Probabilities for Electron in a Box (n=1 & n=2)

  • Thread starter Thread starter Ming0407
  • Start date Start date
  • Tags Tags
    Box Particle
Ming0407
Messages
8
Reaction score
0
What are the probabilities of finding the electron anywhere between x=0 and x=L/4? (n=1 and n=2)


Can you give example to me?
 
Physics news on Phys.org
Do you know the wave function for a particle in a box, or can you try to derive them?
 
http://user.mc.net/~buckeroo/PODB9.gif true? n=1 or n=2 L=L/4? this is answer?
 
Last edited by a moderator:
That is the correct wave function. Can you use that to predict the probability in being in one section of the box? Remember that \left|\psi\left(x\right)\right|^2 is a probability density, so you have to integrate over some interval in x (The right interval is specified in your problem, can you spot it?).
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top