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 (usually<sup>1</sup>) 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:

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

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

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[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).