# Question 28c8a

Dec 11, 2017

$1.9 \cdot {10}^{3}$ ${\text{m s}}^{- 1}$

#### