For a given isotope, one can obtain the binding energy using the semi-empirical mass formula.
For example, $^{237}_{\ \ 93}\rm Np$ has a binding energy of 1782.8 MeV. From this information, how can the likelihood of the isotopes stability be found?
Nuclear Physics – How to Obtain Isotope Stability: A Comprehensive Guide
binding-energyisotopesnuclear-physicsstability
Related Solutions
When people say that the decay rate depends critically on the $Q$ value, they're talking about alpha decays compared to other alpha decays. When you compare alpha decay to emission of other small clusters, the dependence on the atomic number $Z_c$ of the emitted cluster is much more prominent. The reason is as follows.
In the Gamow model of beta decay, we assume that the decay rate is the product of three factors: (1) a hand-wavy probability of preformation of the cluster; (2) the frequency with which a cluster assaults the Coulomb barrier; and (3) the probability of transmission through the barrier. (Re #1, don't take Ohanian too seriously when he says this factor is 0.1 to 1. Actually the literal existence of any cluster bouncing around inside an atomic nucleus would violate the exclusion principle. The whole thing is just a model.)
The critical factor is the tunneling probability $P$, which can be estimated using the WKB approximation, which looks like $\exp(-\int\ldots)$, where the integral is over the classically forbidden region. The integral depends on the Q value, because a higher Q value both shrinks the classically forbidden region and reduces the value of the integrand within that region. However, the height of the Coulomb barrier is proportional to the product $Z_cZ_d$ of the atomic numbers of the cluster and the daughter nucleus. If you want all the gory details, you can google the Geiger-Nuttall equation. But the result turns out to be of the form $\ln P=a-b$, where the $Z$-dependence is dominated by the term $a=(1/\hbar)\sqrt{32Z_cZ_d m_c R ke^2}$. For the alpha decay of uranium, we have $a\approx 74$. In Yval's example, decay by emission of 9Be basically doubles the value of $a$, which reduces the decay rate by a factor of $e^{-74}\approx10^{-32}$.
In the question, Yuval estimated that emission of Be should only be down by a factor of $e^{-2}$ relative to emission of alphas. This was an algebra mistake. We have an expression of the form $e^{-Zu}$, where $u$ is a constant. Changing this expression from $e^{-2u}$ to $e^{-4u}$ doesn't just reduce its value by a factor of $e^{-2}$, it reduces it by $e^{-2u}$, which is a huge factor.
Actually, as pointed out by JoeHobbit, the real mystery isn't why we don't emit larger clusters, it's why we don't emit lighter objects like protons or deuterons. A proton doesn't have to worry about preformation, and its tunneling probability would be much higher. Possibly this is due to the lower $Q$ value for proton emission. This comes into the Geiger-Nuttall equation because $b\propto Z_cZ_d/\sqrt{Q}$. In fact proton emission does occur, but it's only competitive for extremely proton-rich nuclei. There is also neutron emission, which doesn't involve any Coulomb barrier at all; as you'd expect, its half-life is very short (on the order of the assault frequency) when it's energetically allowed.
The semi-empirical mass formula is derived from a fit to the measured masses. If you know the numbers of protons and neutrons then the idea is that the SEMF should give you a good estimate of that mass (there are of course small residuals of the order 0.2 MeV to the fit).
$$M(A,Z)c^2 = (A-Z)m_n c^2 + Zm_pc^2 - AE_b,$$ where $A,Z$ are the mass number and atomic number and $E_b$ is the binding energy per nucleon that is described by the SEMF. So of course if you can measure the mass $M$, then you could easily rearrange this formula to give $E_b$. However, the beauty of the SEMF is it gives a very simple way to attempt to understand what is happening to the binding energy as you change the number of nucleons and the neutron/proton ratio. It also allows you to predict the properties of exotic or short-lived nuclei (for example in the crusts of neutron stars), where you may not have the luxury of a laboratory-based measurement.
In direct response to your question. Yes you can do this and the answers should be very similar. But there are small differences because the SEMF is just a model, and a relatively simple one at that. For instance if you were thinking of the Weizsacker liquid-drop version, then this doesn't predict the larger binding energies of the "magic number" nuclei, that are better describe by the shell model.
Best Answer
From the binding energy given experimentally, using precise QM calculations or using a given formula, one should first check for "stability in particles", if the binding is negative, you will of course not have stability.
Then the next thing, if you have a formula, is to check for each type of stability. For example, to check for stability against a given fission, calculate the binding energy of the fragments, obtain the new energy and compare.
To check for, let's say a beta minus, replace the nucleus $(A,Z)$ in the formula by $(A,Z+1)$ , obtain the binding energy, the new total energy (you can safely neglect the mass of the neutrino and even the one of the electron in most cases) and compare.
For your specific example, this is a bit tricky because a large variety of decay channels are potentially allowed.
Edit
Another way to proceed is to look at a binding energy per nucleon or mass excess against A diagram:
We can see that 237Np is far on the right as a binding energy per nucleon smaller that the most stable elements like 56Fe .
One can then conclude that 237Np can potentially decay to a more stable element to increase it's binding energy (although these decay, that can be alpha decay can have excessively small probability and a time constant excessively long).