Heisenberg Uncertainty Principle Lets find an electron Photon

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

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

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. 0607 E-34 m) The uncertainty of momentum is (0. 145 kg)(0. 6 m/s) = 0. 087 kg m/s And now we use x p > h/4 : x(0. 087 kg m/s) = (6. 626 E-34 Js)/(4 ), x = 6. 0607 E-34 m so x is ± 3. 03 E-34 m So not really very much.

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 10 -19 J) t > h/4 t = 2. 1 x 10 -16 s

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 10 -31 kg x = 0. 070 x 10 -9 m (. 070 x 10 -9 m) p > (6. 626 x 10 -34 Js)/4 p = 7. 5 x 10 -25 kg m/s p = m v (7. 5 x 10 -25 kg m/s) = (9. 11 x 10 -31 kg) v v = 8. 3 x 105 m/s