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Author Topic: How fast would a free electron need to travel to avoid capture by a nucleus?  (Read 4879 times)

Offline jeffreyH

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If we have an atom with a valency of 1, say hydrogen, and strip the electron from the nucleus, what speed would a free electron have to be travelling near the nucleus to prevent a fall into the Coulomb potential well?
« Last Edit: 09/01/2014 19:23:04 by chris »


 

Offline syhprum

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Re: A question about electrons
« Reply #1 on: 09/01/2014 19:27:12 »
I am probably talking out of the back of my head but I will have a go.
The energy required to completely remove an electron is the ionization potential something like 4 ev in the case of Hydrogen knowing the mass of the electron this amount of energy can be converted into a velocity according E=MV^2/2
Is this anywhere near the truth or just a confused ramble ?
 

Offline jeffreyH

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Re: A question about electrons
« Reply #2 on: 09/01/2014 23:42:19 »
I am probably talking out of the back of my head but I will have a go.
The energy required to completely remove an electron is the ionization potential something like 4 ev in the case of Hydrogen knowing the mass of the electron this amount of energy can be converted into a velocity according E=MV^2/2
Is this anywhere near the truth or just a confused ramble ?

Well let's have a go. If we take velocity as c we get E=MC^2/2. C^2 is around 89875517873681764. Lets take the mass of an electron next. This is 9.10938291e-31 so multiplying the two gives 8.1871050654591619976025324e-14. Divide by 2 gives 4.0935525327295809988012662e-14. At light speed this is a very reduced value it would be interesting, if you have gotten this right, to plot the sequence from c right down to the Planck length.

BTW Planck length per second should be the interpretation and no this did not use the mass of the hydrogen atom. That was on purpose.
« Last Edit: 10/01/2014 00:13:59 by jeffreyH »
 

Offline jeffreyH

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If we use 2 Planck lengths for the calculation as in 2 Planck lengths per second we get 4.7589239073369986335843032e-100. Still orders of magnitude but still has an initial value of 4. OK all this makes little sense and is a bit light-hearted but considering in 2GM/c^2 for the event horizon uses a 2 as the multiplier and our denominator was 2 there may be something more significant here if 4 ev was correct above.
 

Offline jeffreyH

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This got me intrigued. Kinetic energy is KE = 1/2 m * v^2. This gives velocity as v = sqt ( 2 * KE / m). For 4 ev we convert to 6.40870932e-19 joules of kinetic energy. Substituting this and the mass of the electron we get 1186194.226133525169667927801475 m/s this is around 0.4 % the speed of light. Can this be the right velocity? What have I done wrong?
« Last Edit: 10/01/2014 01:06:14 by jeffreyH »
 

Offline syhprum

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Although the effect would not be very large at .004c would not the apparent increase of mass of the electron with velocity have to be taken into account.
Jeffery H
As the energy of 4ev is rather a guesstimate is not carrying out the calculation to 30 decimal places rather overkill!
« Last Edit: 10/01/2014 06:38:50 by syhprum »
 

Offline jeffreyH

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Although the effect would not be very large at .004c would not the apparent increase of mass of the electron with velocity have to be taken into account.
Jeffery H
As the energy of 4ev is rather a guesstimate is not carrying out the calculation to 30 decimal places rather overkill!

LOL Sorry cut n paste. I will need to look into the mass increase. Thanks for the pointer.
 

Offline syhprum

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My guess was widely out apparently the ionization energy is 13.6ev
 

Offline jeffreyH

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Still this question has profounder significance. With distance from the nucleus the escape velocity of the electron will change as this has an inverse square nature. It must relate to charge strength over distance. As an attractive force it bears a striking similarity to gravitation. Has anyone tried to devise any mathematics for this concept? This is an electromagnetic attraction but studying it may give some insight into the mechanisms of gravity and could lead to a valid theory of quantum gravity.
 

Offline jeffreyH

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My guess was widely out apparently the ionization energy is 13.6ev

Well this value gives 2187234 m/s but how far away from the nucleus does this need to be? Just to add to this post. If the electron were travelling at c then at points very near the nucleus it would still likely escape but as we successively reduce this velocity at some point the electron would be captured. My personal guess is that even at very near light speed if the electron was close enough to the nucleus it would be captured at a precise distance. This would mean at that range photons would be trapped hence only the more energetic electrons emit photons. This could also supply a mechanism for differences between conduction and valency bands in semi-conductors other than simply the valency itself.
« Last Edit: 10/01/2014 23:26:38 by jeffreyH »
 

Offline jeffreyH

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I have found a limit on the uncertainty principle that rids physics of universe wide uncertainty and is linked to the particle wave function. I found it whilst reading "Erwin Shrodinger and the Quantum Revolution" by John Gribbin. The way physics progressed through the 1920s and some of the steps led me to the answer. It involves strings and I will be putting together a paper on the findings. Uncertainty is still there but limited in scope. Schrodinger's cat is dodging the poison.
 

