Radial Probability: Finding Max Distance From Proton for 2s State Electron

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Homework Help Overview

The discussion revolves around determining the most probable distance from a proton for an electron in a 2s state, focusing on the radial probability function and its maximum. Participants are exploring the mathematical aspects of the problem, particularly the derivative of a fourth-degree polynomial derived from the radial probability expression.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the maximum of the radial probability function by taking its derivative and setting it to zero, but encounters difficulties with a fourth-degree polynomial. Some participants suggest that the derivative simplifies to a cubic equation, while others discuss the merits of using graphing calculators versus solving polynomials analytically.

Discussion Status

Participants are actively engaging with the mathematical problem, with some offering guidance on using graphing tools to find the maximum, while others advocate for solving the polynomial analytically. There is a recognition of differing opinions on the use of technology in solving mathematical problems.

Contextual Notes

There is an underlying tension regarding the appropriateness of using calculators in coursework, with some participants emphasizing the importance of understanding polynomial behavior over reliance on computational tools.

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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 [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?
 
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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.
 
4th degree polynomials can also be solved. A graphing calculator seems a poor way to solve a problem that admits a closed form solution.
 
loom91 said:
4th degree polynomials can also be solved. A graphing calculator seems a poor way to solve a problem that admits a closed form solution.
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
 

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