Question #fe328

1 Answer
Feb 10, 2018

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# 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.