High School Mass Defect: Is my understanding correct?

  • Thread starter Thread starter A M
  • Start date Start date
  • Tags Tags
    Mass Mass defect
Click For Summary
Einstein's mass-energy relationship states that an increase in energy corresponds to an increase in mass, expressed by the equation ΔE=Δmc^2. The mass defect refers to the phenomenon where the mass of an atomic nucleus is less than the sum of its constituent nucleons, calculated as Mass Defect = Zm_p + Nm_n - m_x. This mass defect is directly related to the nuclear binding energy, which is the energy required to separate nucleons within a nucleus. Iron-56 and nickel-62 exhibit the highest binding energy per nucleon, indicating they are the most tightly bound and least massive. The discussion concluded with a correction regarding the units used in binding energy equations, highlighting the importance of maintaining consistency in scientific calculations.
A M
Messages
83
Reaction score
16
TL;DR
I want to write a student article specially for those who don't have a background in nuclear physics. I've been suggested to share my basic understanding & ask if they're correct.
I would be grateful if anyone could explain where my mistakes are:
(Please note that diagrams are designed just to give a simple imagination of the article & make it more understandable; they do NOT correspond precise information.)
Mass – Energy Relationship:
According to Einstein’s special theory of relativity, when the energy of a body increases, so does its mass, and vice versa. If the difference in energy is indicated by ΔE and the difference in mass by Δm, these two quantities are related by his famous equation:
##ΔE=Δmc^2##
When 'c' is the velocity of light (##2.9979×10^8 m/s##).

Mass Defect:
The mass of an atomic nucleus is less than the sum of the individual masses of the free constituent nucleons, this 'missing mass' is known as the mass defect. So for a nucleus (X) with Z protons and N neutrons we can write:
##m_x, m_n ,m_p## -the masses of a nucleus (X), free neutron and free proton
##m_x<Zm_p+Nm_n##
Mass Defect= ##Zm_p+Nm_n-m_x##

As it has been written in the first part, nuclear binding energy is the minimum energy we need to add to a nucleus to separate all of its nucleons.
So for the binding energy of that nucleus (##B_x##) we could write:
##m_x+B_x=Zm_p+Nm_n##
##B_x=Zm_p+Nm_n–m_x##
And we can conclude that the binding energy of a nucleus corresponds/is its mass defect [by ##E=mc^2##].
Iron-56 and nickel-62 have the highest nuclear binding energy per nucleon; meaning that they have the least mass per nucleon. As a matter of fact 'more tightly bound means less massive.'References:
https://www.physicsforums.com/posts/6216782/https://chem.libretexts.org/Bookshe...x)/Miscellaneous/460:_Mass-Energy_Equivalence
 
Last edited:
Physics news on Phys.org
Seems OK except that you're mixing units in your binding energy equations. Because ##E = mc^2## not ##E = m##, there should be a factor of ##c^2## in there, unless you've explicitly stated somewhere that you're using units where ##c = 1##.
 
  • Like
Likes A M
Yes you're right, but to be honest, I've already given up writing that article.
Anyway, thanks for your correction! :smile:
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K