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This question already has an answer here:

The masses of electron, proton and neutron (in SI units) are (approx.):

\begin{equation} m_e=\text{electron mass}=9.109\times 10^{-31}\ \text{kg} \end{equation} \begin{equation} m_p=\text{proton mass}=1.673\times 10^{-27}\ \text{kg} \end{equation} \begin{equation} m_n=\text{neutron mass}=1.675\times 10^{-27}\ \text{kg} \end{equation}

The unified atomic mass unit ($u$) is defined as one twelfth of the mass of Carbon-12.

\begin{equation} u=1.661\times 10^{-27}\ \text{kg} \end{equation}

Now, Carbon-12 contains 6 electrons, 6 protons and 6 neutrons, so the mass of this isotope should be:

\begin{equation} m=\text{Carbon-12 mass}=6\times(m_e + m_p + m_n)=6\times 3.3489 \times 10^{-27}\ \text{kg} \end{equation}

Then:

\begin{equation} \frac{m}{12}=0.5\times 3.3489 \times 10^{-27}\ \text{kg}=1.674 \times 10^{-27}\ \text{kg} \end{equation}

But this value is different from the value of $u$. Shouldn't $m/12$ be equal to $u$?

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marked as duplicate by Emilio Pisanty, Community May 20 '17 at 20:34

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No.

The difference in that calculation is due to the nuclear binding energy of the carbon nucleus, which affects the mass via the good old $E=mc^2$. In short, pretty much all nuclei are a bit lighter than their constituent parts would be, which reflects the fact that you would need to put in energy (a.k.a. mass) to split them up into said constituent parts.

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