Question

Centripetal force=mv^2/r
r=0
Infinite acceleration?

Answer 1

If you could revolve a particle in an extremely tight path about a point, then yeah, the force required would be enormous.

Answer 2

How could you possibly rotate around a path with a radius of 0?

Answer 3

Good question...

Answer 4

What supplies the centripetal force, hmm? Gravity from a mass, or electrostatic attraction from a charge? In either case, the force goes like 1/r^2, so as r->0 the force goes to infinity anyway. No worries.

Answer 5

In an atom the radius is too small so?

Answer 6

@experimentX @mukushla

Answer 7

what the size of you particle?

Answer 8

im talking about an electron revolving in n=1 around an atom,the radius would be almost negligible isn't it? so what would be the centripetal acceleration?
Infinite
?

Answer 9

http://www.physicsforums.com/showthread.php?t=3739

Answer 10

As far as we know the electron is an elementary particle. Elementary particles are defined to have zero radius. They are 0 dimensional objects.

Answer 11

sorry don't know that ..

Answer 12

Even if its infinite force/acceleration where does it come from?
Nature?then the whole world will end :O

Answer 13

lol ..

Answer 14

if the radius is 0 ,then its not circular motion...............

Answer 15

well in an atom electron revolves approximately at the speed of light. And if r= 0 how it could be a circular motion...you need to consider limits

Answer 16

Honourable Gentleman of the Physical keyboard @darvin969 @ghazi , Allow me to humbly share with you the WELL - KNOWN Bohr's calculation of the Hydrogen atom RADIUS, and THE VELOCITY of the electron in this Hydrogen Atom. As I hope you will read the following excellent page-or-two you wil realize
1. The electron moves at LES THAN 1% the speed of light in the atom
2. The radius of the lowest orbit is known and FINITE
3. The force, energy and all else - are finite.
BTW I recommend you
Everything You Always Wanted to Know About the Hydrogen Atom (But Were Afraid to Ask)

The Bohr Model
http://www.pha.jhu.edu/~rt19/hydro/hydro.html

http://en.wikipedia.org/wiki/Bohr_model

A) Do ask simple and fundamental questions
B) DO NOT give a SIMPLISTIC and hubris-like answers - check with scientific knowledge !

Answer 17

@DLS READ the above

Answer 18

Dear @Mikael I didn't say electron moves at the speed of light..is used the word approximately which means 99% of that. and the question is whether at r=0 we call it as circular motion or not? kindly go through the question discussed :)

Answer 19

Firstly Centripetal force is not a force in itself. It is the name given to the force in the situation given to justify circular motion. For electron motion around nucleus the electrostatic force between the nucleus and the electron is the centripetal force.

Answer 20

Also there is difference between r=0 and r tends to zero. The orbital size of the electron is extremely small, but this doesn't means that it is zero.

Answer 21

Dear @ghazi IT IS NOT at all 99% of c.
It is about 0.5% of c - please STUDY th e excellent link I posted

Answer 22

http://www.pha.jhu.edu/~rt19/hydro/hydro.html

Answer 23

Mistake = "well in an atom electron revolves approximately at the speed of light."

Answer 24

\[d_{\text H}=1.1\text [Å]=1.1\times10^{-10}[\text m]\]
\[r_{\text H}=0.56\text [Å]=5.6\times10^{-11}[\text m]\]

Answer 25

Now to the DLS's question:
Surprisingly REAL electronic cloud-of-probability has actually a non-negligible "concentration" in the CENTER \[r = 0\]

Answer 26

Buuuut - it does not follow actuall Newtonian mechanics. The Centripetal force is INvalid notion for quantum distributions.

Answer 27

the radius of the atom is certainly not zero , and the electron is moving much less than the speed of light for most atoms,

Answer 28

Yeah, Unkel , and I was especially surprised about its non-zero weight in the "area" of the Nucleus !

