I Energy released calculation using Binding energy and mass defect

phantomvommand
Messages
287
Reaction score
39
TL;DR Summary
I notice a discrepancy in calculating the energy released when using binding energy and mass defect.
Consider the equation
X (200, 50) + n (1, 0) -> Y (120, 30) + Z (70, 20) + 11 n(1, 0)

Let p be the mass of a proton, n be the mass of a neutron.
BE(X) = [50p + 150n - M(X)] c^2
BE(Y) = [30p + 90n - M(Y)] c^2
BE(Z) = [20p + 50n - M(Z)]c^2

The energy released when using BE (products) - BE (reactants) is thus: [M(X) - M(Y) - M(Z) - 10n] c^2
On the other hand, the mass released using [Mass (reactants) - Mass (products)]c^2 = [M(X) - M(Y) - M(Z)] c^2

There is a difference of 10n * c^2. Which is the correct calculation and why is the other wrong? Thank you!
 
Physics news on Phys.org
How can you leave out the neutrons?
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top