# Question #1549b

Mar 24, 2017

We know that one atomic mass unit is defined as 1/12th the mass of an atom of Carbon-12 isotope.
As such $1 \text{ amu"=1.6606xx10^-27" kg}$
Velocity of light $c$ in vacuum is $= 2.9979 \times {10}^{8} {\text{ ms}}^{-} 1$

Using the Einstein's mass energy equivalence relation we get
$E = m {c}^{2}$
$E = 1.6606 \times {10}^{-} 27 \times {\left(2.9979 \times {10}^{8}\right)}^{2} \text{ J}$

We also know that $1 \text{ eV"=1" electronic charge in C"xx1 " V}$
$= 1.60218 \times {10}^{-} 19 \text{ J}$

Therefore, we get
$E = \frac{1.6606 \times {10}^{-} 27 \times {\left(2.9979 \times {10}^{8}\right)}^{2}}{1.60218 \times {10}^{-} 19} \text{ eV}$
$\implies E = 931.5 \text{ MeV}$, rounded to one decimal place