2
$\begingroup$

Suppose one has a single atom of a radioactive isotope that radiates an Alpha particle.

I'll accept that the timing of when that Alpha particle radiates is completely random, when the atom is perfectly isolated from its external environment.

But it is unclear to me that a single atom is ever perfectly isolated. And it seems like the single atom, perfectly isolated, might be at absolute zero and therefore I'd feel uncomfortable accepting that it is still unstable.

And I would guess that as the ambient temperature of the atom's environment, in photonic terms (suppose the atom is in a blackbody oven), rises to infinity in finite time, the isotope will "pop" like popcorn with 100% certainty inside the finite interval.

However, I'll accept that there is a degree of randomness in terms of when the atom will emit its Alpha particle...within the finite interval.

But that is a guess. What is the relationship between heat and the single atom emitting the Alpha particle in a blackbody oven brought to some "approximately infinite" temperature from absolute zero in finite time?


I think I need to clarify my question so that it gets interpreted at a knowledge level appropriate for learning to occur:

All my question really amounts to is: does photonic radiation modulate the empirical frequency of atomic random walks by producing, on a single atom with no neighbors to whack him, a differential in emission likelihood when you range the temperature from 0 to infinite amplitude (Not infinite frequency). I'm going to say that, from any perspective, "infinity" probably causing the alpha particle to eject is a good null hypothesis, and the idea is to explore where the theory stops and some new behavior begins.

My intuition regarding the notion of heat in a nucleus is not statistical mechanical heat. Nor is it photonic blackbody heat. Also, as a side note, I do not actually consider gamma rays to be photonic radiation due to their ejection from a nucleus.


Notion of Heat in a Nucleus let's lock this down in a standard-compatible way so that this question is not hand-wavy, going forward.

The notion of heat in a nucleus, as I intend to mean it, is defined as follows:

Given a particular nuclear state of an atom at 273 degrees kelvin in a blackbody oven and protected from all particle collisions through some experimental technique, we define this to be one measurement unit.

If I set up many of these atomic units under the same conditions, I expect to observe a certain half-life when the atom in question is an alpha particle emitter.

The definition of "layman's atomic heat", which I want to circumscribe here, is variation against expectations of that emission half life.

For example, if I place an array of these measurement units in orbit, an measure them against an array here at ground level, I would observe a "lower temperature" at ground level, due to general relativity: I will have measured a longer half life at ground level than in orbit, in terms of seconds measured at ground level.

So, therefore, what happens if I have an array of these measurement units at standard temperature... and another array of these units subjected to persistent lasing at a frequency of your choice between 150nm and 11000nm (or perhaps smaller, but not at the gamma wavelength) with respect to the half life observed in the measurement units as the apparent temperature, in terms of photonic radiation, is increased from zero to infinity?


In fact, I view gamma rays existence more or less as evidence that photonic heat manipulates atomic nuclei. But, you have to restrict your test to a single atom in order to make sure no nosy neighbors are poking the atom via statistical mechanics style heat to generate the emission.

My personal interpretation of heat in a nucleus follows a markov model interpretation of the atom:

A random walker who has a probability distribution with respect to where he steps, but takes zero steps in a unit of time is unlikely to encounter a low probability state (say, emission of an alpha particle, beta particle, or gamma photon).

A random walker who takes many steps per unit time would enter an unlikely state more often, and emit his or her particle/ray.

In fact, we use this property to prove general relativity: if we speed up or slow down radioactive material, we can get fewer or more radioactive emissions.

So, in essence, we can do what I am asking (increase the "heat" of a nucleus) merely by putting them in a box, on an airplane, and flying them around.

What I would like to know is: can we do it with a high amplitude laser. Laser implies any wavelength from 150 nm to 11000 nm. Therefore, lasers have absolutely nothing to do with gamma rays, and the term is used incorrectly at that point.

I have no idea how you would make a gamma emitting diode. Perhaps you would hit it with concentrated light emitting diodes, which is essentially what I am asking here.


Things which are new to me:

  1. No photon can penetrate the nucleus unless it has a sufficiently small wavelength

    -- But it is not clear that a photon needs to penetrate a nucleus to effect its interior. Clearly, a nucleus can be moved, rattled, and spun around by interacting with its electron shell. And, it is not clear to me that spin, vibrations, and other photon->shell->nucleus dynamics couldn't do anything to effect the "random walking" of the nucleus interior.

