# Question #459d4

Apr 16, 2017

$l a m \mathrm{da} = 2.4 \cdot {10}^{- 11} \text{m}$

#### Explanation:

Start by looking up the mass of an electron

${m}_{\text{electron}} \approx 9.1094 \cdot {10}^{- 31}$ $\text{kg}$

Now, according to the de Broglie hypothesis, matter can actually behave like a wave.

In this case, your electron will exhibit wave-like behavior and have an associated matter wave. The wavelength of this electron is called the de Broglie wavelength and is given by the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{l a m \mathrm{da} = \frac{h}{p}}}} \to$ the de Broglie wavelength

Here

• $p$ is the momentum of the electron
• $l a m \mathrm{da}$ is its de Broglie wavelength
• $h$ is Planck's constant, equal to $6.626 \cdot {10}^{- 34} {\text{kg m"^2"s}}^{- 1}$

Notice that the equation uses the momentum of the electron, which depends on the velocity of the electron, $v$, and on its mass, $m$, as given by the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{p = m \cdot v}}}$

$p = 9.1094 \cdot {10}^{- 31} {\text{kg" * 3 * 10^(7)"m s}}^{- 1}$

$p = 2.733 \cdot {10}^{- 23}$ ${\text{kg m s}}^{- 1}$

Plug this into the equation that allows you to calculate the de Broglie wavelength to find

$l a m \mathrm{da} = \left(6.626 \cdot {10}^{- 34} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{kg"))) "m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-1)))))/(2.733 * 10^(-23)color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s}}^{- 1}}}}\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{l a m \mathrm{da} = 2.4 \cdot {10}^{- 11} \textcolor{w h i t e}{.} \text{m}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the velocity of the electron.