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?
[Physics] Should the expression for energy conservation in alpha decay include the mass of electrons
energy-conservationnuclear-physics
Related Solutions
Alpha decay is treated as a special case and is usually not included in the term "fission", although it arguably is a fission (especially in the case of light elements where the alpha can represent an appreciable fraction of the original mass).
So, for the purposes of a quiz you want "fission".
This example of calculating an alpha lifetime using a tunneling nuclear model may help.
Please note that the alpha particle after decay has the energy it had in the bound energy level.
the incredible range of alpha decay half-lives can be modeled with quantum mechanical tunneling. The illustration represents the barrier faced by an alpha particle in polonium-212, which emits an 8.78 MeV alpha particle with a half-life of 0.3 microseconds.
Momentum conservation will give the daughter nucleus a momentum .
The daughter nucleus might be in an excited state and then there would be a gamma decay to a lower stable level.
The binding energy curve rises up to iron and then decreases, so your statement " D binding energy per nucleon always increases" is true only if the daughter nucleus has a higher atomic number than iron.
This handbook goes into the energy details in page 4.
Best Answer
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:
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.