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Question: How to express the mass of proton and an electron in \(MeV\)/\({c^2}\)?...

How to express the mass of proton and an electron in MeVMeV/c2{c^2}?

Explanation

Solution

Hint : In order to compute the masses of proton and electron in the given quantity, it is important to understand the Einstein’s equation of mass-energy equivalency, which is given as:
E=mc2E = m{c^2}
where c = speed of light in vacuum in ms−1m{s^{ - 1}}.

Complete step-by-step answer:
The mass of the nucleus formed by the combination of the nucleons like protons and neutrons is slightly lesser than the sum of individual masses of the nucleons. This difference in mass is called mass defect. If one wonders why there is a decrease in mass, it is because when the nucleons are bound, they lose some energy to bind to each other. This lost energy corresponds to the decrease in mass accounted for, due to Einstein's mass-energy equation.
The energy corresponding to the mass defect is called binding energy per nucleon and the value is equal to 931 MeV.
Binding energy is the energy required to separate the nucleons from each other in the nucleus of the atom. The formula for binding energy is based on the Einstein’s Energy-mass equivalence formula:
E=mc2E = m{c^2}
Thus, binding energy per nucleon is equal to 931 MeV.
The masses of nucleons such as protons and neutrons, are expressed in terms of atomic mass units.
One atomic mass unit is equal to one-twelfth of the mass of an atom of the carbon C-12 isotope. This amu expressed in grams is given as:
1amu=1⋅67×10−27g1amu = 1 \cdot 67 \times {10^{ - 27}}g
When the mass in the above equation of Einstein’s energy-mass equivalence is expressed in amu, the result is the energy in MeV. Therefore,
931MeV=1amu×c2931MeV = 1amu \times {c^2}
⇒1amu=931×MeVc2\Rightarrow 1amu = 931 \times \dfrac{{MeV}}{{{c^2}}}
Therefore,
The mass of proton in amu = 1.007276 u
Expressing in the units, MeVc2\dfrac{{MeV}}{{{c^2}}}, we get –
931×1⋅007276=937⋅77MeVc2931 \times 1 \cdot 007276 = 937 \cdot 77\dfrac{{MeV}}{{{c^2}}}
The mass of electron in amu = 0.00054858 u
Expressing in the units, MeVc2\dfrac{{MeV}}{{{c^2}}}, we get –
931×0⋅00054858=0⋅5107MeVc2931 \times 0 \cdot 00054858 = 0 \cdot 5107\dfrac{{MeV}}{{{c^2}}}

Note: The energy units for microscopic units are expressed in terms of electron-volt (eVeV).
One electron-volt is equal to the energy gained by an electron when it is accelerated inside an electric field whose potential difference is equal to 1 volt.
The conversion from electron-volt to joule is as follows:
1eV=1.602×10−19J1eV = 1.602 \times {10^{ - 19}}J