What is the speed of an electron with a wavelength of #"0.1 nm"#? If this electron were brought up to this speed from rest, what potential difference was needed?

1 Answer
Sep 8, 2017

#v = 7.27 xx 10^6 "m/s"#

#|V| = "150.43 V"#

Could we have done this problem with a photon? Why or why not?


An electron, being a mass-ive particle, follows the de Broglie relation:

#lambda = h/(mv)#

where:

  • #h = 6.626 xx 10^(-34) "J"cdot"s"# is Planck's constant.
  • #m# is the mass of the object in #"kg"#.
  • #v# is its velocity in #"m/s"#.

We know that the rest mass of an electron is #9.109 xx 10^(-31) "kg"#. So, its forward/positive velocity (which we call the speed) is given by:

#color(blue)(v) = h/(lambdam)#

#= (6.626 xx 10^(-34) cancel"kg"cdot"m"^cancel(2)"/s")/(0.1 cancel"nm" xx (cancel"1 m")/(10^9 cancel"nm") xx 9.109 xx 10^(-31) cancel"kg")#

#=# #color(blue)ul(7.27 xx 10^6 "m/s")#

In order to have this speed, since it has a mass, it must also have a kinetic energy of:

#K = 1/2 mv^2#

#= 1/2 (9.109 xx 10^(-31) "kg")(7.27 xx 10^6 "m/s")^2#

#= 2.41 xx 10^(-17) "J"#

Recall that an electron's charge is #-1.602 xx 10^(-19) "C/e"^(-)#.

Now consider that #"1 V"cdot"C" = "1 J"#, is a unit of energy, and an electron-volt (#"eV"#, also a unit of energy) is by definition the work done in #"J"# required to push one electron through a potential difference of #"1 V"#.

Work done, #W#, on a mass over (not at) a distance #Deltavecx# is:

#W = vecFDeltavecx#

It follows that we (somewhat) analogously have the relationship:

#W = underbrace(overbrace((1.602 xx 10^(-19) "C")/cancel("1 e"^(-)))^"Electrical 'mass'" xx overbrace("1 V")^"Electrical 'distance'")_"Work done on one electron" xx cancel("1 e"^(-))/("1 eV")#

#=> 1.602 xx 10^(-19) "J"# for every #"1 eV"#

And thus, the magnitude of the energy involved was:

#2.41 xx 10^(-17) cancel"J" xx ("1 eV")/(1.602 xx 10^(-19) cancel"J")#

#=# #ul"150.43 eV"#

And by definition, we thus have that the magnitude of the potential difference was:

#color(blue)(|V| = ul"150.43 V")#.