Question

From the Heisenberg uncertainty principle, how do you calculate Δx for an electron with Δv = 0.300 m/s?

Answer 1

A formulation of the Heisenberg Uncertainty Principle is (Physical Chemistry: A Molecular Approach, McQuarrie):

`\mathbf(DeltavecxDeltavecp_x >= h)`

where:

• `Deltavecx` is the uncertainty in the position.
• `Deltavecp_x` is the uncertainty in the momentum.
• `h = 6.626xx10^(-34) "J"cdot"s"` is Planck's constant.

Recall from physics that momentum is defined as `vecp = mvecv`, where `m` is mass and `vecv` is velocity.

Since `vecp = mvecv`, `Deltavecp_x = m_eDeltavecv_x`, assuming the mass `m_e` of an electron is constant.

Solving for `Deltavecx`, we would start by ignoring the greater-than sign, because if we get a value that satisfies the equal sign, we establish the bare minimum requirement to satisfy this principle.

`color(blue)(Deltavecx) = h/(Deltavecp_x)`

`= color(blue)(h/(m_eDeltavecv_x))`

`= (6.626xx10^(-34) "J"cdot"s")/((9.109xx10^(-31) "kg")("0.300 m/s"))`

`= (6.626xx10^(-34) "kg"cdot"m"^2"/s"^cancel(2)cdotcancel("s"))/((9.109xx10^(-31) "kg")("0.300 m/s"))`

`= (6.626xx10^(-34) cancel("kg")cdot"m"^cancel(2)"/"cancel("s"))/((9.109xx10^(-31) cancel("kg"))("0.300" cancel("m/s")))`

`=` `color(blue)("0.00242 m"`

---

And indeed, the Heisenberg Uncertainty Principle is satisfied:

`DeltavecxDeltavecp_x stackrel(?)(>=) h`

`= stackrel(Deltavecx)overbrace(("0.00242 m"))stackrel(Deltavecp_x)overbrace([(9.109*10^(-31) "kg")("0.300 m/s")])`

`= 6.61 xx 10^(-34) "J"cdot"s" > h` `color(green)(sqrt"")`