Why? Because it seems to me that the nucleus ought to be able to effect itself. Clearly, if an electron shell is moved around, the nucleus moves too. Not only that: the nucleus "knows" it moved, because the motion is resisted by inertia -- information the nucleus must emit to us, the accelerator. (I am speaking in abstract terms right now)

When the nucleus is accelerated to infinity, we have received all inertial information it is possible for the nucleus to emit, relative to us, and the nucleus will never undergo a state transition (a random walk step) relative to us, forever. (this is relativity, at the nucleus).

So, clearly, if we want to get an emission, we could, at the very least, add inertial signaling power (slow it down).

My question amounts to, at a high level: is there some other mechanism which will modulate rate of emission other than, obviously, hitting it with particles or positrons.

Note that hitting it with a positron seems like it will produce the gamma rays needed... and giving an electron enough energy to slam into the nucleus will cause a proton to become a neutron and emit a pair of gamma photons into the nucleus... what happens next I have no idea.

-- (: subnote) that one need not give an electron huge energies for this to occur. Giving a shell electron any energy at all creates increases in the non-negative likelihoods the electron will suddenly "be" inside the nucleus, causing an annihilation event, a neutron to form, a pair of gamma rays to bounce around the nucleus, (and an alpha particle to pop out?). And if nothing pops out, then one would assume that, by some unknown mechanism, likelihood of emission has changed.

Anyhow, on this reasoning alone, one need only pick out a wavelength that matches an electron that the atom is unlikely to lose as it gains energy. Then, slowly add energy to that electron to effectively pump likelihood of electron manifestation inside the nucleus until, pop, the alpha particle zips out. That's my hobbyist take (so I don't have the most confidence in it).


But I am a physics amateur; maybe it is better to consider the atom as a whole: will heating it with infinite amplitude lasers cause alpha particle emissions to be random within some fixed interval, but non random outside the interval?

Update

It appears that likelihoods in the nucleus are continuous, just like electrons. Particles are waves, and their location is not perfectly determined.

While a particles "frequency" is not simple, a particle's wave function is analogous to its frequency. The Wave Function must occupy certain bins, or states, and there is probably some "continuous probability density function" (pdf) that describes transition likelihoods from state to state at, say, 273 degrees kelvin for a single atom, inside a blackbody oven, protected from collisions from all other massive particles other than its own electrons.

Further, at any time, an atom may transition from one wave function to another due to probability rather than energy, no matter how high the energy cost of that transition. This is the principle behind tunneling, it is something we have to correct for constantly in our modern processors, and it is how we charge our cell phones. The effect was worked with empirically by Tesla as an alternative power delivery system, and we've worked with quantum tunneling, therefore, for almost 200 years.

From a mathematical perspective, it is possible to declare a fuzzy, irreversible action (something that is non deterministic) deterministic across an interval via observation of an asymptotic decay of likelihood that is an exponential function of time in the first derivative.

So all you need to do to prove that alpha radiation can be manipulated deterministically is prove that the likelihood of emission changes as a function of (x) exponentially.

Then, you need to supply (x). Relativity accomplishes this task by causing the likelihood of emission to diminish to zero through acceleration. So likelihood to emit an alpha particle will diminish to zero in finite velocity (the speed of light). And note: this can be varied continuously rather than discretely.

What I was looking for was a way to say: how do we increase that likelihood in a similarly continuous fashion.

And it seems like, although the results measured one atom at a time will be discrete, increasing the likelihood of electrons moving into the nucleus with the pressure of normal laser beams or practical blackbody oven will indeed increase that likelihood.

Whether the first derivative of the likelihood is an exponential function of energy absorbed by the atom is, ultimately, the question, as that would allow us to say that, while random in terms of the interval, once the atom absorbs a finite amount of energy, it will have emitted an alpha particle with 100% certainty, regardless of the state transition energies.

And so that is the question: how does atomic likelihood to emit an alpha particle change as a function of the generic photon-ic (something that is of a nature with photons) energy it absorbs (rather than inertial, which we no for sure will knock nuclei apart).

And "no it doesn't change at all and here is why" would be a perfectly great answer.

