Question

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?

Answer 1

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

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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.