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