# What is the de Broglie wavelength of an electron that has been accelerated through a potential of #"10 V"#?

##### 1 Answer

#lambda = "0.388 nm"# .

Well, since an electron is a particle *with mass*, it can be described by the **de Broglie relation**:

#lambda = h/(mv)# where:

#h = 6.626 xx 10^(-34) "J"cdot"s"# is Planck's constant. Remember that#"1 J" = ("1 kg"cdot"m"^2)/"s"# .#m_e = 9.109 xx 10^(-31) "kg"# is the rest mass of an electron.#v# is its speed. We don't need to know that per se.

And since it is also moving with a certain speed, it will have a kinetic energy of:

#K = 1/2 mv^2 = p^2/(2m)# where:

#p = mv# is the linear momentum of the particle.#m# is the mass of the particle.#v# is the speed of the particle.

And so, we can get the *forward momentum* in terms of the kinetic energy:

#p = sqrt(2mK) = mv#

Lastly, note that in

Therefore, the **de Broglie wavelength** of the electron is:

#color(blue)(lambda) = h/sqrt(2m_eK_e)#

#= (6.626 xx 10^(-34) cancel"kg"cdot"m"^cancel(2)"/"cancel"s")/sqrt(2 cdot 9.109 xx 10^(-31) cancel"kg" cdot 10 cancel"eV" xx (1.602 xx 10^(-19) cancel"kg"cdotcancel("m"^2)"/"cancel("s"^2))/(cancel"1 eV"))#

#= 3.88 xx 10^(-10) "m"#

#=# #color(blue)("0.388 nm")#

And so, this is on the order of an X-ray photoelectron.