# Question f4c7c

Nov 21, 2016

$1.47 \cdot {10}^{6} {\text{m s}}^{- 1}$

#### 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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{l a m \mathrm{da} = \frac{h}{p}}}} \to$ the de Broglie wavelength

Here

$l a m \mathrm{da}$ - the wavelength of the electron
$h$ - Planck's constant, equal to $6.626 \cdot {10}^{- 34} \text{J s}$
$p$ - the momentum of the electron

Rearrange to solve for $p$

$l a m \mathrm{da} = \frac{h}{p} \implies p = \frac{h}{l a m \mathrm{da}}$

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

${\text{1 J" = "1 kg m"^2 "s}}^{- 2}$

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")))

$h = 6.626 \cdot {10}^{- 34} {\text{kg m"^2"s}}^{- 1}$

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 $p$

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.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{p = m \cdot v}}} \to$ describes the momentum of the molecule

Here

$m$ - the mass of the electron
$v$ - its velocity

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

$p = m \cdot v \implies v = \frac{p}{m}$

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.