The following has been derived from the wiki page "Extended periodic table" and is one of the many examples d or f block elements being used in synthesizing new heavier elements (or at least this is the pattern I figured out by searching many other examples of the nuclear reactants in the way to synthesize heavy elements):

Attempts to synthesize still undiscovered elements:

\begin{align} \ce{^254_99Es + ^48_20Ca &-> ^302_119Uue^*} \\ \ce{^249_97Bk + ^50_22Ti &-> ^295_119Uue + 4 ^1_0n} \\ \ce{^{nat}_68Er + ^136_54Xe &-> ^{298,300,302,303,304,306}Ubb^* -> no atoms} \\ \ce{^238_92U + ^66_30Zn &-> ^304_122Ubb^* -> no atoms} \end{align}

The question is, why is it like this? How did the scientists reached an agreement on which elements to collide? Was it anything especial about those atoms?

P. S. If you are sure that there is an exception (exceptions omit rules; at least in this case), your answer will be accepted.

  • $\begingroup$ This is for Physics! $\endgroup$
    – Georg
    Commented Dec 21, 2014 at 22:41
  • 1
    $\begingroup$ Something special was asked about the atoms. If the answer is a property of an atom, it must be chemistry. $\endgroup$
    – M.A.R.
    Commented Dec 22, 2014 at 12:56

2 Answers 2


There are different strategies for synthesizing superheavy elements.

Two of the main strategies are referred to as "hot fusion" and "cold fusion".

The first two reactions are good examples of the "hot fusion" strategy. I will focus on these two reactions, as the other two are much older and I don't think they are good examples of current thinking.

Experimentally, a very heavy element target (the Es or Bk in the first two reactions) is bombarded by an ion beam of the second element. The target element is choosen on the basis of being very heavy, availible and not rapidly decaying.

$^{48}_{20}\ce{Ca}$ is choosen for a very special reason. This is not ordinary calcium. The most common isotope of calcium has only 20 neutrons, but this isotope has 28!. It is extremely neutron rich, espeically compare to other relatively stable isotopes in this mass-range of the periodic table. $^{50}_{22}\ce{Ti}$ is also very neutron rich.

Why are neutron rich nuclei chosen? There are two reasons for this:

First, confining many protons in a small space causes electrostatic repulsion. As atomic number increases, the stable nuclei deviate further and further from a 1:1 neutron:proton ratio, because having more neutrons keeps the protons further apart and decreases electrostatic repulsion energy.

Secondly, you will notice the asterisk (*) in the 1st, 3rd and 4th equations. This denotes the compound nucleus which initially forms in the fusion reaction. It is an extremely ($<10^{-21}$s) short lived intermediate state, not the ground state of a nucleus, but more like a highly excited state. It's formation is not considered creating a new element. It is critical to the strategy of forming an element that the intermediate state loss energy through neutron "evaporation". Notice that in the second equation, the step of the neutrons evaporating is shown, while the intermediate state is not shown. Generally speaking, in the hot fusion strategy, 3-5 neutrons will evaporate to cool the intermediate state.

For further reading see:

Future of superheavy element research: Which nuclei could be synthesized within the next few years? and

The discovery of the heaviest elements Rev. Mod. Phys. 72, 733


The stability of heavy nuclei is a function of both atomic number and atomic mass. Therefore, an appropriate number both of protons and of neutrons in the target nucleus must be attained, or else vanishingly short lifetimes can be expected. (See the Wikipedia article on, e.g., the island of stability, and links therein.) The 'source' nuclei are chosen to provide proton and neutron counts that are anticipated to yield heavy nuclei with sufficiently long lifetime so as to be measurable.

Of course, it's not as simple as just choosing the right total number of protons and neutrons in your starting nuclei, as you're not guaranteed that all of the neutrons and protons will stay put in the synthesized nucleus, nor that you won't have rapid radioactive emissions that knock you away from your desired composition. For example, in the various syntheses of dubnium reported at Wikipedia, various numbers of neutrons were ejected as part of the nuclear synthesis 'reaction'.

  • $\begingroup$ This answer is informative, but unfortunately is not the main answer for the question. Yet, thanks for sharing your knowledge. $\endgroup$
    – M.A.R.
    Commented Dec 27, 2014 at 20:25
  • $\begingroup$ Can you provide further clarification as to the type of answer you're seeking? It's unclear what other sorts of considerations to which you may be referring. (I assume you're not looking for 'frivolous' reasons, such as "they like their names" or somesuch.) $\endgroup$
    – hBy2Py
    Commented Dec 27, 2014 at 21:15
  • $\begingroup$ :D Of course not. I just asked for an expansion of your 2nd paragraph, since it's what's been mainly asked. More references, more examples, more description. You agree that for this type of question, a paragraph of answer is not really enough. $\endgroup$
    – M.A.R.
    Commented Dec 27, 2014 at 21:20

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