[CHM 331] Part B Chapter 15 Particle in a Box

[Andrew J. Pounds]

On 02/07/2017 08:09 PM,wrote:
>
> Dr. Pounds,
>
>
> On Part B of the Chapter 15 Particle in a Box problem that deals with 
> P(x) how am I to determine the integral parameters? Like for E1 I want 
> to use 0.5/a but I don't get any of the answer choices when I do it 
> that way. And when I use a/0.5 then I can't cross out the a's in 
> sin(2pix/a). Does "a" have a value that I'm just not getting? So far I 
> have
>
> 2/a[x-(a/2pi)sin(2pix/a)]
>
> It's just the integral parameters that I don't get.
>
>
> Thanks,
>
>
>

So I am assuming that you are talking about problem 15.34.   On part B 
all space would be on the open interval $(-\infty,\infty)$, but we know 
that for this box, since it has a length of $l$, that "all space" will 
be on the closed interval $[0,a]$. If you integrate $\psi^*\psi$ on that 
interval you will get one for the probability.  In other words there is 
a 100% probability of finding the particle in the box.


So, what if we want to find the probability of finding the particle on 
the closed interval [0.32 a, 0.35 a] then I set up the following 
integration.


$\int_{0.32 a}^{0.35 a} \frac{2}{a} \sin^2\left(\frac{3 \pi x}{a}\right) dx$


Mathematica can do this integral easily, or you can use the integral 
tables on page 1016 of your text.   You should get a small positive number.




-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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