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

Oct 10, 2017

$\lambda = \text{0.388 nm}$.

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

$\lambda = \frac{h}{m v}$

where:

• $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$ is Planck's constant. Remember that $\text{1 J" = ("1 kg"cdot"m"^2)/"s}$.
• ${m}_{e} = 9.109 \times {10}^{- 31} \text{kg}$ 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:

$K = \frac{1}{2} m {v}^{2} = {p}^{2} / \left(2 m\right)$

where:

• $p = m v$ 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:

$p = \sqrt{2 m K} = m v$

Lastly, note that in $\text{1 eV}$, there is $1.602 \times {10}^{- 19}$ $\text{J}$, i.e. one needs $\text{1 eV}$ to push one electrons' worth of charge through a potential difference of $\text{1 V}$ in an electric field.

Therefore, the de Broglie wavelength of the electron is:

$\textcolor{b l u e}{\lambda} = \frac{h}{\sqrt{2 {m}_{e} {K}_{e}}}$

= (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"))

$= 3.88 \times {10}^{- 10} \text{m}$

$=$ $\textcolor{b l u e}{\text{0.388 nm}}$

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