Calculating the energy that is released in nuclear fusion

Click For Summary

Homework Help Overview

The discussion revolves around calculating the energy released in nuclear fusion, specifically focusing on the binding energy and mass defect as outlined in an AQA physics exam question. Participants are examining the mass of hydrogen and its conversion to energy in MeV.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the discrepancy between the mass of hydrogen provided in the mark scheme and their own calculations based on the mass of a proton. They question whether the value of 1.00728 u is a standard that should be memorized.

Discussion Status

Participants are actively clarifying the source of the mass value for hydrogen and discussing the implications of the formula sheet's information. Some have noted that the mass of the proton is indeed listed on the formula sheet, which may resolve the original poster's confusion.

Contextual Notes

There is a noted concern regarding the adequacy of the formula sheet in providing necessary information for the calculations, which may affect how participants approach the problem.

Tangent100
Messages
24
Reaction score
0
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.
 
Physics news on Phys.org
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.
 

Attachments

  • proton mass.png
    proton mass.png
    3.1 KB · Views: 512
  • Like
Likes   Reactions: Tangent100
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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 16 ·
Replies
16
Views
8K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
5K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K