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Question: If a proton and antiproton come close to each other and annihilate, how much energy will be released...

If a proton and antiproton come close to each other and annihilate, how much energy will be released?
A. 1.5×10−10J1.5 \times {10^{ - 10}}J
B. 3×10−10J3 \times {10^{ - 10}}J
C. 4.5×10−10J4.5 \times {10^{ - 10}}J
D. 2×10−10J2 \times {10^{ - 10}}J

Explanation

Solution

Even though scientists believed from a very long time that matter can neither be created nor destroyed, they later discovered in the 20th century that matter can be completely annihilated in the sense, completely destroyed without leaving any traces behind. This discovery was possible through Einstein’s energy-mass equivalence: E=mc2E = m{c^2}

Complete Step by Step Answer:
Every object in this universe has energy within them. According to Albert Einstein, this energy possessed by the matter is equivalent to the product of mass by a constant equal to the square of the speed of light in a vacuum i.e. c2{c^2}.
E=mc2E = m{c^2}
Annihilation means the complete destruction of this matter wherein the entire mass of the object is completely converted to its equivalent energy as given in the above equation. Thus, the matter does not even leave a trace and completely, disappears by converting all of it into the invisible energy.
Anti-proton is described as a nucleon that has the opposite charge as that of a proton but the same mass. So, when the proton and antiproton bombard, their masses add up in the sense, the total mass is equal to twice that of the proton itself.
m=2×mpm = 2 \times {m_p}
Mass of proton in atomic mass units, mp=1amu=1.67×10−27kg{m_p} = 1amu = 1.67 \times {10^{ - 27}}kg
⇒m=2mp=2×1.67×10−27kg=3.34×10−27kg\Rightarrow m = 2{m_p} = 2 \times 1.67 \times {10^{ - 27}}kg = 3.34 \times {10^{ - 27}}kg
When the proton and antiproton add up, the combined particles annihilate by converting entire mass into energy by Einstein’s mass-energy equivalence theory:
E=mc2  E = m{c^2} \\\
Substituting,
E=3.34×10−27×(3×108)2  E = 3.34 \times {10^{ - 27}} \times {\left( {3 \times {{10}^8}} \right)^2} \\\
Solving,
E=3.34×10−27×9×1016 ⇒E=30.06×10−27+16J ⇒E=30.06×10−11J ⇒E=3.006×10−10J≈3×10−10J  E = 3.34 \times {10^{ - 27}} \times 9 \times {10^{16}} \\\ \Rightarrow E = 30.06 \times {10^{ - 27 + 16}}J \\\ \Rightarrow E = 30.06 \times {10^{ - 11}}J \\\ \Rightarrow E = 3.006 \times {10^{ - 10}}J \approx 3 \times {10^{ - 10}}J \\\

∴\therefore Option (B) is the correct answer.

Note:
There is an alternative approach to solving this problem.
Energy released per 1 atomic mass unit when it is destroyed, Ea=931MeV{E_a} = 931MeV
By converting to joules, we have,
Ea=931×106×1.602×10−19J{E_a} = 931 \times {10^6} \times 1.602 \times {10^{ - 19}}J
=1.49×10−10J= 1.49 \times {10^{ - 10}}J
Since the mass of the proton is 1 a.m.u and there are 2 proton masses colliding here, the total atomic mass unit = 2 a.m.u
Therefore, the total energy released,
E=2×931MeV=2×1.49×10−10E = 2 \times 931MeV = 2 \times 1.49 \times {10^{ - 10}}
=2.98×10−10≈3×10−10J= 2.98 \times {10^{ - 10}} \approx 3 \times {10^{ - 10}}J