Question

Find the energy equivalent of one atomic mass unit. First in Joules and then in MeV. Using this, express the mass defect of ${}_{8}^{16}O$ in $MeV/{{c}^{2}}$.

Answer 1

Hint: We will use the mass energy equivalence relation given by Einstein to calculate how much energy is equivalent to one atomic mass unit and convert it into MeV. After that we will calculate the mass defect for an oxygen atom and convert it in $MeV/{{c}^{2}}$.

Formula used: Mass energy equivalency:
$E=m{{c}^{2}}$

Complete step by step answer:
As we need to use the SI units, we will first need to convert one atomic mass unit in SI units that is kilograms. When an atomic mass unit is multiplied by the Avogadro’s number it gets converted to grams. So, one atomic mass unit = $\dfrac{1}{6.022\times {{10}^{23}}}$grams = $1.66\times {{10}^{-24}}$grams = $1.66\times {{10}^{-27}}$kilograms. Now using the mass energy equivalency,
$E=1.66\times {{10}^{-27}}{{(3\times {{10}^{8}})}^{2}}=1.494\times {{10}^{-10}}$Joules. Now we have to convert this into MeV. $1MeV=1.6\times {{10}^{-13}}$Joules. The energy of one atomic mass unit converted into MeV will be $\dfrac{1.494\times {{10}^{-10}}}{1.6\times {{10}^{-13}}}=933.75$MeV.
Now we will find the mass defect for oxygen. Oxygen atom nucleus has 8 protons and 8 neutrons. Mass of one proton is 1.00727647 amu and the mass of one neutron is 1.008665 amu. The mass of the oxygen nucleus is 15.9949 amu.
The total mass of constituents is $8\times (1.00727647)+8\times (1.008665)=16.1275$amu. The mass defect will then be $16.1275-15.9949=0.1326$amu. In MeV that will be $0.1326\times 933.75=123.815$MeV. When converted into $MeV/{{c}^{2}}$, it will be $\dfrac{123.815}{{{(3\times {{10}^{8}})}^{2}}}=1.375\times {{10}^{-15}}$. The mass defect will be $1.375\times {{10}^{-15}}$$MeV/{{c}^{2}}$.

Note: The calculations done here, and the values taken here are approximate values. This is done because the values here are either extremely small or extremely large, so approximations are taken. When doing these calculations although the values taken are approximate, it is good practice to take upto 2 decimal places in scientific notations from them.