## What is Binding Energy?

**Definition of Binding Energy: **Binding energy is the amount of energy required to completely separate the protons and the neutrons in the nucleus. Therefore, it is an energy that helps us to tear nucleons apart. Inside the nucleus of an atom, there is what we call a nucleon.

**Nucleon** are protons and neutrons in the nucleus of an atom. The protons and neutrons (nucleons) are held together in the nucleus of an atom by very powerful nuclear forces. The amount of energy we need to separate those protons from neutrons in the nucleus of an atom is what we call binding energy.

Therefore, we need a very huge amount of energy to tear those nucleons apart. This energy is over 10^{6} times more than that required to remove the electrons from an atom. Hence, when we separate the nucleus, it follows that the total mass is less than the mass of the nucleus. We also refer to the binding energy as the difference in mass or mass defect.

From the above explanation, we can now see that the binding energy of a nucleus is proportional to the difference between the total mass of the individual nucleons and the mass of the nucleons. Usually, the total mass of the stable nucleus or nuclide is less than the sum of the masses of its constituent nucleons.

## Binding Energy Formula

According to Albert Einstein, the difference in mass between all the individual nucleons and the mass of the nucleus is a measure of the binding energy. The energy released during the nuclear reaction (binding energy) is given by Einstein’s energy equation which is

E = mc^{2}

We can also write **Einstein’s equation which is also known as the binding energy formula as** ΔE = Δmc^{2}

Where

ΔE = Nuclear energy

Δm = change in mass

c = speed of light

The unit of energy in a nuclear reaction is in a unified atomic unit (u) or the electron volt (eV) or Joules (J). Here is how to convert the units:

**Electron volt to Joule**

1 eV = 1.6 x 10^{-19} J

**Mega electron volt to Joules**

1 MeV = 1.6 x 10^{-13} J

**Atomic Unit to Joules**

1 u = 931 MeV = 1.490 x 10^{-10} J

## Mass Defect and Atomic Energy

**Mass defect** (Δm) can be defined as the mass of the element lost when a given mass of a radioactive atom is split into two parts. The formula for calculating mass defect is

**Mass defect (Δm) = mass of the initial nucleus – total mass of the split part**

**Atomic energy** is the energy released as a result of the splitting of a radioactive nucleus by fission or fusion.

## How to Calculate Binding Energy

Here are a few problems that can help us to understand how to find binding energy:

### Problem 1

In a nuclear reaction, the mass defect is 2.0 x 10^{-6} g. Calculate the energy released given that the velocity of light is 3.0 x 10^{8} m/s.

#### Answer

**The final answer to the above question is 1.8 x 10 ^{8} J**

#### Explanation

**Data**

Mass defect, m = 2.0 x 10^{-6} g = 2.0 x 10^{-9} kg

The velocity of light, c = 3.0 x 10^{8} m/s

**Unknown:**

Energy, E = ?

**Formula**

E = mc^{2}

**Solution**

E = mc^{2} = 2.0 x 10^{-9} x (3.0 x 10^{8})^{2} = 1.8 x 10^{8} J

**Therefore, the energy released is 1.8 x 10 ^{8} joules**.

### Problem 2

In a thermonuclear reaction, the total initial mass is 5.02 x 10^{-27} kilograms and the final mass is 5.01 x 10^{-27} kilograms. The energy released in the process is

#### Answer

**The final answer to the above question is E = 9.0 x 10 ^{-13} J **

#### Explanation

**Data:**

Initial mass, m_{1} = 5.02 x 10^{-27} kg

Final mass, m_{2} = 5.01 x 10^{-27} kg

Speed of light, c = 3 x 10^{8} m/s

**Unknown:**

Mass defect, m = ?

Energy released, E = ?

