This is really a comment, since I don't think there is an answer to your question, but it got a bit long to put in as a comment.
If you Google for "Why is technetium unstable" you'll find the question has been asked many times in different forums, but I've never seen a satisfactory answer. The problem is that nuclear structure is much more complex than electronic structure and there are few simple rules.
Actually the question isn't really "why is technetium unstable", but rather "why is technetium less stable than molybdenum and ruthenium", those being the major decay products. Presumably given enough computer time you could calculate the energies of these three nuclei, though whether that would really answer the "why" question is debatable.
Response to comment:
The two common (relatively) simple models of the nucleus are the liquid drop and the shell models. There is a reasonably basic description of the shell model here, and of the liquid drop model here (there's no special significance to this site other than after much Googling it seemed to give the best descriptions).
However if you look at the sction of this web site on beta decay, at the end of paragraph 14.19.2 you'll find the statement:
Because the theoretical stable line slopes towards the right in figure 14.49, only one of the two odd-even isotopes next to technetium-98 should be unstable, and the same for the ones next to promethium-146. However, the energy liberated in the decay of these odd-even nuclei is only a few hundred keV in each case, far below the level for which the von Weizsäcker formula is anywhere meaningful. For technetium and promethium, neither neighboring isotope is stable. This is a qualitative failure of the von Weizsäcker model. But it is rare; it happens only for these two out of the lowest 82 elements.
So these models fail to explain why no isotopes of Tc are stable, even though they generally work pretty well. This just shows how hard the problem is.
The half-life $t_{1/2}$ is defined so that the amount of stuff you have $N(t)$ at some time $t$ is
$$
N(t) = N_0 \cdot 2^{-t/t_{1/2}}
$$
where $N_0$ is how much stuff you start with and you're familiar with the constant $2$.
This definition has several advantages. First, it's easy to explain, since scientists of all ages are familiar with the constant $2$. Second, it's unambiguous. If you asked about a "quarter-life," do you mean when a quarter of your original material remains $t_{1/4}$, which is after two half-lives, or when a quarter of your original material has decayed? The second option is not $\frac12 t_{1/2}$, because the decay is exponential and not linear; it's $t_{3/4} = t_{1/2}\log_2\frac43$, ugh.
"Full life" is not an option because the number of remaining decay-ers approaches zero only asymptotically.
When you start to do calculus, the same sort $\log_2$ ugliness comes up again.
Another, probably nicer, way to write the decay equation is
$$
N(t) = N_0 e^{-t/\tau}
$$
where $e = \frac1{0!} + \frac1{1!} + \frac1{2!} + \cdots \approx 2.7183$ is the base of the natural logarithms. The constant $\tau$ is referred to by several names, depending on your mood and the experience level of the person you're communicating with: sometimes $\lambda \equiv 1/\tau$ is called the "decay constant," and sometimes $\tau$ itself is referred to as the "$e$-folding time" or simply "the lifetime."
A nice feature of the exponential form is that, if you assume that every decay is detectable (or that a known fraction of the decays are detectable) then you can find the measured activity of your source:
$$
A(t) = -\frac{dN}{dt} = \frac{N_0}{\tau}e^{-t/\tau}
$$
This means that if you can measure what your source's activity is, and measure long enough to watch it change, you have effectively weighed the radioactive part of the source by fixing $N_0$, even if $N_0$ may be only a few millions of atoms.
Best Answer
The Iodine-131 undergoes beta decay. This is what makes it radioactive. The faster it decays the more radiation it gives off per unit time. Yes the material disappears faster than something with a longer half-life, but its speed of decay means that it is hotter.