Question

Question #fe328

Answer 1

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` is Planck'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 `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

`"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 `->` 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.