Answer 29

\[F_=\frac{m_ev^2}r\]
\[F \frac{r_{\text H}}{m_e}=v^2\]\[v=\sqrt{F \frac{r_{\text H}}{m_e}}\]

Answer 30

\[F=\frac{1}{4\pi\epsilon_0}\frac{q_e^2}{r_{\text{H}}}\]
\[v=\sqrt{\frac{1}{4\pi\epsilon_0}\frac{q_e^2}{r_{\text{H}}} \frac{r_{\text H}}{m_e}}\]
\[v=\sqrt{\frac{1}{4\pi\epsilon_0}\frac{q_e^2}{m_e}}\]

Answer 31

Electrons in the lowest orbital of heavy elements do indeed travel at significant fractions of the speed of light, and it is necessary to use relativistic quantum mechanical calculations to calculate their properties.

Indeed, one famous result is the fact that mercury is a liquid at room temperature. The innermost electrons of mercury are traveling so fast they have effectively a much larger mass than ordinary electrons. Since the Bohr radius shrinks with increasing mass, that means the inner electrons of the mercury atom are drawn unusually close to the atom. That, in turn, makes the whole atom smaller, whcih increases the effective nuclear charge on the valence electrons (those in the outermost shell which can participate in chemical bonding).

Since the valence electrons in mercury are already experiencing a pretty high effective nuclear charge, because mercury is at the end of the 5d block, so that the 7s and 5d subshells have just filled, and there are as yet no 7p electrons, that means mercury has unusually weak chemical bonding activity. So weak, in fact, that the bonds between mercury atoms are not sufficiently strong to keep the element solid at room temperature.

The next few elements (thallium, lead, and bismuth) are also low-melting solids, but their bonds are stronger because they have 7p valence electrons. These are much better screened from the nucleus, by the 7s and 5d electrons, so they are more free to participate in chemical bonding.

Answer 32

Maybe I should also mention that the failure of all known theories of forces at distances approaching zero is well known. This is known as the problem of renormalization, and it pops up even more viciously in relativistic quantum mechanics than in the classical physics discussed here.

The reason is that at those very, very small distances, the energies of interaction become enormous. Consequently they support the creation of arbitrary numbers of virtual particles. For example, consider a point less than 1.4 fm (femtometers) away from an electron. At this distance, the electrostatic energy of interaction between an electron and a positron is equal to twice the rest mass of a an electron. So imagine an electron-positron pair spontaneously arises from nothing (which is possible in relativistic field theories), and that then the newly-created electron then flees to infinity (being strongly repelled by the first electron) and the newly-created positron sticks around. This does not violate conservation of energy! It is entirely possible.

Well...not quite, actually, because of the business about the new electron fleeing to infinity. That's a very slow and unlikely process. But what actually does happen is that electron-positron pairs are created, and the new positron moves closer to the electron, and the new electron moves further away, for a while, before they recollide and go back into the void. What you end up with in general is a cloud of positrons and electrons, with the positrons closer to the electron. This is sometimes called a polarization of the vacuum. It acts to reduce the observable charge on the electron, at long distances (like a few picometers!), just like embedding an ion in a polarizable medium like water reduces the observable charge on the ion at large distances.

But here's the thing: the polarization gets stronger and stronger the closer you get to the electron, because the amount of energy available through newly created electrostatic interactions gets higher and higher. In fact, as r->0 the polarization reaches infinity -- you have an infinite number of short-lived positrons infinitely close to the electron, surrounded by a shell (infinitismally far away) of short-lived electrons, an infinity of those, and so forth, in a boggling bollix of unsummable corrections.

This was one of the major early flaws in the quantum theory of electrodynamics, and it was solved (or "solved" -- not everyone agrees it's an acceptable solution) by "renormalization," where we agree that we cannot know the essential "bare" charge on the electron, and it may well be infinity, in order that the observable charge (the bare charge minus the infinite polarization) ends up finite. There are mathematical ways to express this which are less hand-wavy.