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Question: The energy equivalent of \(\mathrm{1\;amu}\) is: A. \(\mathrm{931\;eV}\) B. \(\mathrm{93.1\;eV}\...

The energy equivalent of 1  amu\mathrm{1\;amu} is:
A. 931  eV\mathrm{931\;eV}
B. 93.1  eV\mathrm{93.1\;eV}
C. 931  MeV\mathrm{931\;MeV}
D. 9.31  MeV\mathrm{9.31\;MeV}

Explanation

Solution

1  amu\mathrm{1\;amu} stands for atomic mass unit. It is a very small unit of mass used at atomic levels. Basically, 1  amu=1.66×10−27  kg\mathrm{1\;amu}=\mathrm{1.66\times10^{-27}\;kg}, which is roughly the mass of a proton. Use Einstein mass-energy equivalence to convert the mass unit to equivalent energy, which is given as E=mc2E=mc^2, where mm is the mass and cc is speed of light. To convert energy in joules to electron volts (eV)(eV), divide by 1.6×10−19\mathrm{1.6\times10^{-19}}.

Complete step by step answer:
The given value 1  amu\mathrm{1\;amu} is an atomic mass unit, which is used to refer to infinitesimal masses of subatomic particles like electron, proton and neutron. The most common unit kilogram   kg\mathrm{\;kg}, becomes too absurd to be used at atomic levels. Let us convert the given mass to SI unit kg\mathrm{kg}

1  amu=1.66×10−27⋅1  kg=1.66×10−27  kg\begin{aligned} \mathrm{1\;amu}=\mathrm{1.66\times10^{-27}\cdot1\;kg}=\mathrm{1.66\times10^{-27}\;kg}\end{aligned}
Now, let us use Einstein’s equation for energy mass equivalence, which is given as:

E=mc2E=mc^2
Substitute mass(m)=1.66×10−27  kg(m)=\mathrm{1.66\times10^{-27}\;kg} and speed of light(c)=3.0×108  m/s(c) =\mathrm{3.0\times10^{8}\;m/s} and solve for the energy EE.
E=1.66×10−27  kg⋅(3.0×108  m/s)2&=1.66×10−27  kg⋅(9.0×1016  m2/s2)&=14.94×10−11  J\begin{aligned}E&=\mathrm{1.66\times10^{-27}\;kg}\cdot\left ( \mathrm{3.0\times10^{8}\;m/s }\right )^2\\\&=\mathrm{1.66\times10^{-27}\;kg}\cdot\left ( \mathrm{9.0\times10^{16}\;m^{2}/s^{2} }\right )\\\&=\mathrm{14.94\times10^{-11}\;J}\end{aligned}

Now, to convert the energy to eV\mathrm{eV}, let us divide by 1.6×10−19\mathrm{1.6\times10^{-19}}.
E(eV)=14.94×10−11  J1.6×10−19&=9.33×108  eV&≈931×106  eV&=931  MeV\begin{aligned}E\mathrm{(eV)}&=\dfrac{\mathrm{14.94\times10^{-11}\;J}}{\mathrm{1.6\times10^{-19}}}\\\&=\mathrm{9.33\times10^{8}\;eV}\\\&\approx\mathrm{931\times10^{6}\;eV}\\\&=\mathrm{931\;MeV}\end{aligned}
Hence, the correct answer is option C.

Note:
1.) The student must be careful in reading the question and understanding the concept. amu\mathrm{amu} can be mistaken for a unit of energy, which it is not. Also, this question is related to the equation of energy mass equivalence, which can be tricky to guess.
2.) In the question above and other related questions, unit conversions play a major role. So, it's essential for the student to familiarize with some commonly used units of mass and energy.

Units of mass: SI unit is Kilogram(kg)(\mathrm{kg})
1  amu=1.66×10−27  kg\mathrm{1\;amu}=\mathrm{1.66\times10^{-27}\;kg}
1  mg=1×10−6  kg\mathrm{1\;mg}=\mathrm{1\times10^{-6}\;kg}
1  g=1×10−3  kg\mathrm{1\;g}=\mathrm{1\times10^{-3}\;kg}
1  quintal=100  kg\mathrm{1\;quintal}=\mathrm{100\;kg}
1  lb=0.4536  kg\mathrm{1\;lb}=\mathrm{0.4536\;kg}

Units of Energy: SI unit is Joule(J)(\mathrm{J}). Other units are:
1  eV=1.6×10−19  J\mathrm{1\;eV}=\mathrm{1.6\times10^{-19}\;J}
1  erg=1×10−7  J\mathrm{1\;erg}=\mathrm{1\times10^{-7}\;J}
1  cal=4.18  J\mathrm{1\;cal}=\mathrm{4.18\;J}
1  MJ=1×106  J\mathrm{1\;MJ}=\mathrm{1\times10^{6}\;J}
1  kWh=3.6×106  J\mathrm{1\;kWh}=\mathrm{3.6\times10^{6}\;J}