Is the Circular Orbit in a Multi-Electron Atom Stable?

In summary, in the classical model of a multi-electron atom, the electron's potential is given by -k/r e^-r/a. The circular orbit is unstable unless 0.5*(1+\sqrt{5})>r/a.
  • #1
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Homework Statement


In a classical model of a multi-electron atom, electrons are assumed to move in a modified electrostatic potential $V(r)$, given by;

$$V(r)=\dfrac{-k}{r}e^{-r/a}$$

Show that the effective potential is ;

$$V_e(r)=\dfrac{J^2}{2mr^2}+\dfrac{-k}{r}e^{-r/a}$$

Then show that the circular orbit is unstable unless;

$$ 0.5* (1+\sqrt{5}) \textgreater \dfrac{r}{a} $$

Homework Equations



Take the derivative of the effective potential and using the fact that it is zero at the radius of the circular orbit, express the constant k.

Then take the second derivative and because the orbit is stable the stationary point, at the radius of the circular orbit, must be a minima, hence the second derivative evaluated at the point is greater than 0 for orbit to be stable.

You insert the k from line 1 in order to simplify the equation and some terms cancel.

The Attempt at a Solution


I have tried to solve the problem multiple times and obtained;

$$k=e^{r/a}*\dfrac{J^2}{mr^3}(\dfrac{1}{r^2}+\dfrac{1}{a})^{-1}$$

this lead me to the inequality;

$$ \sqrt{1+\sqrt{2}} \textgreater \dfrac{r}{a}$$Could anyone tell me whether my approach is correct ??

Thank you
 
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  • #2
What did you do to get that inequality?

What you were supposed to do, as the statement of the problem said, was use the expression for k to remove k from the potential. Then take the second derivative of that expression, now with no k in it. Then determine what makes that second derivative greater than 0.

What did you do?
 
  • #3
That is exactly what I have done; I am attaching a figure with my working;
ieic0l.jpg
.
 
  • #4
Check your algebra when you take the first derivative. In particular, check what the derivative w.r.t. r of exp(-r/a) is.
 
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  • #5
DEvens said:
Check your algebra when you take the first derivative. In particular, check what the derivative w.r.t. r of exp(-r/a) is.

Thank you for pointing this out, I cannot believe I made such a basic mistake.
 

Related to Is the Circular Orbit in a Multi-Electron Atom Stable?

1. What is a circular orbit?

A circular orbit is the path that an object takes around another object due to the force of gravity. In a circular orbit, the distance between the two objects remains constant and the speed of the orbiting object remains constant.

2. How is the stability of a circular orbit determined?

The stability of a circular orbit is determined by the balance between the gravitational force of the central object and the centrifugal force of the orbiting object. If these forces are balanced, the orbit will remain circular and stable.

3. What factors affect the stability of a circular orbit?

The stability of a circular orbit is affected by the mass of the orbiting object, the mass of the central object, and the distance between them. The greater the mass of the orbiting object and the smaller the distance between the two objects, the more stable the orbit will be.

4. How can the stability of a circular orbit be increased?

The stability of a circular orbit can be increased by increasing the mass of the central object or increasing the distance between the two objects. This will increase the gravitational force and decrease the centrifugal force, making the orbit more stable.

5. Can a circular orbit ever become unstable?

Yes, a circular orbit can become unstable if external forces, such as the gravitational pull of other objects, act on the orbiting object. This can cause the orbit to become elliptical or even result in the object leaving the orbit altogether.

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