How do you calculate the uncertainty in velocity (in #"m"cdot"s"^(-1)#) of an electron (mass #9.11xx10^-31 kg#) under the conditions where the uncertainty in position is #4.782xx10^-3 m#?
1 Answer
This utilizes the following version of the Heisenberg Uncertainty Principle:
#\mathbf(DeltavecxDeltavecp_x >= ℏ//2)# where:
#h# is Planck's constant,#6.626xx10^(-34) "J"cdot"s"# and#ℏ = h//2pi# is the reduced Planck's constant.#Deltavecx# is the uncertainty in the position.#Deltavecp_x# is the uncertainty in the momentum.
When you solve this equation, simply change to an equal sign and you thus calculate the minimum uncertainty.
#Deltavecp_x ("min") = ℏ/(2Deltavecx)#
Now use the physics formula for the momentum,
#m_eDeltavecv_x = ℏ/(2Deltavecx)#
#color(green)(Deltavecv_x = ℏ/(2m_eDeltavecx))#
Thus, you have all you need to calculate the uncertainty in the velocity:
#color(blue)(Deltavecv_x) = (6.626xx10^(-34) "J"cdot"s")/(4pi(9.11xx10^(-31) "kg")(4.782xx10^(-3) "m"))#
No unit conversions are required, except for using
#= (6.626xx10^(-34) cancel("kg")cdot"m"^(cancel(2)^(1))"/s")/(4pi(9.11xx10^(-31) cancel("kg"))(4.782xx10^(-3) cancel("m")))#
#=# #color(blue)("0.0121 m/s")#
That's a pretty reasonable uncertainty for the velocity, given that the uncertainty in the position is so high (it's at least a million times the radius of an electron).
It makes sense because the Heisenberg Uncertainty Principle states that having high uncertainty on one of these observables means you'll have low uncertainty on the other observable.
Thus, having a fairly high uncertainty in the position means you should have a fairly low uncertainty in the momentum of this electron.
