The answer lies in this diagram
It shows the binding energy per nucleon over the number of nucleons n in a nucleus.
As you can see, it has a maximum around Iron (n=56), which means that left from iron energy can be released by fusion. For elements heavier than iron you need to put in energy to fuse them together, or equivalantly that fission is energetically favorable.
That is why conditions like in a supernova are necessary to produce the heavier elements.
So, fusioning ${}^2H$ to ${}^4He$ releases energy, but the reverse fission process costs energy. There is no net gain, even in a perfect aparatus.
The important argument for this discussion is the Bethe Weizsäcker formula, which describes the binding energy of nuclei. I will try to give a cursory overview of the most important aspects.
Not only heavy elements show fission and fusion. All elements up to iron-56 (one of the nuclei with the highest binding energy per nucleon) can create energy in fusion (and do so in old stars, where there is a sequence of collapsing shells as the more and more heavy fusion products are undergoing further fusion reactions).
When expending energy you can also create heavier elements by fusion (for example, this is one method for creating ultra heavy elements; shoot heavy nuclei with a lot of kinetic energy on something like lead or gold targets).
For fission the discussion runs the other way around. As there is a maximum in the binding energy curve, you will gain energy when splitting heavy nuclei creating lighter ones if you are above the maximum (because the binding energy difference will be released). But with enough energy you can also fission light elements, one can even separate deuterons by irradiating it with intense gamma radiation of sufficient energy.
Why are neutrons a clever method for fission? Because they are electrically neutral and therefore do not have to cross a Coulomb barrier for reaching the nucleus. And furthermore they show nuclear force interactions, therefore releasing binding energy, when caught by a nucleus (this is the process when fissioning uranium-235, a neutron is caught, releasing binding energy, which causes the nucleus to vibrate, these vibrations cause a charge offset while weakening the nuclear forces due to reducing the connected area thus causing the fission).
Why do electrons not work well? Because the cross section for an electron to interact directly with the nucleus is ridiculously small. Even if it has sufficient energy, it will still be much more likely, that it is deflected by Coulomb interaction never even reaching the nucleus (yes even for attractive Coulomb interaction), than deposition relevant energy in the nucleus. But electrons are useful for mapping the charge distribution in the nucleus.
So electrons can cause fission, but very inefficiently.
And finally why radioactive elements? Well, we know why heavy elements show fission. And there is another fact: Heavy nuclei show a proton-neutron imbalance, as nuclear forces are very short range, when a nucleus gains protons at some point it will need more than one neutron per proton to remain stable (as the Coulomb interaction grows quadratically with charge, the nuclear binding energy just linearly with nucleon number). And this allows another fundamental interaction to act, the weak force. This force is, as the name suggests, very weak, so these processes are improbable, but once it is energetically favourable it can happen, that a neutron in the nucleus decays to a proton, so for not entirely unrelated reasons heavy nuclei tend to be at the same time radioactive and release energy on fission.
So why do radioactive products occur in fusion. Because the results of fusion will often be something like unstable excited states (as there is energy released by the fusion for light elements). Another reason is found in the shell model of the nucleus. Certain "magic numbers" of nucleons will bind in stable shells and thus create more stable cores. When building up fusion products, you will not always hit these sweet spots (and thus the product will be unstable), and there will also always be a proton surplus and will thus show $\beta^+$-decay.
One tile of the mosaic is still missing, for understanding critical fission reactions (and why we do them with neutrons). And that is, that usually some free neutrons are released when a nucleus splits, so in a fission reactor we will naturally have neutrons that drive the reaction, instead of having to use a particle accelerator to create an electron beam to make the fission happen. The relevant quantity is the number of neutrons released per fissioned nucleus. If it is smaller than one, the fission reaction will come to a halt (as each neutron can at most split one more nucleus). But if there are more than one, and even enough that loss processes (neutrons not causing a fission, neutrons flying out of the reactor, neutrons reacting with something other than the nucleus in question), then the rate of reaction will grow exponentially. If this process is not somehow limited, you will have a runaway nuclear reaction, leading to a nuclear explosion (or rather some not quite so unpleasant meltdown and dispersal of the material, an actual nuclear explosion requires delicate engineering). If you control this process (for example by absorbing the surplus neutrons with control rods), then you can have a steady state, critical fission reaction creating energy.
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Please note that water is composed by two hydrogens and one oxygen. The stable isotope of oxygen has atomic number $8$, but atomic mass $16$, i.e. it is composed out of eight protons and eight neutrons. Thus in normal water the ratio of hydrogen contribution to the over all in mass is $2/18$.
When deuterium takes the place of hydrogen we have heavy water, and that means that two more neutrons are added to the original mass balance, an effect again of $2/18$ extra mass. So the water is more dense but by this small factor.
In this article the advantages of heavy water in reactors are given.
heavy water seems to solve the problem:
The reason is that deuterium has already bound a neutron to the hydrogen atom, and thus fewer neutrons needed for criticality are absorbed.