0
$\begingroup$

Does any pair of isotopes exist which are in different states at room temperature—either in isolation or as part of an otherwise-identical compound?

$\endgroup$
  • $\begingroup$ Please Elaborate. $\endgroup$ – A.K. Sep 28 '16 at 2:19
  • $\begingroup$ I'll try ... two isotopes of some element that exist in different states of matter from each other at room temperature--all other variables being equal as well? $\endgroup$ – Richard Simões Sep 28 '16 at 2:27
  • $\begingroup$ Typically no, but processes such as the Gridler-Sulfide or COLEX process can greatly enrich an isotope. $\endgroup$ – A.K. Sep 28 '16 at 2:41
  • $\begingroup$ Ordinary water and heavy water differ in boiling point by $1.4^\circ\,\rm C$. Guess $\ce{HF}$ and $\ce{DF}$ might be the very example you are after (not that it would be of any use for anything, though). $\endgroup$ – Ivan Neretin Sep 28 '16 at 5:50
  • $\begingroup$ @Ivan Neretin Would $HF$ and $DF$ have different state? I am sorry I cant find any data about b.p. or m.p. of $DF$. $\endgroup$ – Mockingbird May 5 '17 at 5:24
4
$\begingroup$

It’s not exactly room temperature but it’s probably the closest you can get to two isotopes having different states so that they can be separated by physical means:

At very low temperatures (below $1~\mathrm{K}$) liquefied helium will separate into a superfluidic helium-4 phase and a normal liquid phase containing mostly helium-3. The superfluid and the normal liquid can be considered different phases and are immiscible.

| improve this answer | |
$\endgroup$
2
$\begingroup$

Isotopes differ in the number of neutrons in their nuclei. For very light elements like H the differences in atomic weight are relatively significant. So if transfer of a hydride or proton is a rate-limiting step in some kinetic process, then the mass of the nucleus can have a measurable effect. This 'kinetic isotope effect' can be used to tease out mechanisms of some chemical reactions, esp. the mechanisms of enzyme catalysis. In the realm of physics, the neutron/proton ratio determines the net spin of a nucleus, with 2 consequences. First, nuclear spin determines if that nucleus is a fermion or boson, with consequences esp. at low temperatures (eg. He-3 vs He-4). Another significant effect is seen in the presence of a strong magnetic field. In this case, nuclei with different spin states possess different energies. Absorption of EM radiation (in the RF-uWave frequencies) causes the nuclei in the lower energy state to 'flip' to the higher state. They will spontaneously return to the original state, emitting the rf energy they previously absorbed. Different isotopes can differ significantly in the way their spins couple with EM fields, an effect that NMR spectroscopy, MRI imaging and similar technics make use of.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.