# Question e8623

Jun 20, 2017

$2.4 \cdot {10}^{- 10}$ $\text{m}$

#### Explanation:

According to the de Broglie hypothesis, this electron can actually behave as a wave, in addition to its particle-like behavior.

This implies that your electron will have an associated matter wave. The wavelength of this matter wave, which is what we call the de Broglie wavelength, can be calculated using 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 $6.626 \cdot {10}^{- 34} {\text{kg m"^2"s}}^{- 1}$

As you know, the momentum of a particle depends on its velocity, $v$, and on its mass, $m$.

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

In your case, you know that the electron is moving at 1%# the speed of light, so its velocity will be equal to

$v = \frac{1}{100} \cdot 3 \cdot {10}^{8}$ ${\text{m s}}^{- 1}$

$v = 3 \cdot {10}^{6}$ ${\text{m s}}^{- 1}$

The mass of the electron is equal to

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

This means that the momentum of the electron is equal to

$p = 9.1094 \cdot {10}^{- 31} \textcolor{w h i t e}{.} {\text{kg" * 3 * 10^6color(white)(.)"m s}}^{- 1}$

$p = 2.733 \cdot {10}^{- 24}$ ${\text{kg m s}}^{- 1}$

Plug this into the equation for the de Broglie wavelength to find

$l a m \mathrm{da} = \left(6.626 \cdot {10}^{- 34} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{kg")))"m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-1)))))/(2.733 * 10^(-24)color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s}}^{- 1}}}}\right)$

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

I'll leave the answer rounded to two sig figs.