# Heisenberg's uncertanty

1. May 26, 2015

### mss90

1. The problem statement, all variables and given/known data
particle pass through a slit of width 0.200mm. the wavelength is 633nm. after it pass through the slit they spread out over a range of angles . use uncertainty principle to determine the min. range of angles.

2. Relevant equations
(delta)p*(delta)y = h/4(pi)

3. The attempt at a solution
I have done some calculations using heisenbergs uncertainty principle but nothing like this.
I usually did it in y-direction but it seems this one needs to be done in x direction so I though;
(delta)px*(delta)x = h/4(pi) but would (delta)x be the width of the slit or the uncertanty of the particles in x-direction after passing through the slit? Also, how would (lambda) be integrated into this? The only thing I can think of is to use (delta)E * (delta)t = h/4(pi) and set E = hc/(lambda), however that would just give me the time? or use h/(lambda) * (delta)x = h/4(pi)

Need some hints here..

R

Last edited: May 26, 2015
2. May 26, 2015

### rolotomassi

You are only going the need to consider y direction from the looks of you since you are only confining the position in the y-plane.

The slit size is the uncertainty in the y position of the photon. So from this calculate the uncertainty in the y momentum, call it p.
Then the range in y momenta is from +p/2 to -p/2 . Use the expression for momentum and wavelength and consider the geometry.

3. May 26, 2015

### rolotomassi

Also p = h / lambda