# When nucleons are assembled in form of nucleus is some mass lost? [duplicate]

We are told that $$B.E = [Zm_p + Nm_n - M] c²$$ Here, Z = Atomic no. ,$$m_p$$ = mass of proton, $$m_n$$ = mass of neutron and M = Mass of nucleus.

Because if the mass is not lost then the energies should cancel as the protons and neutrons are what make up the nucleus so, $$m_p + m_n = M$$ Unless some mass is lost there shouldn't be a difference in mass. And we should get zero bindig energy. Please explain.

• Does this answer your question? Mass defect- From where mass is being lost? Feb 5 at 2:48
• Mass can be converted to energy and vice versa. Energy is always conserved. Here, the loss in mass is converted into energy.
– MrQ
Feb 23 at 13:41

At the level of nuclear dimensions mass is not a conserved quantity , but one derived from the four vectors of special relativity.

In special relativity each particle and composite of particles is described by a four vector. The energy momentum vector

$$\vec p=\pmatrix{E \\ p_xc \\ p_y c \\ p_z c}=\pmatrix{E \\ \vec p c}$$

has as a length the invariant mass, $$m_0$$:

$$\sqrt{P\cdot P}=\sqrt{E^2-(pc)^2}=m_0c^2$$

The four vectors of protons and neutrons that make up the nucleus sum up to a single four vector,which has the invariant mass of the single nucleus. The summing up is vectorial. It is only energy and momentum that are conserved, not mass.

The difference in mass between the sum of free nucleons and the mass of the nucleus is similar to the difference in length when one adds two vectors at an angle, the new vector length is less than the sum of the two lengths.