1. Nov 22, 2006

b2386

Question: What is the most probable distance from the proton of an electron in a 2s state?

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

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

$$= \frac{4}{(2a_0)^3}(r^2)(1-\frac{r}{a_0}+\frac{r^2}{4(a_0)^2})(e^{\frac{-r}{a_0}})$$

To find the maximum, we set the derivative to zero. The normalization constant of $$\frac{4}{(2a_0)^3}$$ 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 $$5a_0$$ 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?

2. Nov 22, 2006

OlderDan

Every term in the derivative has a factor of r times the exponential, so it reduces to a cubic. Cubics can be solved, and in the old days we did that sort of thing. Now there are graphing calculators to help you. If you can use the calculator, graph the function and find the max at around 5.23 a_o. If not, look up the cubic solution methods.

3. Nov 24, 2006

loom91

4th degree polynomials can also be solved. A graphing calculator seems a poor way to solve a problem that admits a closed form solution.

4. Nov 24, 2006

OlderDan

There is nothing poor about using tools to solve problems. There is something poor about being ignorantly dependent on them, a deplorable condition that does seem rather pervasive these days. I agree with you that knowledge of the behavior of polynomials is more than worthwhile. But I don't think it is necessary to go back to the potentailly tedious calculations involved every time you need to solve one of them. If use of the tool is acceptable in the context of the coursework the OP is engaged in, perhaps their time is better spent learning more physics