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?

Sep 8, 2017

$v = 7.27 \times {10}^{6} \text{m/s}$

$| V | = \text{150.43 V}$

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:

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

where:

• $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$ is Planck's constant.
• $m$ is the mass of the object in $\text{kg}$.
• $v$ is its velocity in $\text{m/s}$.

We know that the rest mass of an electron is $9.109 \times {10}^{- 31} \text{kg}$. So, its forward/positive velocity (which we call the speed) is given by:

$\textcolor{b l u e}{v} = \frac{h}{\lambda m}$

$= \left(6.626 \times {10}^{- 34} \cancel{\text{kg"cdot"m"^cancel(2)"/s")/(0.1 cancel"nm" xx (cancel"1 m")/(10^9 cancel"nm") xx 9.109 xx 10^(-31) cancel"kg}}\right)$

$=$ $\textcolor{b l u e}{\underline{7.27 \times {10}^{6} \text{m/s}}}$

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

$K = \frac{1}{2} m {v}^{2}$

$= \frac{1}{2} {\left(9.109 \times {10}^{- 31} \text{kg")(7.27 xx 10^6 "m/s}\right)}^{2}$

$= 2.41 \times {10}^{- 17} \text{J}$

Recall that an electron's charge is $- 1.602 \times {10}^{- 19} {\text{C/e}}^{-}$.

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

Work done, $W$, on a mass over (not at) a distance $\Delta \vec{x}$ is:

$W = \vec{F} \Delta \vec{x}$

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")

$\implies 1.602 \times {10}^{- 19} \text{J}$ for every $\text{1 eV}$

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")

$=$ $\underline{\text{150.43 eV}}$

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

$\textcolor{b l u e}{| V | = \underline{\text{150.43 V}}}$.