Using slope=0 technique to find most likely spherical shell

Question

In this PDF

http://riedo.gatech.edu/Teaching/Modern_Physics/hw/HW3_2010_MP_SOL.pdf

problem#1, the instructor solves the question of which spherical shell (what radius $r$) has the greatest probability for the 3p hydrogen electron to sit in.

I understand everything except the one step where she throws out a factor of $r^4$.

The wave function for the 3p electron is:

$\dfrac{4}{(81 \sqrt 6)}(6 - r)(r)e^{-r/3}$

(I neglect the $a_0$ factors as, like other constants, they would not affect the calculation of where the points are where $d(probability\_of\_finding\_electron)/dr = 0$.)

Indeed the first steps are to throw out all unnecessary factors: we can dump all constant factors like $a_0$ and $81 \sqrt 6$ as well as the $4\pi$ in the spherical shell surface area expression $4 \pi r^2$

I.e. the probability for each particular shell is

$(4 \pi r^2)(\Delta r) \cdot \Psi^{*} \Psi$

And dumping all the constant factors including $\Delta r$ leaves us with:

$r^2 [(6-r)(r)e^{-r/3}]^2$

=

$r^4 [(6-r)^2 e^{-2r/3}]$

as the expression to be minimized.

Now the instructor then throws out the $r^4$, saying we can neglect solutions at $r=0$.

But doesn't this actually change the expression when you take $d(probability\_of\_finding\_electron)/dr$?

In other words, the zeroes of:

$\frac{d}{dr}(r^4 [(6-r)^2 e^{-2r/3}])$

are different from

$\frac{d}{dr}([(6-r)^2 e^{-2r/3}])$?

Would you please tell me why it's okay to throw out the $r^4$?

Answer 1

The instructor hasn't thrown out $r^4$. If you do the calculations properly you will get the desired result. What the instructor is telling is that while simplifying and taking commons out for $\frac{\text{dP(r)}}{\text{dr}}=0$ you get,$$(\text r^3)(6-\frac{\text r}{\text a_0})(f(\text r))=0$$which gives, $\text r=0$ and $6\text a_0$ from the first two terms. But these two values correspond to minimum probability positions as obtained from the expression for $\text P(\text r)$ by setting $\text P(\text r)=0$. So, for maximum probability positions the first two terms of the above equation cannot be zero as they are zero for minimum probability positions. Hence, you have to check the solutions from $f(\text r)=0$.

Answer 2

Firstly, you cannot simply 'neglect' $a_0$ (as the *.pdf actually shows).

In some test books the following substitution is used: $\rho=\frac{r}{a_0}$, so the $a_0$ doesn't have to be 'carried around'.

As your expression for $P(\rho)$ is made of three factors, each factor can be evaluated for extrema individually. $\rho=0$ is an obvious minimum from the first factor, $\rho=\infty$ is another from the exponential term.

The middle factor is evaluated by deriving it to $\rho$, setting to $0$ and that gives:

$\frac{2}{3}\rho^2-10\rho+24=0$.

Or $\rho=3$ and $\rho=12$.

Since as $P(\rho)$ is highest for $\rho=12$, we find that the most probable radial position is $r=12a_0$.