Question #94aa0

All you have to do here is to use the equation that establishes a relationship between the de Broglie wavelength of the electron`,`lamda`, and its **momentum**,`p#.

`color(blue)(ul(color(black)(lamda = h/p))) ->` the de Broglie wavelength

Here

• `p` is the momentum of the electron
• `lamda` is its de Broglie wavelength
• `h` is Planck's constant, equal to `6.626 * 10^(-34)"kg m"^2"s"^(-1)`

The momentum of the electron depends on its mass and on its velocity as shown by the equation

`color(blue)(ul(color(black)(p = m * v)))`

Here

• `m` is the mass of the electron
• `v` is its velocity

before plugging in your values, make sure that the units match those used in the expression of Planck's constant.

Planck's constant uses the units `"kg m"^2 "s"^(-1)`, which can be written as

`"kg m"^2 "s"^(-1) = color(blue)("kg m s"^(-1)) * "m"`

Notice that you have `color(blue)("kg")` as the units for the mass of the electron and `color(blue)("m s"^(-1))` as the units for its velocity, so you can safely plug in the values as they were given and solve for `p`.

`p = 9.11 * 10^(-31)color(white)(.)"kg" * 4.3 * 10^6color(white)(.)"m s"^(-1)`

`p = 3.917 * 10^(-24)` `"kg m s"^(-1)`

This means that the electron will have a de Broglie wavelength equal to

`lamda = (6.626 * 10^(-34) color(red)(cancel(color(black)("kg m s"^(-1)))) * "m")/(3.917 * 10^(-24)color(red)(cancel(color(black)("kg m s"^(-1))))) = color(darkgreen)(ul(color(black)(1.7 * 10^(-10)color(white)(.)"m")))`

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