Particle through a slit Uncertainty of position

darkfall13
Messages
30
Reaction score
0

Homework Statement



An air rifle is used to shoot 1.0-g particles at 100 m/s through a hole of diameter 2.0 mm. How far from the rifle must an observer be to see the beam spread by 1.0 cm because of the uncertainty principle? Compare this answer with the diameter of the Universe (2x10^26 m)

Homework Equations



\Delta{x}\Delta{p} \geq \frac{\bar{h}}{2}

The Attempt at a Solution



If we solve for \Delta{x} and expand p to mv and plug in we receive \Delta{x} = 5.275\cdot{10}^{-34}. So I believe I'm on the wrong track, how do we use the hole diameter (if even needed) and what else should I look at to begin this problem? Fraunhofer Diffraction? de Broglie waves?

Thank you!
 
Physics news on Phys.org
I realize this is an old thread, but I have the exact same question... so bump!
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top