# What speed do subatomic particles travel around the nucleus?

1. Dec 30, 2004

### Gamish

What speed do subatomic particles travel around the nucleus? Do the speeds vary with different paricles, IE Electron, Proton, Neutron? Are the speeds differnet with its idenical positron? I do know that electrons and protons travel around the nuclueus at differnt speeds, this we call "heat", in which photons are sometimes released at the according frequency.

Last edited: Dec 30, 2004
2. Dec 30, 2004

### Gokul43201

Staff Emeritus
Gamish,

Protons make up the nucleus and so, do not revolve around it.

Older models of the atom (Rutherfold, Bohr) had the electrons revolving around the nucleus, but we know that this is only a very crude representation. The behavior of an electron (or any particle, for that matter) is determined from something known as its wavefunction. The wavefunction of an electron in a stable, non-interacting atom does not evolve in time, so we can discard the picture of revolving electrons when we treat the atom quantum mechanically.

Heat has nothing to do with the "speeds of revolution". Heat, in a material medium comes from the kinetic energy of the atoms/molecules themselves. And when electrons are involved, it is the free electrons that contribute to the heat capacity. These are the electrons that are not considered to be bound strongly to any one nucleus.

Yes, electronic transitions do result in an emission or absroption of photons and vice versa. However, excited electronic states are very short lived, and they end up transfering energy into the free electrons or phonons (vibrational modes of the atoms). It is through these that heat propagates along a medium.

3. Dec 31, 2004

### Gamish

OK, you did not answer my question, I am assuming that you do not know the answer. If you did not know, electrons do travel around a nuclues, I am asking at what speed. Lets say the speed of the electron on one atom of hydrogen, at 50 degreas C.

4. Jan 1, 2005

### elas

Electrons form a shell, would you ask 'what speed does the earth's crust travel around the core'? The question cannot be answered in the form you desire because there is no fixed orbital speed, all speeds up to the speed of light are possible depending on the degree of activity.
Gokul43201 has the answer in proffessional form, I hope my amatuer approach will add some enlightenment.

5. Jan 1, 2005

### Gamish

. Actually, no, according to special relativity, no object with rest mass can reach the speed of light (c). As an objects mass increases near speeds of c, more energy will be put towered the moving of the object rather than the actual speeding up of the object.

Also, can you please tell me what variables are involved that determine the speed at which electrons travel around the nucleus of it's atom?

6. Jan 1, 2005

### dextercioby

$$\langle\hat{v}\rangle_{|n,l,m\rangle}=\frac{1}{m}\langle\hat{p}\rangle_{|n,l,m\rangle}=\frac{1}{m}\langle n,l,m|\hat{p}|n,l,m\rangle$$

So it suffices to know the quantum state of the electron in one certain representation and in the same representation the form of the momentum operator.

Daniel.

7. Jan 2, 2005

### elas

Also, can you please tell me what variables are involved that determine the speed at which electrons travel around the nucleus of it's atom?

Spin (and therefore rotation) are slowed down in experiments that show the wave nature of electrons, as this is done by cooling; we know that temperature affects speed of rotation. Distance from the centre also plays a part in determining rotation speeds when electron is viewed as a particle. Variations in shell density probably cause variations in shell rotation even within a single atom, leading to (three dimewnsional) wave rotation.

You are of course, correct about c.

8. Jan 3, 2005

### tozhan

i think i heard a while ago that an electron takes about 150 attoseconds (150 quintillionths ($$10^{-18}$$) of a second) to 'orbit' a nucleus. if we take the diameter of an atom to be $$10^{-10}$$ metres we can VERY roughly make a guess at its speed.

$$\frac{\pi10^{-10}}{150^{-18}}=2.1*10^{8}ms^{-1}$$

does this seem reasonable?? It would be about about 1.4 times its rest mass if you apply special relativity to it.

i only have very patchy idea of classical physics so all you quantum guys can tell me what it really is!

also, the Pauli exclusion principle says electrons can orbit in pairs when the electrons are of opisite spin, but i though all naturally occuring electrons were spin-left? i also dont understand lone pairs, is this the same thing?

9. Jan 3, 2005

### dextercioby

1.The result u obtained for the speed of 'orbit' is absurd and it comes from the fact u used a wrong value for the period of orbit.I suggest u take a good look into the Bohr's theory of atom and then u can come up with a decent value for the period.The velocity for the first Bohr orbit should be 10 times less than u found and then u can conclude that relativistic effects are not essential.

