Question

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?

Answer 1

`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?

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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")`.