Composed with thumbs on the app, pls forgive any typos

$\endgroup$
15
  • $\begingroup$ Does this answer your question? Randomness of radioactive decay $\endgroup$
    – Jon Custer
    Dec 21 '20 at 0:28
  • $\begingroup$ @JonCuster Not really. Basically, I have a bound proof that, with temperature as the agent, establishes that radiation is non random within the time it takes (the interval) of the oven warming to sufficiently high temperature to kick the alpha particle out. But I think that, holding temperature constant, the randomness probably holds. My question is: how low is the temperature that would spit out that alpha particle with 100% certainty? It must be lower than the temperatures that would destabilize the heavy atom entirely... $\endgroup$
    – Chris
    Dec 21 '20 at 0:33
  • 1
    $\begingroup$ Why do you think temperature has much if any effect on processes in the nucleus? Conditions in a nucleus in a nuclear explosion are well understood, and that is a highly non equilibrium at temperatures far exceeding any black body you can make. $\endgroup$
    – Jon Custer
    Dec 21 '20 at 0:42
  • 2
    $\begingroup$ If you want to use "personal definitions" of standard terms (including heat, photonic radiation, and gamma rays), then choose new terms and define them, or choose a different venue. As it stands, this question should be closed as per these guidelines. $\endgroup$ Dec 21 '20 at 16:06
  • 2
    $\begingroup$ @Chris It's up to you. Just be aware that when you include phrases like "My personal interpretation of heat in a nucleus is not statistical mechanical heat" you are explicitly disconnecting the thread from standard physics, and thus from the scope of this site. $\endgroup$ Dec 21 '20 at 16:28
1
$\begingroup$

It is a bit hard to parse your question because you ask at least 6-7 sub-questions that are related, but not exactly the same. So I will try to answer the following question:

"How does atomic likelihood to emit an alpha particle change as a function of the generic photon-ic (something that is of a nature with photons) energy it absorbs (rather than inertial, which we no for sure will knock nuclei apart)."

There are two ways to increase the decay rate with photons, as you guessed yourself.

  1. Increase the temperature of the atoms, which results in an increased emission rate of alpha particles which is still random.
  2. Stimulated emission, which results in non-random emission of alpha particles.

The simplified energy barrier for an alpha particle looks as follows:

https://www.open.edu/openlearn/science-maths-technology/scattering-and-tunnelling/content-section-5.2

The alpha tunneling rate is given from particle-in-a-box quantum mechanics by the equation below. See here for an interactive demonstration.

$$\log(\tau) = A - B \frac{Z}{\sqrt{E_{\alpha}}}$$

Here $\tau$ is the decay time, $Z$ is final number of protons and $E_{\alpha}$ is the energy of the emitted alpha particles. If $\tau$ is in seconds and $E_{\alpha}$ in MeV, then $A=-46.83$ and $B=-1.454$.

The effect of temperature is to add a thermal kinetic energy of $\frac{3}{2}k_{B} T$, so that the energy barrier is slightly lowered $E_{\alpha} \rightarrow E_{\alpha}-\frac{3}{2}k_{B} T$. One can verify from the above equation that, because $E_{\alpha}$ is of order 5 MeV (or $5\cdot 10^{10}$ K), photons need to heat the atom up to $\sim 10^{8}$ Kelvin to produce a 1% change in the alpha decay time. Such temperatures are actually possible with lasers at the National Ignition Facility (NIF), where they study nuclear fusion, rather than alpha decay.

The other option is to try to initiate stimulated emission. To do this, you need to produce an electric field comparable to the barrier height (Megavolts) divided by the nuclear barrier distance (femtometers). In other words, $\mathrm{E}\approx E_{\alpha}/r_0$, where $r_0$ is the nuclear confinement radius. If the electric field is this large, it will lower the barrier and allow the alpha particle to tunnel away much more easily. Since $r_0$ is typically of order 1 femtometer, you need an electric field of order $5\cdot 10^{19}\, V/cm$ for stimulated emission. To make an electric field with lasers at even 1% of this magnitude to drive stimulated alpha emission, you would need a laser fluence of $3.3\cdot 10^{32} \,\mathrm{W}/\mathrm{cm}^2$. It seems the record for highest laser fluence from the Guinness book of records is only $2.2\cdot 10^{22} \,\mathrm{W}/\mathrm{cm}^2$ at the moment.

An important aspect of stimulated emission is that the emission is no longer random, but is closely tied to the time-dependence of the driving electric field. So the emission could be periodic in time (positive and negative) rather than exponential, for example. You could also get the reverse process of alpha particles returning back into the nucleus.

