atom

Afrasastrasz SG

my chemistry teacher posed my class this question .none can answer.

y does electron not attract to the nucleus of atom instead spin around.

after this he say,

y does the moon not crash into the earth instead just spin around

how to answer

i said centrifugal forces but he ask me wat about it i too lazy to expplain

teraexa SG

It's actually about centripetal forces (not centrifugal).

In the case of an electron around a nucleus, the centripetal force resulting from the electronic attraction keeps the electron at a stable orbit around the nucleus. (ok in reality it's not an orbit model but i shan't confuse you here)

Same for the earth and moon. The gravitational force between the earth ad moon is equivalent to the centripetal force and hence the orbit is stable. (even though in reality the orbit is somewhat decaying but i shan't confuse you again)

Afrasastrasz SG

It's actually about centripetal forces (not centrifugal).

oh nice answer thanks!

UltimaOnline SG

For those interested in a University level discussion on this topic, visit :

http://www.advancedphysics.org/forum/showthread.php?t=7892

Some excerpts (the full discussion is 7 pages long).

>>> Using the Bohr atom which is an approximation of the true state of affairs. the electrostatic attraction F = k(q1*q2)/r^2 is opposed by the centrifugal force F = mV^2/r so that at some distances from the nucleus k(q1*q2)/r^2 = mV^2/r.
This is why the electron only orbits at distances from the nucleus that are related. They are quantised. At intermediate distances the forces are unbalanced. <<<

>>> If the electron were sitting still at a certain distance away from the nucleus, then the coulomb force would pull the electron into the nucleus. Since the electron is moving with a given velocity, call it V, then the coloumb force causes centripetal acceleration towards the nucleus. Remember, in centrifugal motion there is no centripetal velocity, only acceleration. Therefore, this combination of the tangential velocity and centripetal acceleration causes the electron to orbit around the nucleus.

Also, consider the p-orbital in an atom. If we solve schrodinger's 1-D equation for the n = 2 orbital, we find that the electron is allowed to exist anywhere except very close to the nucleus or very far away from it (at the end of the orbital radius).

Remember, however that this is only the one dimensional equation. If you solved the 1-D schrodinger equation for n=1, or for your hydrogen atom for example, we find that the electron can exist anywhere except the end of the orbital radius. So from the result it would seem that the electron is actually allowed to move closer and closer to the nucleus. This is not the case since the waveform for the n = 1 solution allows the electron to exist in a hollow sphere, or a 2-manifold. If you were to solve the 3-D schrodinger equation for n = 1, then you would get a wave form similar to the solution for 1-D at n=2; in other words if you solve the 3-D equation, you would find that the probability of finding the electron as you approach the nucleus decreases to zero exponentially. <<<

>>> look at the quantum world of probability.
The electron is a wavepacket with a wave function.
the probability of finding the electron at a given location in the volume around the nucleus. You use spherical polar coordinates. the probability of finding the electron in volume element dV located at r,theta, phi.
We are only interested in r, so we integrate over all possible values and theta and phi are equal to unity. The probability for r - Pr = r^2lRn,t(r)l^2
here's a graph of the result for n = 1
Note that there are probabilities for the electron to be at an r other than a0, but small. The probability of it being at a0 is 0.55. <<<

>>> First point: Classical or non classical orbit.

If you want to talk about a classical orbit you must have negligible quantum effects. Let's call Delta x and Delta p the incertitude on the radius r of the orbit and on the momentum.

The orbit is classical if:

Delta x << r
Delta p<< p

So we have:

(Delta x)/r *(Delta p)/p << 1

But the uncertainties relations impose:

(Delta x)/r *(Delta p)/p >> h/pr

Now using the Bohr formula: pr=nh

One gets:

(Delta x)/r *(Delta p)/p >> 1/n

ie: The orbits are classical for large quantum number n

Now a simple evaluation of r0, the radius of the fundamental.

V=-e^2/r0

The kinetic energy can never be zero due to the uncertainty principle:

T>= Tmin= h^2/2mr0^2

So if one decreases r0 the potential energy decreases but the kinetic energy increases.

The minimal mechanical energy is thus:

Emin=h^2/2mr0^2-e^2/r0

It reaches a minimum when: r0=h^2/me^2 = Bohr radius.

The main points:

r0 is not the dimension of the confining potential (size of the box).
The uncertainty principle sets a minimum on T that strongly affect the stability behaviour of the system. <<<

The full discussion is 7 pages long :

http://www.advancedphysics.org/forum/showthread.php?t=7892

Audioboxing SG

I would say that a more correct answer would be the law of inertia, which states that if an object is moving, it continues to move at uniform velocity (same direction and speed), untill an external force is applied.

Now, if the force is always perpendicular to the velocity of the electron, the speed of the electron cannot change. So what must change is the direction. If you go into detail, you will find the electron to move in a circle in such circumstances.

So it doesn't crash, because of this argument! At least for the moon, because electrons are very small particles that need another set of laws to correctly describe.

With regards,

Audioboxing

New lurker in the homework section. :P

-------------------------------

Details, if you are interested:

For simplicity we assume for the orbit, the accelaration towards the center would be always perpendicular to the velocity of the orbiting object.

This kind of accelaration causes a very small velocity component towards the center in a very small amount of time.

The resultant velocity thus changes only it's direction, and not it's magnitude because the velocity imparted from the attractive force towards the center is very small.

The changes in direction are the same at every position of the orbit, which follows from similar arguments.

Hence the orbit is circular, which implies that the distance of the object to the center does not change (It doesn't crash!).

For a certain force and a certain velocity, only a certain circle with a certain radius may be formed.

In fact, the electron around the nucleus will spiral in and crash(!) within fractions of a second, if one based the entire argument on inertia and centripetal forces only.

Because the velocity of the electron will be too low compared to the velocity imparted by the force towards the center! Too fully understand the motion of electrons around the nucleus, you need quantum mechanics, which is mentioned by the above poster!