Quantum mechanics probability detection

1. Dec 6, 2013

hvthvt

1. The problem statement, all variables and given/known data
A one-dimensional infinite well of length 200 pm contains an electron in its third excited state. We
position an electron-detector probe of width 50 pm so that it is centered within the well. (a) What is the probability of detection by the probe? (b) If we insert the probe as described 1000 times, how many times should we expect the electron to materialize on the end of the probe (and thus be detected)?

My question is: since the width is 50 pm, the interval is 50 pm, which is "relatively big". (E.g. 2.0pm would be small)
Should I integrate the formula which is given for probability OR is 50 pm small enough to see it as constant? Meaning that the probability would be 20% ?

2. Relevant equations

p=ψ2dx

3. The attempt at a solution

(√2/L * sin(xnπ/L))2 dx

2. Dec 6, 2013

ShayanJ

Its you who decides 50pm is small enough or not because it depends on the accuracy that you want!

3. Dec 6, 2013

hvthvt

Oh really, haha :) I guess that my answer would be wrong on the test if I would not integrate

4. Dec 6, 2013

ShayanJ

Ohhh...looks like I should correct myself.Its your professor who decieds 50pm is small enough or not.And if he thought so,he would write it in the question.So yeah,you should integrate!

5. Dec 6, 2013

hvthvt

So how should I integrate this?
What are the limits? 75 and 125? because 50 is in the center of the 200.
So∫2/L*sin(3∏/200pm*x) integrating from 7 to 125. How should I work this out?
I should take y=3pix/L ??

6. Dec 6, 2013

ShayanJ

The integral is:
$\frac{2}{200}\int_{75}^{125} \sin^2{(\frac{3 \pi x}{200})} dx=\frac{1}{100}\int_{75}^{125}\frac{1-\cos{(\frac{3\pi x}{100})}}{2}dx=\frac{1}{200}[\frac{1}{2}x|_{75}^{125}-\frac{100}{6 \pi}\sin{(\frac{3\pi x}{100})}|_{75}^{125}]$
Where I have used the identity $\sin^2{x}=\frac{1}{2}(1-\cos{2x})$

7. Dec 6, 2013

hvthvt

Ooooh.. That works well. Thank you very much !!!!