Einstein's mass-energy equation
In a nutshell
The nucleons which make up a nucleus have a smaller mass when bound in a nucleus than when the nucleons are separated. This is because when the nucleons are bound in a nucleus, some energy is released. This energy comes from the mass defect between bound and unbound nucleons.
Equations
Description | Equation |
Einstein's mass-energy equation | ΔE=Δmc2 |
Constants
name | symbol | value |
speed of light in a vacuum | | 3×108ms−1 |
mass of proton | | 1.673×10−27kg |
mass of neutron | | 1.675×10−27kg |
unified atomic mass unit | | 1.661×10−27kg |
Variable definitions
Quantity Name | Symbol | Derived Unit | SI BASE Units |
| | | kgm2s−2 |
massdefect | | | |
Mass defect
The mass of a nucleus is less than the mass of the individual nucleons within the same nucleus.
Example
Helium-4 is made up of two protons and two neutrons. The helium nucleus is lighter than the mass of two protons and two neutrons individually.
| 1 | Mass of individual nucleons | 2 | Mass of nucleons bound in nucleus | |
This difference in mass is called the mass defect and comes from when nuclei form. When nucleons come together to form a nucleus, they lose a small amount of mass and release it in the form of energy.
This is related by arguably the most famous equation in physics. Einstein's mass-energy equation:
ΔE=Δmc2
If all of the nucleons in the nucleus were to be pulled apart, the amount of energy required to do this is the same amount of energy that was released when the nucleus was formed. This is due to the conservation of energy and the amount of energy required to fully separate a nucleus is given by what is called the binding energy.
Note: The binding energy is not the same for different nuclei .
To calculate the mass defect of a nucleus, the following equation can be used:
Δm=((npmp+nnmn)−(A×u))
Note: The mass of protons and neutrons not bound in the nucleus are calculated using the rest masses. The mass of protons and neutrons which are bound in the nucleus are calculated using the unified atomic mass u.
Example
Calculate the binding energy for a 612C nucleus.
Firstly, write down the known values:
np=6nn=6mp=1.673×10−27kgmn=1.675×10−27kgu=1.661×10−27kg
Next, write down the equations needed and rearrange if necessary:
Δm=((npmp+nnmn)−(A×u))ΔE=Δmc2
Then, substitute the values into the equation for mass defect:
Δm=(6×1.673×10−27+6×1.675×10−27)−(12×1.661×10−27)Δm=1.56×10−28kg
Then substitute the mass defect into the mass-energy equation:
ΔE=1.56×10−28×(3×108)2ΔE=1.404×10−11J
Convert from J to eV:
ΔE=1.404×10−11÷1.6×10−19ΔE=87750000eV
Make sure to round to the lowest number of significant figures of the values given by the question:
binding energy,E=8.775×107eV=87.75MeV
The binding energy for a carbon-12 nucleus is 87.75MeV.
Binding energy per nucleon
The binding energy per nucleon is a useful way of comparing the binding energy of different nuclei. It can be calculated using the formula:
binding energy per nucleon=nucleon number(A)binding energy(B)
Graphically the binding energy per nucleon plotted against the nucleon number looks like this:
Note: Iron is the most stable element in terms of binding energy. Everything to the left of iron releases energy via nuclear fusion. Everything to the right releases energy via nuclear fission.
The binding energy per nucleon graphs can be used to calculate the energy released in a fission or fusion reaction.
Example
The graph below shows the energy released during the fission of uranium-235. It splits into two daughter isotopes of rubidium-92 and caesium-140. Calculate the total energy released in one uranium-235 fission reaction.
Firstly, write down the known values:
BU=7.4MeVBCs=8.2MeVBRb=8.8MeVZU=235ZCs=140ZRb=92
Next, write down the equations needed and rearrange if necessary:
Ereleased=Eafter−EbeforeEreleased=(ERb+ECs)−EU
Then, substitute the values into the equation:
Ereleased=(92×8.8+140×8.2)−(235×7.4)Ereleased=218.6MeV
Make sure to round to the lowest number of significant figures of the values given by the question:
energy released,Ereleased=220MeV
The energy released in a single fission of a uranium-235 nucleus is 220MeV.
Pair production and annihilation
During a high energy event, high energy photons can be converted into matter and antimatter. This is called pair production and an equal amount of matter and antimatter is created in the creation of particles.
The minimum amount of energy the photon can have is equivalent to the rest masses of both the particle and antiparticle. One photon creates two particles, so the equation used is:
Eγ=2mc2
During annihilation, a particle and anti-particle annihilate one another releasing two identical photons travelling in opposite directions. Their masses get converted into energy.
Two particles create two photons, so the equation to calculate the energy of one of the photons is:
2Eγ=2mc2Eγ=mc2