2.Pauli's exclusion principle applies to all fermions without discrimination.Electrons have 2 possbile 'orientations' for their spin:up (m_{s}=+1/2) and down (m_{s}=-1/2),so there are two types of electrons judging after their spin projection on Oz eigenvalues.

Daniel.

10. Jan 3, 2005

### tozhan

being out by a factor of 10 isnt bad for me ;-) my sketchy mathes is always a little wrong! can anyone come up with a real answer? or shall i search google?

Tom

11. Jan 3, 2005

### dextercioby

The first orbital speed for the H atom in the theory of Bohr is
$$v(n=1)=\alpha c\sim\frac{c}{137}$$
,where $\alpha$ is the famous fine structure constant of Sommerfeld and it has the approximate value of 1/137,number which can be obtained if u plug in the constants in its definition
$$\alpha=:\frac{e^{2}}{4\pi\epsilon_{0}\hbar c}$$

Daniel.

12. Jan 4, 2005

### Gokul43201

Staff Emeritus
Thanks Dexter, for the clarification.

In my hurry, I did a terribly shoddy job. I started out with the hydrogen atom, but eventually decided to just write down how to find the velocity operator in a general 1-d case, but forgot to remove the subscripts that I had started out with.

What I simply meant to suggest was that

$$<\hat {v}_x> = \frac {\hbar}{im} \int_{- \infty}^{\infty} \psi }^* \frac {\partial \psi }{\partial x }~dx$$

What I was hoping to show was that information about "velocity" can be got from the wavefunction, and that systems such as those being discussed can not be treated classically. I suspected that Gamish may be looking for a classical answer to his question (since he brought up "heat" and so on). But now, I think, by "heat" he merely meant radiation.

Last edited: Jan 4, 2005
13. Jan 4, 2005

### Gamish

Example

Can you give me an example using that equasion, im not good with understanding tex formatting. And can you please explaine what all those variables are?

14. Jan 4, 2005

### dextercioby

Gokul gave a particular realization of the formula i posted in the post no.6 of this thread.

$<v_{n,l,m}>$ should have been put under the form
$\langle \hat{v}_{x}\rangle _{|n,l,m\rangle}$ and represents the average of the operator v_{x} computed with the Q system in the pure state $|n,l,m\rangle$.\psi and \psi star represent the wave functions of the Q system (probably H atom he,meant,but in that case,the formula is totally wrong;let's assume he didn't consider the H atom,though the indices "n","l","m" make us think that way) in the coordinate representation.

The partial derivative wrt to 'x' is a part of the formula giving the 'x' component of the momentum operator in the position representation.As i said before (when spaeking about the H atom),if the wave functions depend on all 3 variables,then Gokul's formula is incomplete,as the integral must be evaluated over all three coordinates.But let's stick to this simple unidimensional case.

U virtually have to find and use a wavefunction of the system psi(x) and with it compute 2 quantities:its derivative wrt to "x" and its complex conjugate,multiply the 2 results and integrate over the real axis.Then simply multiply with the reduced Planck's constant and divide through the product 'im'.U'll find the average of the velocity operator in the Q state u've chosen.

Daniel.

Last edited: Jan 4, 2005
15. Jan 5, 2005

### da_willem

In Quantum mechanics (QM) all you can know about a partice is it's state. Where in Classical mechanics you're solving the equation of motion (Newtons F=ma eg) to find the position of a particle as a function of time, in QM you solve (in the nonrelativistic case) the Schrödinger equation. This equation:

$$i\hbar\frac{\partial \Psi}{\partial t}=\hat{H}\Psi$$

gives you, when you know the phsyical circumstances ($H$), the state $\hat{\psi}$. This state does not directly give you the position of a particle as a function of time. You can only use it to find a 'probabality density' for the particle's position, from wich an 'expectation value' (mean) for the particles position can be found. In QM this is all we can know about a particles position, the probablility that upon measurement you find it here or there!

Now for a molecule this Schrödinger equation can only be a analytically solved for a hydrogen atom (and some similar atoms), a system of a proton and an electron. The relevant part of the state you need to say something about the electrons position or velocity is called the wavefunction. It's 'square' (actually its modulus) gives you the aforementioned probability distribution. The result in the form of a wavefuncion you can find in almost any textbook on QM. It is denoted by $$\psi_{n,l,m}$$ and is actually a whole collection of wavefunction depending on what numbers of n, l and m you are interested in. They represent the different orbitals, or excited states, of the hydrogen molecule. Note that n is the same variable as in Bohrs theorem of the hydrogen atom and is called the 'principle quantum number'.

