# Question #f4c7c

##### 1 Answer

#### Explanation:

The first thing to do here is to figure out the **momentum** of the electron based on its associated *de Broglie wavelength*, which is equal to

#color(blue)(ul(color(black)(lamda = h/p))) -># thede Broglie wavelength

Here

Rearrange to solve for

#lamda = h/p implies p = h/(lamda)#

Before plugging in the value given to you, make sure that the **units** you have for Planck's constant and for the de Broglie wavelength *match*.

In this case, you need to use the fact that

#"1 J" = "1 kg m"^2 "s"^(-2)#

to write Planck's constant as

#h = 6.626 * 10^(-34) "kg m s"^color(red)(cancel(color(black)(-2))) * color(red)(cancel(color(black)("s")))#

#h = 6.626 * 10^(-34)"kg m"^2"s"^(-1)#

Notice that the wavelength is expressed in *meters* and that you have *meters squared* as a unit in the expression of Planck's constant, which means that you're ready to plug in the value and find

#p = (6.626 * 10^(-34)"kg m"^color(red)(cancel(color(black)(2)))"s"^(-1))/(5.00 * 10^(-9)color(red)(cancel(color(black)("m")))) = 1.3252 * 10^(-24)"kg m s"^(-1)#

Now, the **momentum** of the electron depends on its *mass* and on its *velocity*.

#color(blue)(ul(color(black)(p = m * v))) -># describes themomentumof the molecule

Here

You will use the electron's mass to find its *speed*, which in this particular context can be used to replace velocity.

#p = m * v implies v = p/m#

Plug in your values to find

#v = (1.3252 * 10^(-24)color(red)(cancel(color(black)("kg")))"m s"^(-1))/(9.00 * 10^(-31)color(red)(cancel(color(black)("kg")))) = color(darkgreen)(ul(color(black)(1.47 * 10^6"m s"^(-1)))#

The answer is rounded to three **sig figs**.