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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 - PowerPoint PPT Presentation
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Heisenberg Uncertainty PrincipleLet’s find an electronPhoton changes the momentum of electronx p h/ (smaller , bigger p)xp > h/4
x – uncertainty of position p - uncertainty of momentum
Et > 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
Heisenberg Uncertainty Principlexp > h/4
Et > h/4Strange quantum effects:• Observation affects reality• Energy is not conserved (for t)• Non determinism • Quantum randomness• Quantum electrodynamics
Example: What is the uncertainty in the position of a 0.145 kg baseball with a velocity of 37.0 ± 0.3 m/s. (x = 6.0607E-34 m)
The uncertainty of momentum is (0.145 kg)(0.6 m/s) = 0.087 kg m/s
And now we use xp > h/4:x(0.087 kg m/s) = (6.626E-34 Js)/(4), x = 6.0607E-34 mso x is ±3.03E-34 m
So not really very much.
t = 2.1 x 10-16 s
Et > h/4
(2.5 x 10-19 J)t > h/4
t = 2.1 x 10-16 s
For what period of time is the uncertainty of the energy of an electron 2.5 x 10-19 J?
v = 8.3 x 105 m/s
xp > h/4
p = mvm = 9.11 x 10-31 kgx = 0.070 x 10-9 m(.070 x 10-9 m)p > (6.626 x 10-34 Js)/4p = 7.5 x 10-25 kg m/sp = mv(7.5 x 10-25 kg m/s) = (9.11 x 10-31 kg)vv = 8.3 x 105 m/s
you know an electron’s position is ±.035 nm, what is the minimum uncertainty of its velocity? (4)