**Formula**

Step 1: Mass Defect, m = m_{1} – m_{2}

Step 2: Energy, E = mc^{2}

**Solution**

Step 1

m = m_{1} – m_{2} = 5.02 x 10^{-27} – 5.01 x 10^{-27} = 1.0 x 10^{-29} kg

Step 2

The energy released, E = mc^{2} = 1.0 x 10^{-29} x (3 x 10^{8})^{2} = 9.0 x 10^{-13} J

**Therefore, the energy released is 9.0 x 10 ^{-13} joules**

### Problem 3

In the fusion of hydrogen isotopes into helium, the decrease in mass is about 0.65%. Calculate the energy obtainable when 1.0 grams of hydrogen is used.

#### Answer

**The final answer to the above question is 5.85 x 10 ^{11} J**

#### Explanation

**Data**

Since 1.0 g of hydrogen is used, we can start by converting grams into kilograms

1.0 g = 1.0/1000 = 0.001 kg

Also, mass defect, m = (0.65 / 100) x 0.001 = 6.5 x 10^{6} kg

Speed of light, c = 3 x 10^{8} m/s

**Unknown**

Energy, E = ?

**Formula**

E = mc^{2}

**Solution**

E = mc^{2} = 6.5 x 10^{6} x (3 x 10^{8})^{2} = 5.85 x 10^{11} J

**Therefore, the energy obtainable is 5.85 x 10 ^{11} joules**

### Problem 4

A possible fusion reaction is _{1}^{2}H + _{1}^{2}H —> _{1}^{3}H + _{1}^{1}H + Q, where Q is the energy released as a result of the reaction. If Q = 4.03 MeV, calculate the atomic mass of _{1}^{3}H in atomic mass units. [ _{1}^{2}H = 2.0141u; _{1}^{1}H = 1.00783u; 1u = 931 MeV ].

#### Answer

**The final answer to the above question is 3.01604 u**

#### Explanation

_{1}^{2}H + _{1}^{2}H —> _{1}^{3}H + _{1}^{1}H + Q

Let x be the atomic mass of _{1}^{3}H in u

1 u = 931 MeV

Therefore, 4.03 MeV = 4.03 u / 931 = 0.004329 u

Substitute given values into the nuclear equation to obtain

2.01410 + 2.01410 —> x + 1.00783 + 0.0004329

4.0282 = x + 1.01216

**x = 4.0282 – 1.01216 = 3.01604 u**

### Problem 4

The radioactive nuclei _{84}^{210}Po emit an α-particle to produce _{82}^{206}Pb. Calculate the energy in MeV released in each disintegration.

[ Take the masses of _{84}^{210}Po = 209.936730 u; _{82}^{206}Pb = 205.929421 u; _{2}^{4}He = 4.001504, and that 1 u = 931 MeV]

#### Answer

**The final answer to the above question is Q = 5.404 MeV**

#### Explanation

_{84}^{210}Po —> _{2}^{4}H + _{82}^{206}H + Energy (Q)

1 u = 931 MeV

When we insert our values into the above nuclear reaction, we will have

(209.9367300) x 931 = (4.001504 + 205.929421) x 931 + Q

When we make Q subject of the formula, we will end up with

Q = 195451.0956 – 195445.6912 = 5.404 MeV

### Problem 5

What is the binding energy of helium _{2}^{4}He

[Atomic mass of proton = 1.00783 u, the atomic mass of neutron = 1.00867 u]

#### Answer

**The final answer to the above question is 0.033 u**

#### Explanation

**Data**

Atomic energy of proton = 1.00783 u

The atomic mass of neutron = 1.00867 u

Also, the atomic mass of _{2}He = 4 u

Now, the number of protons = 2

The number of neutrons = 4 – 2 = 2

Mass of protons = 2 (1.00783 u) = 2.01566 u

Mass of neutrons = 2 (1.00867 u) = 2.01734 u

The total mass of neutron = 2.01566 + 2.01734 = 4.033 u

Therefore,

Binding energy = Difference between the mass of nucleon and that of nucleus = mass of nucleon – mass of the nucleus

