Why do we take difference of masses of nucleus in mass defect rather than the masses of the complete atom including the electrons? For instance, in this lecture, the Q value for beta decay has been calculated by taking the difference of masses of nuclei which is (Mass of atom)-(Mass of electrons).
The only reason to ignore electron masses when computing $Q$-values is to introduce MeV-scale errors.
Read your source material carefully, to make sure you aren't double-counting electron masses either. I prefer the Nuclear Wallet Cards, which define (emphasis mine):
Mass
The mass of a nucleus with Z proton and N neutrons in a neutral-atom state is: $$ \text{Mass}(Z,N)=Z*\text{Mass(hydrogen atom)}+N*\text{Mass(free neutron)} - BE(Z,N) $$ where BE(Z,N) is the Binding Energy, that is, the energy needed to dissociate the nucleus into free nucleons. Note in this product, masses are given in energy units.The atomic mass unit (amu) is defined so that 1 amu is equal to the mass of a 12C atom divided by 12.
Mass excess (Δ)
The mass excess $Δ(Z,N)$ is defined as: $$ Δ(Z,N)=( \text{Mass}(Z,N) (\text{amu}) - A ) × \text{amu} $$where Mass(Z,N) (amu) is mass in atomic mass units and amu=Mass(6,6)/12. The mass excess as well as many other related quantities can be obtained using QCalc
That is, in the the various NNDC databases, the tabulated masses are for neutral atoms, rather than for bare nuclei. There are several reasons for this choice:
In general, it's much easier to measure the mass of a neutral atom than to produce an ensemble of completely-ionized nuclei and measure their masses. The neutral atom data are more complete and more reliable.
For heavy nuclei with many electrons, the electron binding energies make up a non-negligible correction to the atomic mass excess. Computing the nuclear mass is not just a matter of subtracting the electron masses.
On this second point, consider the K-edge x-rays as a direct measurement of the innermost electron's binding energy: in heavy atoms, that's up to 100 keV. Counting up the electrons in filled K-, L-, and M-shells, you can easily get up to an MeV. The scale for mass excesses across the periodic table is a few tens of MeV.
Furthermore, if you're looking at beta decays between adjacent nuclei, you are generally looking at the decays of neutral atoms to nearly-neutral atoms: that energetic internal electron structure is substantially unchanged by the decay. (The jargon is that most of the electrons are "spectators.") Computing a bare nuclear mass with the accuracy needed for computing $Q$-values is not only a challenge, it's also usually irrelevant.
So if you are interested in the weak decays among mass-7 nuclei, you would construct the table
| isotope | mass excess (MeV) | half-life |
|---|---|---|
| lithium-7 | $14.9071$ | stable |
| beryllium-7 | $15.7689$ | 53 days |
and conclude that the electron-capture decay of neutral beryllium
$$ \begin{align} \rm ^7Be &\to \rm{}^7Li + \nu_e \\ \rm\left(^7{Be}^{4+} + 4e^-\right)_\text{ground state} &\to \rm\left(^7{Li}^{3+} + 3e^-\right)_\text{ground state} + \nu_e \end{align} $$
has $Q$-value $15.8 - 14.9 = 0.9\rm\,MeV$. The stoichiometry is that the initial and final states each have seven baryons and four leptons, but the weak interaction transfers one unit of charge between the lepton sector and the baryon sector.
However the "equivalent" decay by positron emission,
\begin{align} \rm ^7Be &\to \rm{}^7{Li}^- + e^+ + \nu_e \\ \rm\left(^7{Be}^{4+} + 4e^-\right)_\text{ground state} &\to \rm{}\left(^7Li^{3+} + 4e^-\right)_\text{ground state} + e^+ + \nu_e \end{align}
is forbidden, because the final state has an electron-positron pair whose additional mass makes the overall $Q$-value negative. (This particular example has the interesting physical consequence that $\rm^7Be$ does not occur on Earth in neutral matter, but its bare nuclei $\rm^7{Be}^{4+}$ are stable components of galactic cosmic rays.)
Likewise in the mass-3 system,
| isotope | mass excess (MeV) | half-life |
|---|---|---|
| hydrogen-3 | $14.9498$ | 12.32 years |
| helium-3 | $14.9312$ | stable |
the very small $Q$-value $\rm 18.6\,keV$ for the decay
$$ \rm \left(^3H^+ + e^-\right)_\text{ground state} \to \rm \left(^3{He}^{2+} + e^- \right)_\text{ground state} + e^- + \bar\nu_e $$
is obtained by comparing the mass excesses for the neutral atoms.
