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	$\Delta E=\Delta m c^2$

Constants

name	symbol	value
$speed \ of \ light \ in \ a \ vacuum$	$c$	$3\times 10^8 \,m\,s^{-1}$
$mass \ of \ proton$	$m_p$	$1.673\times10^{-27} \, kg$
$mass \ of \ neutron$	$m_n$	$1.675\times10^{-27} \, kg$
$unified \ atomic \ mass \ unit$	$u$	$1.661\times10^{-27} \, kg$

Variable definitions

Quantity Name	Symbol	Derived Unit	SI BASE Units
$energy$	$E$	$J$	$kg\,m^2\,s^{-2}$
$mass\, defect$	$\Delta m$	$kg$	$kg$

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.

Physics; Nuclear physics; KS5 Year 12; Einstein's mass-energy equation

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:

$\Delta E=\Delta m c^2$

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:

$\Delta m=((n_pm_p+n_nm_n)-(A\times u))$

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 $_6^{12}C$ nucleus.

Firstly, write down the known values:

$n_p=6 \newline \\[0.1in]n_n=6 \newline \\[0.1in] m_p=1.673\times10^{-27} \, kg \newline \\[0.1in]m_n=1.675\times10^{-27} \, kg \newline \\[0.1in]u=1.661\times10^{-27} \, kg$

Next, write down the equations needed and rearrange if necessary:

$\Delta m=((n_pm_p+n_nm_n)-(A\times u)) \newline \\[0.1in]\Delta E = \Delta m c^2$

Then, substitute the values into the equation for mass defect:

$\Delta m=(6\times1.673\times10^{-27} +6\times1.675\times10^{-27})-(12\times 1.661\times10^{-27}) \newline \\[0.1in]\Delta m=1.56 \times 10^{-28} \, kg$

Then substitute the mass defect into the mass-energy equation:

$\Delta E = 1.56\times 10^{-28} \times (3 \times 10^8) ^2 \newline \\[0.1in]\Delta E = 1.404\times 10^{-11} \, J$ 

Convert from $J$ to $eV$ :

$\Delta E = 1.404\times 10^{-11} \div 1.6\times 10^{-19} \newline \\[0.1in]\Delta E = 87750000 \, eV$

Make sure to round to the lowest number of significant figures of the values given by the question:

$binding \ energy, E=8.775\times 10^ 7 \, eV = 87.75 \, MeV$

The binding energy for a carbon-12 nucleus is $\underline{87.75 \, MeV}.$ 

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:

$binding \ energy \ per \ nucleon=\frac {binding \ energy (B)}{nucleon \ number (A)}$

Graphically the binding energy per nucleon plotted against the nucleon number looks like this:

Physics; Nuclear physics; KS5 Year 12; Einstein's mass-energy equation

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.

Physics; Nuclear physics; KS5 Year 12; Einstein's mass-energy equation

Firstly, write down the known values:

$B_U=7.4 \, MeV \newline \\[0.1in]B_{Cs}=8.2 \, MeV \newline \\[0.1in]B_{Rb}=8.8 \, MeV\newline \\[0.1in]Z_U=235 \newline \\[0.1in]Z_{Cs}=140 \newline \\[0.1in]Z_{Rb}=92 \newline \\[0.1in]$

Next, write down the equations needed and rearrange if necessary:

$E_{released} =E_{after}-E_{before} \newline \\[0.1in]E_{released} =(E_{Rb}+E_{Cs})-E_{U}$

Then, substitute the values into the equation:

$E_{released} =(92\times 8.8+140\times 8.2)-(235 \times 7.4) \newline \\[0.1in]E_{released} =218.6 \, MeV$

Make sure to round to the lowest number of significant figures of the values given by the question:

$energy \ released, E_{released}=220 \, MeV$

The energy released in a single fission of a uranium-235 nucleus is $\underline{220 \, MeV}.$ 

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:

$E_{\gamma}=2mc^2$

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:

$2E_{\gamma}=2mc^2 \newline \\[0.1in]E_{\gamma}=mc^2$