As an example I will try to answer your question for the ground state (n=1, l=m=0) of a hydrogen atom. The wavefunction as a function of the radial coordinate r is

$$\psi_{1,0,0}=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}$$ with $$a=\frac{4 \pi \epsilon_0 \hbar^2}{me^2}=0,529E-10 m$$ (Bohr radius in meters)

As said before the 'square' gives you the probability density:

$$|\psi_{1,0,0}|^2=\frac{1}{\pi a^3}e^{-2r/a}$$

And tells you the elektron has the greatest probability to be found at r=o wch is at the nucleus! Furtermore the wavefunction is spherically symmetric. But all this is not really relevant for finding out the velocity, so let's move on.

As said before you cannot find the position of the electron, simply because i has no position. You can find the expectation value for the position though. Suppose you know a particle can be found with probability at x=1, probability (1/4) at x=2 and probability (1/4) x=4 you can find the expectation value by calculating the following sum:

$$<x>=\sum_i P(x_i)x_i = (1/2)*1+(1/4)*2+(1/4)*4=2.$$

But now we don't deal with a discrete spectrum of probabilities ((1/2), (1/4) and (1/4)) but a continuum of probabilities. So instead of summing over the probabilities you will have to integrate:

$$<x>=\int |\psi_{1,0,0}|^2 x dx$$

Where the probabilities P are replaced by the probability density $|\psi_{1,0,0}|^2$. This integral has to be evaluated over all space.

Also the electron has no definite speed, but only a probability to be measured that speed, or that speed. So I will have to dissapoint you, again all we can find out about the velocity is it's expectation value <v>. Now:

$$<v>=\frac{d<x>}{dt}=\int \frac{\partial |\psi_{1,0,0}|^2}{\partial t} x dx = \frac{-i \hbar}{m} \int \psi_{1,0,0}^* \frac{\partial \psi_{1,0,0}}{\partial x} dx$$

Where the last step follows after some juggling with the integral and using the Scrödinger equation. [I don't know have far your mathematics skills reach but $|\psi_{1,0,0}|^2 = \psi_{1,0,0}^* \psi_{1,0,0}$ is called the 'modulus' and a star denotes 'complex conjugation'. In case your wavefunction is real and this means just respectively squaring and doing nothing.]

This is (for the groundstate of hydrogen) the result dextercioby and gokul pointed out and hopefully you now know more or less know where it came from.

16. Jan 5, 2005

### da_willem

But anyway. Using the equation to find the velocity of an electron in a hydrogen atom will not work. Because the atom as a whole stands still. <x> of the electron is zero. And the expectation value of the velocity will be zero as well, because it moves with the same speed but opposite sign in the x and -x direction. With a real wavefunction like the of the electron in a hydrogen atom this can also be seen in a different (mathematical) way. The formula for <v> involves an 'i'. But as the wavefunction is real and ofcourse <v> is real thi leaves only one possibility: the integral vanishes and <v>=0.

17. Jan 12, 2005

### vluth

In the Bohr model of the Hydrogen atom (which gives some right answers, but is known to be essentially incorrect) electrons _do_ spin around the nucleus. In the simplest case of a hydrogen atom with a single electron spinning around a single proton, the electron moves at about 1/137 of the speed of light, which is MUCH faster than sound. Sound travels at about 1100 feet per second, while light travels at 186,000 miles per second.. You do the math.

As a side note, with the Bohr model, the 'inner' electrons in atoms with greater atomic numbers would be moving faster...

18. Jan 12, 2005

### dextercioby

Neat trick u used... But hopefully u're aware of the tedious proof that
$$\langle \hat{v}_{x}\rangle _{|\psi\rangle} =\frac{d\langle \hat{x}\rangle _{|\psi\rangle}}{dt}$$

,independent of the description...

As in Gokul's case,please remove the indices (1,0,0).They remind of the H atom and it's not the case with your example/formula.

Daniel.

PS.I didn't trust your formula,so i did the proof by my own.It wasn't nice at all. :yuck: Juggling with the Heisenberg and Schroedinger descriptions...Anyways,i'm a theorist,so i shouldn't complain.

19. Jan 13, 2005

### da_willem

I didn't know the proof was that tedious. But Griffiths only 'postulated' it, so that should say something. And about the (1,0,0), My post was about the hydrogen atom, wich has a fairly easy ground state wavefunction, so it's okay they remind you about it.