# Question fe328

Feb 10, 2018

Here's what I got.

#### Explanation:

What you're looking for here is the de Broglie wavelength of the electron, which is calculated using the equation

$l a m {\mathrm{da}}_{\text{matter}} = \frac{h}{m \cdot v}$

Here

• $l a m {\mathrm{da}}_{\text{matter}}$ is the de Broglie wavelength
• $h$ is Planck's constant, equal to $6.626 \cdot {10}^{- 34} \text{J s}$
• $m$ is the mass of the electron
• $v$ is its velocity

You didn't provide the mass of the electron, so you will not be able to find a numerical answer. However, you can use that equation to find the de Broglie wavelength of the electron as a function of its mass $m$.

Now, before plugging in the value you have for the velocity of the electron, take a second to rewrite the units for Planck's constant as

$\text{J" * "s" = ("kg" * "m"^2)/ "s"^(2) * "s" = ("kg" * "m"^2)/"s}$

This means that you have

$h = 6.626 \cdot {10}^{- 34} \quad {\text{kg m"^2"s}}^{- 1}$

So, plug in the velocity of the electron into the equation to find

lamda_"matter" = (6.626 * 10^(-34) quad "kg m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-1)))))/(m * 3.3 * 10^(6) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))#

$l a m {\mathrm{da}}_{\text{matter" = ((2.0079 * 10^(-34) quad"kg")/m) " m}}$

To find a numerical value, simply plug in the mass of an electron in kilograms $\to$ you can find it listed here.

The final answer must be rounded to two sig figs, the number of sig figs you have for the velocity of the electron.