# How do you calculate the mass defect for C-14? How do you calculate the binding energy?

Apr 11, 2018

Well, you take the theoretical mass of combining all the particles, and subtract from it the actual isotopic mass. That mass accounts for the binding energy.

You'll need:

• ${m}_{p} = \text{1.00727647 amu/proton}$
• ${m}_{n} = \text{1.00866492 amu/neutron}$
• ${m}_{e} = 5.48530 \times {10}^{- 4} \text{amu/electron}$
• m_(""^(14) "C") = "14.003241989 amu/atom"
• There are $\left(\text{1.60217662 J")/("1 eV}\right)$.
• The speed of light is $2.99792458 \times {10}^{8} \text{m/s}$.
• Avogadro's number is $6.0221413 \times {10}^{23} {\text{mol}}^{- 1}$.

We know that $\text{_6^(14) "C}$ has $6$ protons and $8$ neutrons (and $6$ electrons), so the theoretical mass (that is, if no mass became released energy due to forming the atom) is:

$\text{6 protons" cdot "1.00727647 amu/proton" + "6 electrons" cdot 5.48530 xx 10^(-4) "amu/electron" + "8 neutrons" cdot "1.00866492 amu/neutron}$

$=$ $\text{14.11626936 amu/atom}$

The mass defect is then

$\textcolor{b l u e}{{m}_{\text{defect") = m_"theoretical" - m_(""^(14) "C}}}$

$= \text{14.11626936 amu/atom" - "14.003241989 amu/atom}$

$=$ $\textcolor{b l u e}{\text{0.11302737 amu}}$

$\text{amu}$ is numerically the same as $\text{g/mol}$, so we can then get to the binding energy. This binding energy is found as follows...

The mass that becomes energy is:

$m = \frac{0.11302737 \cancel{\text{g")/cancel"mol" xx "1 kg"/(1000 cancel"g") xx (cancel"1 mol}}}{6.0221413 \times {10}^{23}}$

$= 1.87686348 \times {10}^{- 28} \text{kg}$

So, the binding energy involved in $\text{J}$ per atom is:

${E}_{\text{binding}} = m {c}^{2}$

$= {\left(1.87686348 \times {10}^{- 28} \text{kg")(2.99792458xx10^(8) "m/s}\right)}^{2}$

$= {1.68680000}^{- 11} \text{J/atom}$

We may want it in $\text{kJ/mol}$ and $\text{MeV/nucleon}$.

$\textcolor{b l u e}{{E}_{\text{binding") = (1.68680000 xx 10^(-11) cancel"J")/cancel"atom" xx "1 kJ"/(1000 cancel"J") xx (6.0221413 xx 10^23 cancel"atoms")/("1 mol}}}$

$= \textcolor{b l u e}{1.015815 \times {10}^{10} \text{kJ/mol}}$

$\textcolor{b l u e}{{E}_{\text{binding") = (1.68680000 xx 10^(-11) cancel"J")/cancel"atom" xx cancel"1 eV"/(1.60217662 xx 10^(-19) cancel"J") xx ("1 MeV")/(10^6 cancel"eV") xx cancel"1 C-14 atom"/("14 nucleons}}}$

$=$ $\textcolor{b l u e}{\text{7.520127 MeV/nucleon}}$

The actual value is $\text{7.520319 MeV/nucleon}$.