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

1 Answer
May 19, 2016

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"")#