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)