Question

If a proton and anti-proton come close to each other and annihilate, how much energy will be released?

Answer 1

Step 1: Calculating the total mass :-

An antiproton is defined as a nucleon with the same mass but the opposite charge to that of a proton. Because of this, when a proton and an antiproton collide, their combined masses are twice as large as a proton.
$\mathrm{m}=2\times {\mathrm{m}}_{\mathrm{p}}$

where $\mathrm{m}$ is combined mass

${\mathrm{m}}_{\mathrm{p}}$ is mass of proton

Mass of proton in (amu) ${\mathrm{m}}_{\mathrm{p}}=1\mathrm{amu}=1.67\times {10}^{-27}\mathrm{kg}$

So, $\mathrm{m}=2{\mathrm{m}}_{\mathrm{p}}=2\times 1.67\times {10}^{-27}\mathrm{kg}=3.34\times {10}^{-27}\mathrm{kg}$

Step 2: Conversion of mass into energy :-

When the proton and antiproton combine, the combined particles annihilate by transforming all of their mass into energy:

According to Einstein's mass-energy equivalence theory,$\mathrm{E}={\mathrm{mc}}^2$

where $\mathrm{m}$ is mass

$\mathrm{c}=3\times {10}^8{\mathrm{ms}}^{-1}$, is speed of light

Therefore,

$$\begin{gathered} \mathrm{E}=3.34\times {10}^{-27}\mathrm{kg}\times {(3\times {10}^8{\mathrm{ms}}^{-1})}^2 \\ \Rightarrow \mathrm{E}=3.34\times {10}^{-27}\times 9\times {10}^{16}\mathrm{J} \\ \Rightarrow \mathrm{E}=30.06\times {10}^{-27+16}\mathrm{J} \\ \Rightarrow \mathrm{E}=30.06\times {10}^{-11}\mathrm{J} \\ \Rightarrow \mathrm{E}=3.006\times {10}^{-10}\mathrm{J}\approx 3\times {10}^{-10}\mathrm{J} \end{gathered}$$

So, Energy released will be $3\times {10}^{-10}\mathrm{J}$.