# In nuclear physics, what is meant by the term 'mass defect', and what is its relation to nuclear energy?

Apr 15, 2016

Lost mass in nuclear fission and fusion reactions appears as energy in accordance with $E = m {c}^{2}$, where $c = 3 \times {10}^{8}$ $m {s}^{-} 1$, $E$ is the energy and $m$ is the mass.

#### Explanation:

In both nuclear fission and fusion, the total mass of the products after the event is less than the total mass before it. The difference is called the 'mass defect'.

(At a deeper level, the reason that energy can be produced both when small nuclei join together and when large nuclei break apart is explained by the 'binding energy per nucleon' curve, but that explanation is too detailed to go into here.)

That lost mass is converted into energy. The equation $E = m {c}^{2}$ describes the change, and given that $c = 3 \times {10}^{8}$ $m {s}^{-} 1$, the amount of energy $\left(J\right)$ is $9 \times {10}^{16}$ times as great as the mass defect $\left(k g\right)$.

Apr 16, 2016

See below

#### Explanation:

We know that nucleus of an atom consists of protons and neutrons, collectively called nucleons. (Hydrogen-1 nucleus being exception which has only one proton).

If we compare the combined mass all the nucleons with sum of individual masses of all these nucleons, we will find that
the combined mass is less than sum of individual masses.

This is known as mass defect or sometimes also called mass excess.
It represents the energy that was released when the nucleus was formed, called binding energy of nucleus.
The relation between the mass defect and binding energy is the Einsteins mass-energy equivalence relation ship
$\text{Energy", E="mass} , m \times {c}^{2}$

where $c$ is velocity of light in vacuum, and is 299,792 km/ s, normally taken as $3 \times {10}^{8} m {s}^{-} 1$

In light of new input assuming $A ,$ Mass number or number of nucleons and $Z$, Atomic number, number of protons in the nucleus as nuclear constants.

$\therefore$Mass defect $\Delta m = \text{Atomic number"xx"mass of Hydrogen-1 atom"+("Mass number"-"Atomic number")xx"mass of a neutron"-"mass of Atom}$

$= Z \times {m}_{H} + \left(A - Z\right) {m}_{n} - M$
${m}_{H} = 1.007825 a . m . u .$
${m}_{n} = 1.008665 a . m . u .$
$1 a . m . u . \equiv 1 / 12 \text{ of C"^12"atom} = 1.6606 \times {10}^{-} 27 k g$
$M$ can be found from published tables.

1 a.m.u. with the help of

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Regarding nuclear constant*, the nearest relation to the number posted by you, I find that it looks like you are referring to Ground state radius of Hydrogen atom. It is called Bohr's Radius of Hydrogen atom and it is written as a_@=5.29177 xx 10^ -11 mâ‰ˆ 0.529 Ã….

*This constant is not a nuclear constant but an atomic constant.