Question

Calculate the energy released when an electron annihilates a positron.

Answer 1

In order to obtain the answer for this question, the students should be aware of the Einstein’s mass-energy equivalence relation:
Energy, $E = m{c^2}$
where m = mass and c = a constant whose value is equal to the speed of light i.e. $3 \times {10^8}m{s^{ - 1}}$.

Complete step by step answer:
In classical physics, the quantity mass is conserved, which means that the mass cannot be destroyed.
However, this statement is violated when it comes to Quantum physics, where we study about a process known as annihilation.
Annihilation is a process in which an electron collides with its antiparticle and get completely converted into energy. In other words, the matter is completely destroyed without leaving any traces, since all of the matter has been converted to energy.
This energy is given by the Einstein’s theory of energy-mass equivalence,
$E = m{c^2}$
Step I
The equation of electron annihilation into positron is given as: ${}_{ - 1}^0\beta + {}_{ + 1}^0\beta \to hv + hv$
Where ${}_{ - 1}^0\beta$is an electron,${}_{ + 1}^0\beta$is the positron, and $hv$is the amount of energy emitted in the form of gamma radiation.

Step II
Calculation of energy released in the process of annihilation due to the change of mass is:
\Delta m = 2 \times 9 \times {10^{ - 31}}$$$$kg
$\Rightarrow \Delta m = 18 \times {10^{ - 31}}kg$
Now energy released in this process is:
$\Delta E = \Delta m{c^2}$
$= (18 \times {10^{ - 31}} \times 3 \times {10^8})J$
$= 162 \times {10^{ - 15}}J$
So, Energy in the unit of joule
$\Delta E = 162 \times {10^{ - 15}}J$………….(i)
Dividing equation (i) by $1.6 \times {10^{ - 19}}$, we get
Now, Energy in the unit of electron volt as
$\Delta E = \left( {\dfrac{{162 \times {{10}^{ - 15}}}}{{1.6 \times {{10}^{ - 19}}}}} \right)eV$
$\Rightarrow \Delta E = 1.01 \times {10^6}eV$
$\Delta E = 1.01{\text{M}}eV$

Therefore, the energy released during the annihilation process is $1.01{\text{M}}eV$.

Note: This process of annihilation of matter is applied in the nuclear bombs, where the energy released by the bomb is equal to the mass of the bomb times the square of speed of light as per Einstein's equation explained here.