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
Jan 23, 2018

The minimum uncertainty in position is 2.193 xx 10^(-10) "m".


Quantum particles like electrons and protons work nicely with the Heisenberg Uncertainty Principle:

DeltaxDeltap_x >= ℏ//2

where Deltax is the uncertainty in position and p is the forward linear momentum. ℏ = h//2pi is the reduced Planck's constant and h = 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 Deltap_x = mDeltav_x.

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.