Question

How do you Calculate the Δx for each of the following?

a.) an electron with Δv= 0.100 m/s
b.) a baseball with mass 145 g with Δv = 0.100 m/s

Answer 1

The version of the Heisenberg Uncertainty Principle as defined with uncertainties `Deltavecx` and `Deltavecp_x` is:

`\mathbf(DeltavecxDeltavecp_x >= h)`

Recall from physics that momentum is `vecp_x = mvecv_x` for mass `m` and velocity `vecv`, and that Planck's constant is `h = 6.626xx10^(-34) "J"cdot"s"`, or `"kg"cdot"m"^2"/s"`.

So, if you want to calculate the uncertainty in position:

`DeltavecxmDeltavecv_x >= h`

`color(green)(Deltavecx >= h/(mDeltavecv_x))`

You could resolve the inequality by simply solving for the minimum uncertainty of the x position. Then you would get:

1. `color(blue)(Deltavecx) = h/(m_eDeltavecv_x) = (6.626xx10^(-34) "kg"cdot"m"^2"/s")/((9.109xx10^(-31) "kg")("0.100 m/s")) = color(blue)(7.27xx10^(-3) "m")`

2. `color(blue)(Deltavecx) = h/(m_"ball"Deltavecv_x) = (6.626xx10^(-34) "kg"cdot"m"^2"/s")/(("0.145 kg")("0.100 m/s")) = color(blue)(4.57xx10^(-32) "m")`

The physical interpretation is...

• You know to an OK certainty as to what the position of the baseball is. The radius of an average baseball is about `7.4xx10^(-2) "m"`, so your uncertainty is less than `pm 10%`.
• For an electron, the uncertainty in velocity is quite low (an electron normally moves very quickly, and `"0.100 m/s"` is sluggish for an electron), and thus, the uncertainty for its position is quite high.

Think about the radius of an electron, which is somewhere around `10^(-16) "m"`. That means `7.27xx10^(-3) "m"` is approximately `10^13` times as long as the radius of an electron. That's a horrendous uncertainty!