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b2386

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Answer: The most probable radius is where the radial probability is maximum. We are given [tex]R_{2,0}(r)=\frac{1}{(2a_0)^{3/2}}2(1-\frac{r}{2a_0})e^\frac{-r}{2a_0} so we have[/tex]

[tex]P(r) = r^2(R_{2,0}(r))^2=r^2[\frac{1}{(2a_0)^{3/2}}2(1-\frac{r}{2a_0})e^\frac{-r}{2a_0}]^2[/tex]

[tex]= \frac{4}{(2a_0)^3}(r^2)(1-\frac{r}{a_0}+\frac{r^2}{4(a_0)^2})(e^{\frac{-r}{a_0}})[/tex]

To find the maximum, we set the derivative to zero. The normalization constant of [tex]\frac{4}{(2a_0)^3}[/tex] may be ignored.

At this point, I take the derivative and set it to zero. However, I end up with a 4th degree polynomial.

I know the answer is [tex]5a_0[/tex] or close to that, but when plugging that value into my derivative, I cannot get the solution to equal zero as needed. Can someone please offer some help on where my problem lies?