Mass defect and binding energy are related to each other. By finding the value of mass defect we can easily calculate binding energy of nucleus. Mass defect and binding energy are represented symbolically as ‘∆m’ and B.E respectively. On increasing the value of mass defect, the value of binding energy also increases. These terms are only related to nucleus of element not the entire element.

## Mass Defect (∆m)

During nuclear reactions like fission and fusion, the sum of masses of daughter nuclei(particles) is always greater than the mass of parent nuclei. This difference in masses between daughter particles and parent nuclei is called mass defect(∆m).

Mass defect is converted into energy during the reaction.

### Mass Defect Formula

Let us suppose a nucleus having mass ‘M’. Also, let ‘m_{p}’ be a mass of proton and m_{n} be a mass of neutron present inside it.

Then according to the definition, ∆m is given by formula,

Mass defect (∆m) = [Zm_{p} + (A-Z)m_{n} – M]

Where,

Z = number of protons = atomic number

A = mass number of nucleus/element

(A–Z) = Difference gives number of neutrons

Number of proton in a nucleus is always equal to atomic number or number of electrons of an element.

Proton and neutron are present inside nucleus and the sum of these two is equal to atomic mass/mass number of nucleus or element. Whereas electron is present outside the nucleus and do not contribute for the atomic mass.

Mass defect is expressed in unit amu or Kg. 1 amu is equal to 1.66 x 10^{-27} Kg.

**Packing Fraction**

It is defined as mass defect per nucleon. It is given by,

Packing Fraction = ∆m/A = [Zm_{p} + (A–Z)m_{n} – M]/A

Where, A = mass number/atomic weight

The value of packing fraction may be positive, zero or negative. It is **unitless** quantity.

## Binding Energy (B.E)

In a nucleus, protons and neutrons are held together by **strong nuclear** force. The amount of energy required to separate protons and neutrons far apart is called binding energy. It is also defined as energy equivalent to mass defect(∆m).

### Binding Energy Formula

Binding energy is same as calculating energy from defect mass using Einstein’s famous mass energy relation E=mc^{2}. In terms of binding energy, it is given by

B.E = ∆mc^{2}

Where, ∆m = mass defect and c = speed of light

**Case I**

If ∆m is in **amu** then,

B.E = ∆m x 931 MeV

If mass defect is 1 amu then, it’s binding energy or energy is equivalent to 931 MeV

**Case II**

If ∆m is in **Kg** then,

B.E = ∆m x c^{2 }(In Joule)

Also,

B.E = [Zm_{p} + (A–Z)m_{n} - M] x c^{2}

**Binding Energy Per Nucleon**

Nucleon is the term used to collectively called the mass of proton and neutron in nucleus or mass of nucleus in whole. Binding energy per nucleon is given by,

B.E per nucleon = binding energy/mass number = B.E/A = ∆mc^{2}/A

### Binding Energy Curve

It is the curve plotted between binding energy per nucleon and mass number (A). The curve plotted for the elements of periodic table shows that binding energy per nucleon first increases, attains maximum value and then decreases slowly. This curve is used to predict the stability of different elements.

Image source: wikimedia commons

Average binding energy per nucleon for heavy nucleus is approximately 8Mev. For Iron (fe), it has maximum value of about 8.8Mev.

It’s value is more for medium nuclei. That’s why elements with medium mass range 30 ≤ A ≤ 170 are more stable. In stable nucleus, the ratio of neutron to proton number is 1.3 to 1.4.

## Mass Defect & Binding Energy Problems

**Problem 1**

Calculate the binding energy of an alpha particle. Given,

Mass of He = 4.00126amu, mass of proton (m_{p}) = 1.00727 amu and mass of neutron (m_{n}) = 1.00866 amu.

__Solution__

alpha particles are He-nuclei.

From formula, we have

Mass defect = Zm_{p} + (A–Z)m_{n} – M

Or, ∆m = 2 x 1.00727 + 2 x 1.00866 - 4.00126 (no. of proton is 2 and number of neutron is also 2 in He nucleus)

∆m = 0.03021amu

Binding energy = ∆m x 931 (since ∆m is in amu)

B.E = 0.03021 x 931

B.E = 28.5 MeV __Ans__

**Problem 2**

Calculate binding energy per nucleon of He nucleus which has binding energy 28 MeV

__Solution__

Binding energy (B.E) = 28 MeV

Mass number/atomic mass of He (A) = 4

Now,

B.E per nucleon = B.E/mass number

= 28/4

= 7 MeV

The binding energy per nucleon is 7 MeV.

**Problem 3**

Mass of α-particle is less than that of its constituents by 0.03 amu. Calculate its binding energy per nucleon.

__Solution__

Mass of He-nuclei (A) = 4 (since α-particle are He-nuclei)

Mass defect (∆m) = 0.03

Now,

B.E per nucleon of α-particle = B.E/A

= ∆m x 931/A

= 0.03 x 931/4

= 7 MeV __Ans.__

**Problem 4**

Calculate the binding energy per nucleon of Li-atom. Given,

Mass of Li = 7.0163 amu

Mass of proton (m_{p}) = 1.0076amu

Mass of neutron (m_{n}) = 1.0089 amu

__Solution__

Total number of neutron(A–Z) = 7-3 = 4

We have,

Binding energy per nucleon = ∆m x 931/A

= [Zm_{p} + (A–Z)m_{n} – M] x 931/A

= [3 x 1.0076 + 4 x 1.0089 – 7.0163] x 931/7.0163

= 0.0421 x 931/7.0163

= 5.59 __Ans.__

**Problem 5**

_{2}He^{4} + _{7}N^{14} → _{8}O^{17} + _{1}H^{1} + Q

Calculate the value of Q if Mo_{2} = 17.004533 amu, M_{N} = 14.007514 amu and M_{He} = 4.003837 amu.

__Solution__

From the above reaction

Mass defect (∆m) = mass of reactants – mass of product

= (14.007514 + 4.003837) – (17.004533 + 1.008142)

= - 1.324 x 10-3 amu

Again,

Energy released (Q) = ∆m x 931 MeV

= – 1.324 x 10-3 x 931

= –1.23 MeV __Ans.__

**Problem 6**

Calculate the binding energy per nucleon of _{18}Ar^{40} whose mass is 39.9751 amu.

Given mass of proton (m_{p}) = 1.0078254 amu

Mass of neutron(m_{n}) = 1.008665 amu

__Solution__

Mass of argon(M) = 39.9751 amu

Total mass of protons(Zm_{p}) = 18 x 1.0078254 = 18.1408572

Total number of neutron (A–Z) = 40-18 = 22

Total mass of neutron(A–Z)m_{n} = 22 x 1.008665 = 22.19063

Mass defect (∆m) = [Zm_{p} + (A–Z)m_{n} – M]

= [18.1408572 + 22.19063 – 39.9751]

= 0.3563872 amu

Also,

B.E per nucleon = ∆m x 931/A

= 0.3563872 x 931/39.9751

= 8.3 amu __Ans.__

**Problem 7**

Calculate the packing fraction of _{7}N^{14} having mass 14.003 amu

__Solution__

Actual mass of N-nucleus(M) = 14.003

Mass number of N-nucleus(A) = 14

Now,

Packing fraction = M – A/A

= 14.003 – 14/14

= 2.1 x 10^{-4} __Ans.__

## References

https://en.wikipedia.org/wiki/Nuclear_binding_energy

https://www.britannica.com/science/mass-defect