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

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SUMMARY

The discussion focuses on the stability of circular orbits in a multi-electron atom modeled by the effective potential \( V_e(r) = \frac{J^2}{2mr^2} - \frac{k}{r} e^{-r/a} \). It is established that circular orbits are unstable unless the condition \( 0.5(1+\sqrt{5}) > \frac{r}{a} \) is satisfied. Participants emphasized the importance of correctly calculating the first and second derivatives of the effective potential to determine stability, highlighting a common mistake in differentiating the exponential term.

PREREQUISITES
  • Understanding of classical mechanics and electrostatic potential
  • Familiarity with effective potential in multi-electron systems
  • Knowledge of calculus, specifically derivatives and their applications in physics
  • Experience with inequalities and stability analysis in dynamical systems
NEXT STEPS
  • Review the derivation of effective potential in multi-electron atoms
  • Study the implications of the second derivative test for stability in classical mechanics
  • Learn about the mathematical treatment of exponential functions in differentiation
  • Explore the physical significance of parameters like \( J \), \( m \), and \( a \) in atomic models
USEFUL FOR

Physics students, researchers in atomic and molecular physics, and anyone studying the dynamics of multi-electron systems will benefit from this discussion.

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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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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?
 
That is exactly what I have done; I am attaching a figure with my working;
ieic0l.jpg
.
 
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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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.
 

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