five very small questions

[Baby Astronaut]

1. If heated, will radioactive decay occur quickly? For instance, will the half-life of an element's decay be shortened? (while under enough of a heat increase)

 

2. Triple-slit experiment. Caused me to wonder....in a double-slit experiment, how far apart can the slits be and the particle-wave still manages to enter through both?

 

3. If you hovered airborne over a radioactive spot, so that you were hit by the radiated particles, then you exited the area in a few hours, would the particles that hit cause you to give off radiation later?

 

4. How far does a radiated particle travel in air? i.e. does it get slowed by collisions with air particles?

 

5. Does a collapsed wave function eventually uncollapse? (if so, quickly or slowly)

[npts2020]

1) AFAIK temperatures have no (or at most miniscule) effect on half life of a radioactive material. Am not sure what kinds of experiments on this have been done at extreme temperatures.

2) Not sure. Somebody else here can answer this with more authority than I.

3) Yes but it would not likely be measurable without extremely sensitive instruments. It also depends on the type of radiation.

4) You have to remember there is more than one kind of radiation. Alpha particles will not travel as far through a typical medium as beta particles which are usually less penetrating than gammas, neutrons are a different matter. Enough air (or any medium) will stop any kind of radiation but it may not be practical to use light-years thickness of air as shielding.

5) I don't know much about this.

[Baby Astronaut]

Two more very small (in diameter) but technically massive questions.

 

6. Before 1967, what did Einstein and others call black holes in discussions about their prediction by relativity?

 

7. In Wikipedia's entry for Pauli exclusion principle...

 

It states that no two identical fermions may occupy the same quantum state simultaneously.

What does that actually mean? I had originally understood it as no two electrons can be in the exact same position. i.e. no total overlap.

 

 

3) Yes but it would not likely be measurable without extremely sensitive instruments. It also depends on the type of radiation.

Basically what I mean is a person's exposed to radiation, not on shoes or direct contact but just through air. Will some of the radiation leave with them?

 

I mean nuclear waste radiation, the typical stuff we often hear about.

 

4) You have to remember there is more than one kind of radiation. Alpha particles will not travel as far through a typical medium as beta particles which are usually less penetrating than gammas, neutrons are a different matter. Enough air (or any medium) will stop any kind of radiation but it may not be practical to use light-years thickness of air as shielding.

With their high energy, it'd seem a radiated particle would travel quite far. Yet also it seems a person would generally be safe if a good distance away from a radioactive pile laying in the middle of some open field, say hundreds of yards distant?

 

Not really sure of how radiated particles would move though.

[Horza2002]

2. The doible slits are typically around 10nm apart if I remember rightly...but that depends on the wavelength of light your using (have a look for Youngs double split equation)

 

3. I don't think they would, unless the radiation emmited caused some of the atoms in tha body to become unstable in which case you might still be radioactive afterwards. I think people stay "radioactive" because they have become contaminated by whatever material was decaying in the first place.

 

4. Again it depnds on what type of radiation. High energy alpha particles would still be stopped before a few centimeters of air while high energy gamma would need a few meters of lead to stop them.

 

7. From a chemists point of ciew, it means that no two electrons in a single atom can have exactly the same set of quantum numbers (n, l, ml and ms). If they did they would be occupying exactly the same region of space with is not allowed. A Fermion is a class of particle (I think with a spin = 1/2).

[Baby Astronaut]

2. The doible slits are typically around 10nm apart if I remember rightly...but that depends on the wavelength of light your using

You've answered the other three very nicely, I'd just like more info on this one. If the slits were positioned double that, say 20nm apart, would the particle not be split into two?

 

3. I don't think they would, unless the radiation emmited caused some of the atoms in tha body to become unstable in which case you might still be radioactive afterwards. I think people stay "radioactive" because they have become contaminated by whatever material was decaying in the first place.

Exactly what I sought. Thanks.

 

 

 

Ok, #3, 4, and 7 fully answered by npts2020 and Horza2002.

 

Still to go...

#

1

2 (partly)

5

6

Anyone care to give them a whirl?

