An electron traveling at 5.6 xx 10^5 "m/s" has an uncertainty in its velocity of 2.64 xx 10^(5) "m/s". What is the uncertainty in its position?
1 Answer
The minimum uncertainty in position is
Quantum particles like electrons and protons work nicely with the Heisenberg Uncertainty Principle:
DeltaxDeltap_x >= ℏ//2 where
Deltax is the uncertainty in position andp is the forward linear momentum.ℏ = h//2pi is the reduced Planck's constant andh = 6.626 xx 10^(-34) "J"cdot"s" .
The speed it is traveling at is actually irrelevant, but we can rewrite this in terms of the uncertainty in speed. A nonrelativistic treatment gives no change in the rest mass of the electron, so that
DeltaxcdotmDeltav_x >= ℏ//2
Therefore:
Deltax >= (ℏ//2)/(mDeltav_x)
As a result, the minimum uncertainty in the position is:
color(blue)((Deltax)_"min") = ((6.626 xx 10^(-34) "J"cdot"s")/(4pi))/(9.109 xx 10^(-31) "kg" cdot 2.64 xx 10^5 "m/s")
= color(blue)(2.193 xx 10^(-10) "m")
As it turns out, the relative uncertainty in momentum was:
(mDeltav_x)/(mv_x) xx 100% = (Deltav_x)/v_x xx 100%
= (2.64 xx 10^5 "m/s")/(5.6 xx 10^5 "m/s") xx 100%
= 47%
So the uncertainty in momentum is very large. That means we should have expected the uncertainty in position to be very small... and it is.