For an insight into the nuclear phenomenon it is essential to determine accurately the atomic mass of an element. The instrument used for this purpose is called mass spectrometer.
The unit in which atomic and nuclear mass are measured is called atomic mass unit (a.m.u.).
One atomic mass unit is defined as $\frac{1}{12}$th of the mass of an atom of ${}_6^{12}\mathrm{C}$ isotope.
As Avogadro number:
$$ N_A = 6.023 \times 10^{23} $$
Therefore:
$$ \text{Mass of } 6.023 \times 10^{23} \text{ atoms of } {}^{12}\mathrm{C} = 12\,\mathrm{g} $$
$$ \text{Mass of one atom of } {}^{12}\mathrm{C} = \frac{12}{6.023 \times 10^{23}}\,\mathrm{g} $$
By definition of $1\,\mathrm{a.m.u.}$:
$$ 1\,\mathrm{a.m.u.} = \frac{1}{12} \times \left(\frac{12}{6.023 \times 10^{23}}\right)\,\mathrm{g} $$
$$ \Rightarrow 1\,\mathrm{a.m.u.} = 1.66 \times 10^{-27}\,\mathrm{kg} $$
Clearly one a.m.u. represents the average mass of a nucleon and is represented by $u$.
In terms of this unit:
$$ \text{Mass of electron } (m_e) = 0.00055\,u $$
$$ \text{Mass of proton } (m_p) = 1.0073\,u $$