[swansont]

1. If heated, will radioactive decay occur quickly? For instance, will the half-life of an element's decay be shortened? (while under enough of a heat increase)

 

No, other than possibly electron capture. There are a few reports where people claim to have seen a T dependence, but AFAIK they have not been replicated.

[Horza2002]

1. Radioactive decay is a spontaneous, random process. Spontaneous means that no outside infulence (e.g. volume, temperature, pressure) will affect the rate of decay. Random means that you cannot predict which atom will decay exactly whn. So no, increasing the temperature does not alter the rate of decay.

 

2. The one thing you need to remember when dealing with quantum ideas is that common sence and everyday experiences do not apply. The double split experiment was used to show that partciles (which people beleived where little blobs of matter) where actually waves (like ripples in a pond). What this experiment shows is that the "particle" goes through both slits *AT THE SAME TIME*. This gave the most credible evidence at the time for what is know known as wave-particle duality. Now, you can treat "particles" as waves or particles depending on the circumstance. In electronics, electrons are usually treated as particles (blobs of matter) where as in chemistry electrons are treated as waves....that is simply because they work best for their own circumstances. Sorry thats a bit long winded.

 

5. Sorry not sure about that one....not entirely sure what a collpased wavefunction is (im an organic chemist so not exactly my area of expertise).

 

6. I dont understand what you mean by this question.

 

Hopefully thats helped a little

[Baby Astronaut]

No, other than possibly electron capture. There are a few reports where people claim to have seen a T dependence, but AFAIK they have not been replicated.

I'm thinking like water boiling, the measuring scientist's not going to know which molecules convert to steam and rise first, but they can determine the "half-life" of the process as a whole.

 

Heat produces more action over a period of time, I'm guessing. The exact process of decay might just be currently unmeasurable, making it seem random, but it's possible we'd find that whatever causes it might be sped up with heat, in lab tests. By super exciting the atom, it might in turn excite all its subatomic processes.

 

A better test might be to see if a half-life gets "delayed" by keeping the material in nearly 0K temperature?

 

2. The one thing you need to remember when dealing with quantum ideas is that common sence and everyday experiences do not apply. The double split experiment was used to show that partciles (which people beleived where little blobs of matter) where actually waves (like ripples in a pond). What this experiment shows is that the "particle" goes through both slits *AT THE SAME TIME*.

I'd just like to know: would that still occur if more distance were put between the slits? I'm thinking a wave's going to expand and therefore should hit both, but at a certain distance apart it won't.

 

 

6. I dont understand what you mean by this question.

"Black Hole" was coined by John Wheeler in 1967. Yet physicists already had long been in discussion over whether they existed -- since relativity predicted them. If no one had called it a black hole previously, what name did they use?

Edited January 12, 2010 by Baby Astronaut

[Horza2002]

2. Yes your right, if you move the slits further apart, they will get to a point where u get more classical behaviour (as in the particles will go through one slit or the other). If I remember rightly, once the gap is larger than the wavelenght of the wave, it becomes clasical again....but I don't know y that is.

 

I suppose I should mention that when the slits are close together, if you attempted to measure which slit the particles went through, then you would see that you find out which slit it goes through BUT the interferance pattern resulting would be different from when you were not measuring the slits. Just a thought you might find interesting.

[D H]

5. Doesn't make sense. Wave function collapse is an irreversible process. (Some physicists who are not in the Copenhagen interpretation camp argue that the concept of wave function collapse doesn't make sense, period.)

 

6. Collapsar or frozen star.

[Baby Astronaut]

5. Doesn't make sense. Wave function collapse is an irreversible process.

I meant some time after the collapse process is done, will it begin to un-collapse by itself? Also I just found info on Wikipedia that might be referring to exactly that.

 

http://en.wikipedia.org/wiki/Wave_function_collapse#History_and_context

However, when the wave function collapses -- process (1) -- from an observer's perspective the state seems to "leap" or "jump" to just one of the basis states and uniquely acquire the value of the property being measured, ei, that is associated with that particular basis state.

After the collapse, the system begins to evolve again

according to the Schrödinger equation or some equivalent wave equation.

