# Question #fe328

##### 1 Answer

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

#lamda_ "matter" = h/(m * v) #

Here

#lamda_ "matter"# is the de Broglie wavelength#h# isPlanck's constant, equal to#6.626 * 10^(-34)"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

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

#"J" * "s" = ("kg" * "m"^2)/ "s"^(2) * "s" = ("kg" * "m"^2)/"s"#

This means that you have

#h = 6.626 * 10^(-34) quad "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)))))#

#lamda_"matter" = ((2.0079 * 10^(-34) quad"kg")/m) " m"#

To find a numerical value, simply plug in the mass of an electron in **kilograms** **here**.

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