1. Heisenberg Uncertainty Principle
Let’s find an electron
Photon changes the momentum of electron
x  
p  h/ (smaller , bigger p)
xp > h/4
x – uncertainty of position
p - uncertainty of momentum
Et > h/4
E - uncertainty of energy
t - uncertainty of time
if x = 3.5 ± 0.2 m, x = 0.4 m – it’s the range

2. Heisenberg Uncertainty Principle
xp > h/4
Et > h/4
Strange quantum effects:
Observation affects reality
Energy is not conserved (for t)
Non determinism
Quantum randomness
Quantum electrodynamics

3. Example: What is the uncertainty in the position of a 0
Example: What is the uncertainty in the position of a kg baseball with a velocity of 37.0 ± 0.3 m/s. (x = E-34 m)
The uncertainty of momentum is (0.145 kg)(0.6 m/s) = kg m/s
And now we use xp > h/4:
x(0.087 kg m/s) = (6.626E-34 Js)/(4), x = E-34 m
so x is ±3.03E-34 m
So not really very much.

4. For what period of time is the uncertainty of the energy of an electron 2.5 x 10-19 J?
Et > h/4
(2.5 x J)t > h/4
t = 2.1 x s
t = 2.1 x s

5. you know an electron’s position is ±
you know an electron’s position is ±.035 nm, what is the minimum uncertainty of its velocity? (4)
xp > h/4
p = mv
m = 9.11 x kg
x = x 10-9 m
(.070 x 10-9 m)p > (6.626 x Js)/4
p = 7.5 x kg m/s
p = mv
(7.5 x kg m/s) = (9.11 x kg)v
v = 8.3 x 105 m/s
v = 8.3 x 105 m/s