Could they mean after the wave function collapses, it all eventually goes back to a superposition again (i.e. the reverse of collapsing)?

 

 

Ok small parts of two questions remain.

 

#

1 (have replications of temperature dependence failed or just never been attempted?)

5 (does a collapsed wave function eventually return to its undisturbed state?)

[StrontiDog]

1. If heated, will radioactive decay occur quickly? For instance, will the half-life of an element's decay be shortened? (while under enough of a heat increase)

 

3. If you hovered airborne over a radioactive spot, so that you were hit by the radiated particles, then you exited the area in a few hours, would the particles that hit cause you to give off radiation later?

 

4. How far does a radiated particle travel in air? i.e. does it get slowed by collisions with air particles?

 

I am qualified to take a shot at these three, so I will.

 

Number 1: The answer is no.

 

For the most part, radioactive decay is caused by an imbalance of charge between the neutrons and protons in the nucleus of the isotope. Very basically you can call a neutron a chargeless proton or vice versa. They can be either. In the nucleus, they switch back and forth between being a P⁺ and N^o in a completely random manner. Heat--even core of the sun-type heat--has no effect at this level of matter. Neither does absolute zero. No effect at all.

 

 

As a matter of fact, if you shoot a neutron and it never collides with anything, it will decay into a proton by emitting an electron (okay, it's a negative beta particle, but we only know that because we know where it came from, it's indistinguishable from an electron). A neutron on its own even has a half life of about 10.3 minutes. http://hyperphysics.phy-astr.gsu.edu/HBASE/particles/proton.html

 

 

For number three: I'm going to have to make the assumption that by 'radiated particle' you mean the ionizing agent, which can be a particle or a photon. If this is what you mean, the answer is YES if you were hit by neutrons (or protons, both of which fall under the category of 'radiation.' )

 

Both can 'activate' certain elements in your body (oxygen and sodium, primarily) which will decay with varying half lives of their own. It literally turns an element into an isotope and 'makes' it radioactive.

 

A quick and dirty check of an individual who has been exposed to a lot of neutrons is to have them tuck a gamma-sensitive instrument (Geiger probe or Sodium Iodide) under their arm and there are rules-of-thumb for dose estimates, based on the dose rates observed and the time since exposure. All of these activated isotopes have short half lives, though. You won't be radioactive for long.

 

Alpha, photon and beta radiations will not do this.

 

For number four: It depends on the energy of the 'radiation' and what type it is. Air attenuation actually has Half Value Layer measurements (linear distance through which a particle or photon travels before half of them have been absorbed). For example, 10 keV photons are attenuated 35% when passing through 50 cm of air.http://www.irs.inms.nrc.ca/papers/PIRS629r/node16.html

 

Charged particles are a lot more prone to running into those pesky air molecules, but they all have values of one sort or another. Air is still matter, after all.

 

Since I'm not qualified to answer any of the others, I won't try.

 

Bill Wolfe

Edited February 6, 2010 by swansont
turned volume down

[swaha]

1. no.

4. it should for i think otherwise it would travel infinitely long distance.

5. not yet got the collapse of wave function.

[swansont]

Number 1: The answer is no.

 

For the most part, radioactive decay is caused by an imbalance of charge between the neutrons and protons in the nucleus of the isotope. Very basically you can call a neutron a chargeless proton or vice versa. They can be either. In the nucleus, they switch back and forth between being a P⁺ and N^o in a completely random manner. Heat--even core of the sun-type heat--has no effect at this level of matter. Neither does absolute zero. No effect at all.

 

 

As a matter of fact, if you shoot a neutron and it never collides with anything, it will decay into a proton by emitting an electron (okay, it's a negative beta particle, but we only know that because we know where it came from, it's indistinguishable from an electron). A neutron on its own even has a half life of about 10.3 minutes. http://hyperphysics.phy-astr.gsu.edu/HBASE/particles/proton.html

 

Careful here — neutron decay also emits an antineutrino, and neutrons and protons are not interchangeable. As your link shows, a proton us made up of uud quark set, while a neutron is udd.

