Q:

Should the expression for energy conservation in alpha decay include the mass of electrons?

Let $M_P$, $M_d$, $m_\alpha$, $m_e$ and $Q$ the mass of the parent nucleus, daughter nucleus, alpha particle, electron and the disintegration energy, respectively. I understand that applying conservation of energy in alpha decay we get that
$$M_{P}c^2=M_{d}c^2+m_{\alpha}c^2+Q \tag{1}$$
where $Q$ is the sum of the kinetic energy of the alpha particle plus the kinetic energy of the daughter nucleus.
But reading a book written by my teacher, he says that the energy conservation law carry us to the equation
$$M_{P}c^2=M_{d}c^2+m_{\alpha}c^2+2m_{e}c^2+Q \tag{2}$$
My question is, which equation is the correct? And if the answer is (2), how do I derive it?

A:

In an alpha decay no electrons are created or destroyed. There is a small correction needed for the Coulomb term when the alpha escapes without carrying two electrons with it, but that is at chemical, not nuclear energy scales and is (usually1) sorted out by chemical means in fairly short time scales.
So, no you do not figure the mass of any electrons into the energy balance equation (or equivalently, you have the same number before and after).
That said, as written you are treating the mass of atoms on one hand (the parent and the daughter) and the mass of a bare alpha particle on the other. The distinction is that the mass of the daughter does not include the two "spare" electrons. That represents a inconsistency in how the work is done.
Correct ways to do this would include:

To use the bare nuclear masses for the parent and the daughter as well as for the alpha. No electrons needed.
To use the mass of He-4 instead of $m_\alpha$.2
Use the mass of the -2 state of the daughter ion (probably the best way).
Include the two spare electrons explicitly as your teacher has done.

Given the nature of the date tabulations that are easy to find, number 2 and 4 are the "easy" choices.

Now, you've asked how you "derive" this. Well, it's just an expression of the conservation of energy, so you add up the energy contributions in the initial state and those in the final state, and set them equal to one another.

An interesting side note here is that electrons usual make up about 1 part in 4000 or less of a atom's mass and can be neglected for most purposes, but what we have here is a subtraction of nearly equal values, so precision counts.

[1] There are some situations in which ionization can remain for "long" time scales (many milliseconds at least), but these conditions generally require considerable effort to create in the laboratory (low impurity environments and the like).
[2] 4.002602 u rather than 4.001506 u.

