# 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

#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"# isPlanck's constant.#m# is themassof the object in#"kg"# .#v# is itsvelocityin#"m/s"# .

We know that the rest mass of an electron is **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

Now consider that *energy*, and an electron-volt (*energy*) is by definition the **work done** in *electron* **through a potential difference** of

**Work done**, *over* (not at) a distance

#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")# .