 

For number three: I'm going to have to make the assumption that by 'radiated particle' you mean the ionizing agent, which can be a particle or a photon. If this is what you mean, the answer is YES if you were hit by neutrons (or protons, both of which fall under the category of 'radiation.' )

 

Both can 'activate' certain elements in your body (oxygen and sodium, primarily) which will decay with varying half lives of their own. It literally turns an element into an isotope and 'makes' it radioactive.

 

A quick and dirty check of an individual who has been exposed to a lot of neutrons is to have them tuck a gamma-sensitive instrument (Geiger probe or Sodium Iodide) under their arm and there are rules-of-thumb for dose estimates, based on the dose rates observed and the time since exposure. All of these activated isotopes have short half lives, though. You won't be radioactive for long.

 

Alpha, photon and beta radiations will not do this.

 

 

Alpha radiation can certainly activate material, as can high-enery photons.

[John Cuthber]

"A better test might be to see if a half-life gets "delayed" by keeping the material in nearly 0K temperature? "

From the point of view of the energies of nuclear reactions, everything (except, perhaps a supernova) is at nearly 0K.

[StrontiDog]

"A better test might be to see if a half-life gets "delayed" by keeping the material in nearly 0K temperature? "

From the point of view of the energies of nuclear reactions, everything (except, perhaps a supernova) is at nearly 0K.

 

We're talking about the trading of charge (wanna call it quarks, go ahead) between the nucleons. The nucleons. Don't mess with this stuff, it isn't really up for debate. Temperature is not powerful enough to matter at this scale. Think Higgs bosons and what they are actually made of.

 

I doesn't make any difference. The effect is still random, but predictable on a large scale. Think what temperature does, and then think weak and strong nuclear forces. It's not brain surgery, folks. It's simple thermal dynamics. Kid's stuff. Like it or not, that's the way it is. Sorry.

 

Temperature isn't capable of influencing the actions on the Quantum Level. There just isn't enough energy there to make a difference.

 

Bill Wolfe

[John Cuthber]

We're talking about the trading of charge (wanna call it quarks, go ahead) between the nucleons. The nucleons. Don't mess with this stuff, it isn't really up for debate. Temperature is not powerful enough to matter at this scale. Think Higgs bosons and what they are actually made of.

 

I doesn't make any difference. The effect is still random, but predictable on a large scale. Think what temperature does, and then think weak and strong nuclear forces. It's not brain surgery, folks. It's simple thermal dynamics. Kid's stuff. Like it or not, that's the way it is. Sorry.

 

Temperature isn't capable of influencing the actions on the Quantum Level. There just isn't enough energy there to make a difference.

 

Bill Wolfe

 

ROFL

It's not font size that's important.

The whole point I was making was that thermal energies are too small to affect nuclear reactions.

However at high enough temperatures (for example in the sun) nuclear reactions do take place (but very slowly ,which is why the sun is still going after billions of years).

To really influence the rates you need to get up to much higher temperatures.

It's not brain surgery- so it's a pity you didn't understand it.

[StrontiDog]

. . . .However at high enough temperatures (for example in the sun) nuclear reactions do take place (but very slowly ,which is why the sun is still going after billions of years).

To really influence the rates you need to get up to much higher temperatures.

It's not brain surgery- so it's a pity you didn't understand it.

 

Temperature and pressure determine fusion rates, sure. That ain't radioactive decay, though. And that was the original question. In a way, fusion is the opposite of decay.

 

I can't find any reference to slower rates of nuclear decay in the core of a star, can you tell me where you found the basis for it?

 

You're not talking about gravitational time dilation are you? The sun isn't massive enough to make much difference along those lines, but I suppose that to an outside observer, the nuclear decay rate would appear to be slowed.

 

Sorry, still can't see any effect that temperature (high or low) would have on radioactive half life.

 

I use bigger fonts because I have lousy eyesight. And all this time I thought that size didn't matter.

 

BW

Edited February 10, 2010 by StrontiDog

[timo]

Could they [the Wikipedia article] mean after the wave function collapses, it all eventually goes back to a superposition again (i.e. the reverse of collapsing)?

