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?

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.