Question

If a proton has a speed of `(5.30 pm 0.01) xx 10^4 "m/s"`, what is the uncertainty in position?

Answer 1

The uncertainty in position is given by `Deltax`. And so, refer to Heisenberg's Uncertainty Principle:

`DeltaxDeltap_x >= ℏ/2`

Since you were given a speed of `(5.30 pm 0.01) xx 10^4 "m/s"`, you know the uncertainty is `Deltav_x = 0.01 xx 10^4 "m/s"`. That is the standard way to report values in the literature:

`"Avg"" "pm" ""STDEV"" " "units"`

And so, in doing that, you know the uncertainty in the momentum:

`Deltap_x = mDeltav_x`

`= 1.673 xx 10^(-27) "kg" cdot (0.01 xx 10^4 "m/s")`

`= 1.673 xx 10^(-25) "kg"cdot"m/s"`

Therefore, the uncertainty in the position is likely "big" (relative to `Deltap_x`). It must be, so that the uncertainty product is greater than or equal to `ℏ/2`. Solve for the uncertainty in `x` to get:

`color(blue)(Deltax) >= ℏ/2 cdot 1/(Deltap_x)`

`>= (6.626 xx 10^(-34) "J"cdot"s")/(2 cdot 2pi) cdot 1/(1.673 xx 10^(-25) "kg"cdot"m/s")`

`color(blue)(>= 3.152 xx 10^(-10) "m")`, or `"0.3152 nm"`, or `0.0003152` `mu"m"`.

And so, if a proton were to move around with that much uncertainty in speed due to observing it using a bacterial microscope (which is for things on the order of `mu"m"`), it would be somewhat blurry, but we could probably tell it was there, since the uncertainty is less than what the microscope lets us see.