$\endgroup$
2
  • $\begingroup$ Ha! Thanks! I will modify my question to address this exactly, with no modification of the content related to this answer, and see if I can figure out what the other questions im asking are $\endgroup$
    – Chris
    Dec 21 '20 at 19:42
  • $\begingroup$ And fascinating re the reversibility $\endgroup$
    – Chris
    Dec 21 '20 at 19:56
4
$\begingroup$

On the level of individual atoms, heat exists as infrared photons. For a photon to penetrate the nucleus instead of zipping by the atom without interacting requires it to have a wavelength of order ~diameter of nucleus, which implies a very energetic photon (as in gamma ray). Infrared photons have far too little energy (too long a wavelength) to accomplish this, and so will never get the chance to make the acquaintance of the nucleus itself.

This means that you cannot trigger an atom to eject an alpha particle from its nucleus by heating it up.

$\endgroup$
8
  • $\begingroup$ I don't understand your answer. If you put it in an oven whose temperature is order MeV or higher, the photons will have enough energy to modify the nucleus. $\endgroup$
    – kaylimekay
    Dec 21 '20 at 5:16
  • $\begingroup$ does that make sense when, without gamma rays, this emission can easily be manipulated by putting the atom on an airplane and flying it around. Would light blasting the atom as a whole not cause dynamical changes in the motion of the nucleus? $\endgroup$
    – Chris
    Dec 21 '20 at 15:10
  • $\begingroup$ Genuinely curious... because there are lots of things that we think are modular, but in practice wind up being integrated. The Atom seems pretty vulnerable to that human blind spot. $\endgroup$
    – Chris
    Dec 21 '20 at 15:14
  • $\begingroup$ and the +1 is mine 🙂 $\endgroup$
    – Chris
    Dec 21 '20 at 15:14
  • $\begingroup$ @Chris do you have a reference stating that the emission of an atom is changed by flying it in a plane? I genuinely have not heard of that effect. I know you get more irradiation (alpha, uv, etc.) at higher elevation, but that effect is not tied to an airplane's motion. $\endgroup$
    – KF Gauss
    Dec 21 '20 at 17:07
2
$\begingroup$

There is a basic misunderstanding in your question between classical physics and quantum mechanics.

Thermodynamic variables, as temperature, and observables as heat can be defined using classical statistical mechanics of many particles.

Individual nuclei and their decay belong to the quantum mechanical framework in addition to requiring orders of magnitude energy input to change an energy level in what makes them bound. To destroy a nucleus you need gamma ray, MeV, and higher energies.

One cannot have a gamma ray oven, because it will be destroyed by the gamma rays.

One might have a gamma ray laser in the future , and gamma ray beams to make colliders are in the plan.

For a quantum entity , as a nucleus, to make a quantum transition , the appropriate energy has to be applied, in order to change energy levels, a quantized transition. Since a decay, be it an alpha one, releases energy according to the calculable QM probability, extra energy cannot be involved in the decay, in order to change the quantum mechanical probabilities. If enough extra energy is supplied the nucleus can be destroyed, made to fission, and as in cosmological times turn into a plasma given enough energy , which is studied currently in ion collisions at cern.

$\endgroup$
5
  • $\begingroup$ Are you willing to consider stimulated emission? $\endgroup$
    – kaylimekay
    Dec 21 '20 at 6:21
  • $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/mod5.html#c3 . One cannot have control of alpha particles as one can have on photons (same energy and direction ..to create and alpha wave) in any case the meaning of the nucleus decay lifetime would not change.I suppose ion beams of fixed energy can be made and on could build an experiment irradiating a lot of nuclei and see if they changed to the decay fragment. One has to ask a nuclear physicist. $\endgroup$
    – anna v
    Dec 21 '20 at 7:09
  • $\begingroup$ relevant questions here physics.stackexchange.com/questions/296237/… and physics.stackexchange.com/questions/367861/… $\endgroup$
    – anna v
    Dec 21 '20 at 7:13
  • $\begingroup$ I don't mean coherent stimulated emission. I just mean something like this: Suppose there are two states $A$ and $A'$ that are nearly degenerate so that decays $A'\to A$ are suppressed by phase space. Now suppose another state $B$ whose energy is much higher than $A$ and $A'$. Now if we excite the system from $A'$ to $B$ it will have about equal probability to decay to either $A$ or $A'$ and either decay will happen relatively quickly (assuming no selection rules suppress it, etc). $\endgroup$
    – kaylimekay
    Dec 21 '20 at 7:19
  • $\begingroup$ These will still be independent probabilites, There is not stimulation, just calculabe in principle QM probabilities. $\endgroup$
    – anna v
    Dec 21 '20 at 7:27

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.