be calculated by subtracting the combined mass of reactants from products, converting the result to MeV. A key point of confusion is the mass of the hydrogen atom, which is stated as 1.00728 u in the mark scheme, leading to questions about its source. It is revealed that the mass of the proton is indeed provided on the formula sheet, though initially overlooked by the user. The conversation concludes with an acknowledgment of the importance of thoroughly reviewing provided materials for necessary values.
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Question 1. c) (ii) on this AQA paper... http://filestore.aqa.org.uk/subjects/AQA-PHYA5-1-QP-JUN13.PDF

The solution is here http://filestore.aqa.org.uk/subjects/AQA-PHYA5C-W-MS-JUN13.PDF

So I know that the binding energy will be the mass defect so the value of energy on the left hand side is bigger compared to the right hand side. To find the energy, in MeV, that is released it would be the change in mass that is in MeV. So I'd subtract the combined mass in u on the left from the combined mass of u on the right, then convert the final value to MeV from u.

The problem is the left hand side, the hydrogen atom. The mark scheme states that it is 1.00728 u.

I don't get that... From the formula sheet, http://filestore.aqa.org.uk/subjects/AQA-PHYA4-5-INS-JUN12.PDF, the mass of a proton is 1.67(3)x10^-27. Naturally, to combine to u, I would do 1.673/1.661... but that is not 1.00728 but 1.00722!

Could someone explain where they got their value from? Am I meant to take Hydrogen to be 1.00728 for granted, and so it is simply just the mass in u of hydrogen that I was suppose to know all the way along by memory? Seeing that they done no calculation to get that value, it makes me feel so. Thank you.
 
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Note that 1.00728 u has 6 significant figures. So, to arrive at this number from the mass in kg, you would need to know the mass of the proton in kg to 6 significant figures as well as the conversion factor from kg to amu to 6 significant figures.

But you are right, the formula sheet does not give sufficient information. Probably just an oversight of whoever made up the exam.
 
Oops... A closer inspection of the formula sheet does show that the mass of the proton is given as 1.00728 u.
 

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TSny said:
Oops... A closer inspection of the formula sheet does show that the mass of the proton is given as 1.00728 u.
OHH okay I had no idea they could of placed it there! I was looking on the right hand side of this formula sheet for numbers not the left... Okay cool thanks for your help.
 

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