IPA Potential Energy Approximation

[TerraForce469]

Homework Statement

The IPA potential-energy function $U(r)$ is the potential energy "felt" by an atomic electron in the average field of the other $Z-1$ electrons plus the nucleus. If one knew the average charge distribution $p(r)$ of the $Z-1$ electrons, it would be a fairly simple matter to find $U(r)$.

One might guess that $p(r)$ is spherically symmetric and given by $$p(r)=p_0e^{-\frac{r}{R}}$$ where $R$ is some sort of mean atomic radius.

(a) Given that $p(r)$ is the average charge distribution of $Z-1$ electrons, find $p_0$ in terms of $Z$, $e$, and $R$.

(b) Use Gauss's law to find the electric field $E$ at a point $r$ due to the nucleus and the charge distribution $p$.

(c) Verify that as $r$ approaches $0$ and $r$ approaches $∞$, $E$ behaves as required by the independent particle approximation and the IPA energy levels.

Homework Equations

Gauss' law:

$\int E\ d(4πr^2)=\frac{Q}{ε_0}$

The Attempt at a Solution

I can't even be sure how to come up with an expression for $p_0$... but here goes:

$p_0=\frac{Ze}{4/3πR^3}$

Is this correct, or at least the right direction?

Please, any help on how to approach this problem?

EDIT: there are two uses of $e$ in this problem; exponential (shown in $p(r)$) and electron charge (solved with part (a)).

 

[TerraForce469]

Wait, it is the average charge distribution of the $(Z-1)$ electrons, so it probably should include this term instead of $Z$.

Skipping over part (a), and assuming $p_0$ is constant, the charge due to the electron charge distribution should be $Q=p(r)r$ where $r$ is the distance from the nucleus. So from Gauss' law I have

$E(4πr^2)=\frac{p(r)r}{ε_0}$

Solve for $E$ to obtain whatever it is. Would that be the correct expression for $E$?

 

[rude man]

Better start with (a).

Given a total charge Q uniformly distributed within a sphere of radius R, what is the potential at R?

 

[TerraForce469]

Wouldn't it be... $-\frac{kQ}{R}$? But this is assuming $E=-\frac{kQ}{r^2}$.

But how would that help me find $p_0$?

 

[rude man]

Wouldn't it be... $-\frac{kQ}{R}$? But this is assuming $E=-\frac{kQ}{r^2}$.

But how would that help me find $p_0$?

It's +kQ/R actually.

Well, if p(r) = p0 exp-(r/R) and p(r) = kQ/R when r= R, isn't it pretty obvious?

 

[TerraForce469]

It's +kQ/R actually.

Well, if p(r) = p0 exp-(r/R) and p(r) = kQ/R when r= R, isn't it pretty obvious?

That is the expression for potential. I am looking for $p(r)$ which is the charge density.

 

[rude man]

OK. Charge density is usually written with a ρ (greek letter 'rho'), not p.

So, ρ(r) = ρ₀exp(-r/R). So if you knew the total charge Q , which you do, what relates Q to ρ(r)? How about Q = ∫ρ(r)dV?

 

[TerraForce469]

OK. Charge density is usually written with a ρ (greek letter 'rho'), not p.

So, ρ(r) = ρ₀exp(-r/R). So if you knew the total charge Q , which you do, what relates Q to ρ(r)? How about Q = ∫ρ(r)dV?

Yeah I know, but it's more convenient to type and it was already implied in this context.

So the electron cloud has $(Z-1)$ electrons, meaning total charge = $(Z-1)e$. Average charge density should be $ρ= \frac{(Z-1)e}{4/3πR^3}$.

However the picture we are looking at is that of an electron cloud surrounding the nucleus (at least I think it is), so the given expression for $ρ(r)$ doesn't even make sense to me, much less how to find $p_0$, except that it exponentially decreases as r increases.

Indeed $Q = ∫ρ(r)dV$ is the correct expression to evaluate the total charge at a given distance, but since it only has an $r$ dependence how would $dV$ be transformed?

 

[rude man]

Well, what would be dV of a thin spherical shell of thickness dr, with all the shells making up your sphere?

You can't deal with averages, you have to do an integration.

Don't worry abou the expression for charge density making sense. It doesn't. An electron isn't even considered a particle in quantum mechanics any more! (It's now a "wave function"!). But this is an exercise intended for you to become familiar with basic calculus and classical physics.

 

[TerraForce469]

Well, what would be dV of a thin spherical shell of thickness dr, with all the shells making up your sphere?

You can't deal with averages, you have to do an integration.

Don't worry abou the expression for charge density making sense. It doesn't. An electron isn't even considered a particle in quantum mechanics any more! (It's now a "wave function"!). But this is an exercise intended for you to become familiar with basic calculus and classical physics.

Yes, in part (b) when I apply Gauss' I will have to integrate the charge density with respect to $dV$ in order to obtain the total charge. And if I obtain the correct expression then my results should be consistent with part (c).

Nevertheless I still cannot be too sure for the expression of $ρ_0$. This is the part that's been bugging me.

 

[rude man]

ρ(r) has ρ₀ in it so when you evaluate Q = ∫ρdV the only unknown left is ρ₀.

 

[TerraForce469]

Alright, so I set $(Z-1)e=∫ρ_0e^{-\frac{r}{R}}(4πr^2dr)$ from 0 → ∞ because I want to evaluate the total charge distribution.

I get $ρ_0=\frac{(Z-1)e}{8πR^3}$.

Is this correct by far?

 

[rude man]

Providing you performed the integration correctly - straight A!

 

[TerraForce469]

Providing you performed the integration correctly - straight A!

Thank you so much for your help!