Offline syhprum

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Is there not a simple SR formula relating particle energy and velocity one often sees quotes of LHC protons moving at 99.99999% of light speed
I agree with your figure 2187234 m/s or .00729% c as any increase of apparent mass is irrelevant at this velocity
The electrons can only orbit the nucleolus at well defined levels as dictated by QM the similarity to gravity does not apply
« Last Edit: 11/01/2014 08:37:36 by syhprum »
 

Offline lightarrow

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If we have an atom with a valency of 1, say hydrogen, and strip the electron from the nucleus, what speed would a free electron have to be travelling near the nucleus to prevent a fall into the Coulomb potential well?
~2.2*103 m/s Km/s, if I computed well.
The ionization energy of hydrogen is about 1.3*106 joule/mol:
http://en.wikipedia.org/wiki/Hydrogen
it means that is the minimum kinetic energy an electron shoud have not to bind with the proton (if, paradoxically, it would bind having a greater energy than that, the atom would ionize immediately, releasing the electron again).
From Ekinetic = 1/2 m v2, m = electron mass, you find the answer.

I have used the non-relativistic equation for kinetic energy because I have verified, a posteriori, that the computed speed for the electron is negligible with respect to light speed. If instead the result were comparable with light speed, I would have discovered that my approximation were false, and I would have re-done my computation using the (a bit more complex) relativistic equation. Physicists' trick  :)

--
lightarrow
« Last Edit: 11/01/2014 19:29:14 by lightarrow »
 

Offline Bored chemist

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Well let's have a go. If we take velocity as c
Why?
That's the speed that we know is wrong (because an electron has mass so it can't travel at c).
 

Offline syhprum

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Lightarrow, both JeffreyH and myself make it ~2.2*10^3 Km/s , are two of us wrong or are you ?
« Last Edit: 11/01/2014 15:58:39 by syhprum »
 

Offline jeffreyH

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Well let's have a go. If we take velocity as c
Why?
That's the speed that we know is wrong (because an electron has mass so it can't travel at c).

That was just a liberty I was taking to get to a particular value.
 

Offline lightarrow

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Lightarrow, both JeffreyH and myself make it ~2.2*10^3 Km/s , are two of us wrong or are you ?
the majority wins  :)
I have corrected my post.

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lightarrow
 

Offline jeffreyH

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Does anyone have any references for the work of Boltzmann that Planck used to combine Wein's law and the Rayleigh-Jeans law? I really need Boltzmann's original method of quantization.
 

Offline evan_au

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The energy of the electron can be determined from the wavelength of the light it would emit falling from infinity down to the 1s orbital, ie 91.2nm.
The energy released by falling from closer distances is given by the Lyman series, in ultraviolet wavelengths around 100nm.
If you want to find the energy falling into the second shell, this is given by the Balmer series.
The general answer is provided by the Rydberg formula; set n'=∞.
See: http://en.wikipedia.org/wiki/Hydrogen_spectral_series#Physics

Energy=hc/λ
http://en.wikipedia.org/wiki/Photon#Physical_properties
 

Offline jeffreyH

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I have just found a correlation between the wave function of photons and the wave generated via the virtual particles in the Higgs field. If we consider the photon as a string then we have a string thickness that is proportional to amplitude across the Higgs field. The uncertainty is then restricted by the amplitude. We can view the amplitude by moving the slit in the single-slit experiment.

BTW The string thickness is related to the velocity as well. In the case of the electron this is why the mass appears different while the charge is the same and opposite to the proton. We can only detect the electron as a particle at the tip and the rest of the wave is interference.
« Last Edit: 12/01/2014 05:49:31 by jeffreyH »
 

Offline jeffreyH

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Is there a value for the radius of the electron orbital in a hydrogen atom? I need this for a calculation. I have 5.2910−11m. Is this right? Alternatively 0.0529 nm.
« Last Edit: 12/01/2014 06:01:18 by jeffreyH »
 

Offline jeffreyH

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I don't believe this I can now also describe how stars evolve and what causes the heating mechanism.
 

Offline jeffreyH

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The energy of the electron can be determined from the wavelength of the light it would emit falling from infinity down to the 1s orbital, ie 91.2nm.
The energy released by falling from closer distances is given by the Lyman series, in ultraviolet wavelengths around 100nm.
If you want to find the energy falling into the second shell, this is given by the Balmer series.
The general answer is provided by the Rydberg formula; set n'=∞.
See: http://en.wikipedia.org/wiki/Hydrogen_spectral_series#Physics

Energy=hc/λ
http://en.wikipedia.org/wiki/Photon#Physical_properties

Just remembered this thread. Thanks for the info. I now need to do some reading.
 

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