The above expression will become

**E = 4.033 u – 4 u = 0.003 u**

### Problem 6

The mass of a proton is 1.0074 u and that of a neutron is 1.0089 u. Determine the energy evolved in stabilizing the nucleus of nitrogen of a mass number 14 with 7 protons and 7 neutrons. [ Take the speed of light, c = 3 x 10^{8} m/s; 1 u = 1.67 x 10^{-27} kg ]

#### Answer

**The final answer to the above question is 1.715 x 10 ^{-11} J**

#### Explanation

**Data**

Mass of proton = 1.0074 u

The mass of the neutron = 1.0089 u

Additionally, we have the mass of the nucleus = 14 u

Speed of light, c = 3 x 10^{8} m/s

The number of protons = 7

Number of neutron = 7

We also have 1 u = 1.67 x 10^{-27} kg

To find the mass of the protons, we say

Mass of the protons = 7 x 1.0074 = 7.0518 u

The mass of neutrons = 7 x 1.0089 = 7.0623 u

Total mass of nucleon = mass of proton + mass of neutron = 7.0518 + 7.0623 = 14.1141 u

The mass of the nucleon in kg = 14.1141 x 1.67 x 10^{-27} = 2.3570547 x 10^{-26} kg

Mass of the nucleus in kg = 14 x 1.67 x 10^{-27} kg

Therefore, the **mass defect**, m = mass of nucleon – mass of nucleus = 2.3570547 x 10^{-26} – 2.338 x 10^{-26}

m = 1.90547 x 10^{-28} kg

Energy evolved, E = mc^{2} = 1.90547 x 10^{-28} x (3.0 x 10^{8})^{2} = 1.715 x 10^{-11} J

### Problem 7

Calculate in Joules the binding energy for ^{9}_{4}Be

[Take the atomic mass of ^{9}_{4}Be = 9.01219 u; mass of the proton = 1.00783 u, mass of the neutron = 1.00867 u; unified atomic mass unit, u = 931 MeV; 1eV = 1.6 x 10^{-19} J]

#### Answer

**The final answer to the above question is 9.3 x 10 ^{-12} J**

#### Explanation

Number of protons = 4

The number of neutrons = 9 – 4 = 5

Mass of the protons = 4 (1.00783) = 4.03132 u

The mass of the neutrons = 5 (1.00867) = 5.04335 u

Hence,

Mass of the nucleon = mass of the proton + mass of the neutron = 4.03132 + 5.04335 = 9.07467 u

Now, we will calculate the mass defect

Mass defect = mass of nucleon – mass of nucleus = 9.07467 – 9.01219 = 0.06248 u

Therefore,

The binding energy = 0.06248 x 931 x 10^{6} x 1.6 x 10^{-19 }= 9.3 x 10^{-12} J

### Problem 7

Deuteron and tritium fused to form a helium nucleus according to the equation.

^{2}_{1}He + ^{3}_{1}He —> ^{4}_{2}He + ^{1}_{0}n + Q

Calculate in Joules the energy released.

[^{3}_{1}He = 3.01605 u; ^{2}_{1}He = 2.0141 u; ^{4}_{2}He = 4.00260 u; ^{1}_{0}n = 1.00867 u; 1 u = 931 MeV, and 1 eV = 1.6 x 10^{-19} J ]

#### Answer

**The final answer to the above question is Q = 2.812 x 10 ^{-12} J**

#### Explanation

Total mass of the reactants = 2.01410 + 3.01605 = 5.03015 u

The total mass of products = 4.00260 + 1.00867 = 5.01127 u

Subsequently, our mass defect will become

Mass defect = mass of the reactant – mass of the product = 5.03015 – 5.01127 = 0.01888 u

Therefore, the energy released Q is

Q = 0.01888 x 931 x 1.6 x 10^{-19}

The above expression will now become

**Q = 2.812 x 10 ^{-12} Joules**

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