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Calculating the energy that is released in nuclear fusion

  1. Jun 27, 2016 #1
    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.
  2. jcsd
  3. Jun 27, 2016 #2


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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.
  4. Jun 27, 2016 #3


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    Oops... A closer inspection of the formula sheet does show that the mass of the proton is given as 1.00728 u.

    Attached Files:

  5. Jun 27, 2016 #4
    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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