Under a measurement a system goes into a state (just think of a state as a wave function if you don't know the difference) compatible with the measurement result. That's pretty much all that happens. In particular, the system is not forced to stay in that state forever after. The state will evolve normally according to the equation of motion (Schödinger equation) - it's just not considered the reverse of collapsing. As an example: Assume I'd measure the position very exact. The electron will then be in a state compatible with the statement that it is in the location just measured. That does of course not prevent the electron from moving away later.

 

As a remark: Saying that a state is a superposition has no meaning unless you specify what it should be a superposition of: Is the wave function [math]f(x) = C \sin x [/math] a superposition or not?

[StrontiDog]

Alpha radiation can certainly activate material, as can high-enery photons.

 

Quite correct on both counts. I was looking at the scenario and alpha wouldn't reach or penetrate the 'hovering vehicle', so I discounted it. And while high energy photons can create metastable isomers, these are generally of such short half life that he wouldn't 'take it with him.'

 

Oversimplification is sometimes the same as being wrong.

 

Guilty.

 

The whole 'likely to see from nuclear waste' thing pretty much limits the likelyhood to neutrons and the occasional proton as activation agents. Photon activation of any scale would probably take a lethal dose, while the other two would be survivable.

 

Once again, though, you're right.

 

Bill Wolfe

[swansont]

Under a measurement a system goes into a state (just think of a state as a wave function if you don't know the difference) compatible with the measurement result. That's pretty much all that happens. In particular, the system is not forced to stay in that state forever after. The state will evolve normally according to the equation of motion (Schödinger equation) - it's just not considered the reverse of collapsing. As an example: Assume I'd measure the position very exact. The electron will then be in a state compatible with the statement that it is in the location just measured. That does of course not prevent the electron from moving away later.

 

As a remark: Saying that a state is a superposition has no meaning unless you specify what it should be a superposition of: Is the wave function [math]f(x) = C \sin x [/math] a superposition or not?

 

If you are in a superposition of two eigenstates and you collapse the wave function, you'll be in a stationary state of the system. The system stays in that state until undergoes an interaction.

 

I would say that [math]f(x) = C \sin x [/math] is not a superposition. [math]C_n \sin(n\pi x) [/math] might be solutions, and you might be in the superposition of the n=1 and n=2 states. If you collapse the wave function, you put it in one state. But that's not synonymous with the particle having only one position

[timo]

If you are in a superposition of two eigenstates and you collapse the wave function, you'll be in a stationary state of the system. The system stays in that state until undergoes an interaction.

Your post sounds as if you implicitly or explicitly assume that the collapse is due to a measurement in energy. I don't think that is necessarily the case - think the measurement of position on a particle in a box. The measurement of an observable that does not commute with the Hamiltonian does collapse the system to a state compatible with the measured eigenvalue of that observable (called eigenstate and not necessarily unique). This state then is not an eigenstate to the time evolution operator exp(-iH) and hence the measured observable will change its value under time evolution.

 

I would say that [math]f(x) = C \sin x [/math] is not a superposition.

Why not? It is a superposition of an infinite number of monomials, for example.

 

[math]C_n \sin(n\pi x) [/math] might be solutions, and you might be in the superposition of the n=1 and n=2 states. If you collapse the wave function, you put it in one state. But that's not synonymous with the particle having only one position

[math]C_n \sin(n\pi x) [/math] might be eigenstates to the observable measured, true. In that case, I would not call sin(x) a superposition of states. But they might as well not be eigenstates to the observable measured. In that case, sin(x) can only (if at all) be constructed as a superposition of 2+ eigenstates to the observable measured. In effect, the question whether a state is a superposition can only be answered if you have (defined) a basis for your vector space. And to rephrase my point: As long as you haven't specified the observable you measure (whose eigenstates form the basis) it's pointless to talk about a state being a superposition of (eigen-)states or not.

[Baby Astronaut]

I use bigger fonts because I have lousy eyesight. And all this time I thought that size didn't matter.

 

BW

If you use Firefox: Ctl_+ (plus sign) increasingly enlarges all text on a webpage. Ctl_- (minus sign) reduces it.