If an electron with a position uncertainty of #"1 Å"# has been accelerated through a potential difference of #"6 V"#, what is the uncertainty in the de Broglie wavelength?

I got #5# #Å#.

Well, first, the Heisenberg Uncertainty Principle applies to electrons:

#DeltaxDeltap >= ℏ//2#

where #ℏ = h//2pi# is the reduced Planck's constant, #x# is position, and #Deltap# is the uncertainty in momentum.

The first thing we can obtain is the inequality requirement for #Deltap# to double check the validity of what we actually get.

(No, this is not #Deltap# itself.)

#Deltap >= (ℏ//2)/(Deltax)#

#>= (6.626 xx 10^(-34) "kg"cdot"m"^2cdot"s"^(-1)//4pi)/(cancel"1 Å" xx (1 xx 10^(-10) "m")/(cancel"1 Å"))#

#>= 5.27 xx 10^(-25)# #"kg"cdot"m/s"#

One charge accelerated through a potential difference of #"6 V"# has attained kinetic energy #"6 eV"#. Since energy is a state function, we can choose #DeltaK# such that #K_i = "0 eV"# and #K_f = "6 eV"#.

At first glance, it may seem odd that we were told what the energy even was, but remember that the Uncertainty Principle is an inequality in general.

It is good to use it to verify that the value of #Deltap# that we get from #DeltaK# is valid.

Since #K = 1/2mv^2 = p^2/(2m)# for mass-ive particles,

#DeltaK = K_f = (Deltap)^2/(2m)#

so that the uncertainty in forward momentum, with the rest mass of #9.109 xx 10^(-31) "kg"# for an electron, is actually:

#Deltap = sqrt(2mK_f)#

#= sqrt(2 cdot 9.109 xx 10^(-31) "kg" cdot 6 cancel"eV" cdot (1.602 xx 10^(-19) "kg"cdot"m"^2"/s"^2)/(cancel"1 eV"))#

#= 1.32 xx 10^(-24) "kg"cdot"m/s"#

As it turns out, it really is larger than #5.27 xx 10^(-25) "kg"cdot"m/s"# like it should be.

So, the uncertainty here can be used to find that of the de Broglie wavelength. The wavelength itself is given by

#lambda = h/p#,

so the uncertainty is then (noting that since #h# is obviously a constant, there is no such thing as an uncertainty in #h#)

#color(blue)(Deltalambda) = h/(Deltap)#

#= (6.626 xx 10^(-34) "kg"cdot"m"^2"/s")/(1.32 xx 10^(-24) "kg"cdot"m/s")#

#= 5 xx 10^(-10) "m"#

#=# #color(blue)("5 Å")#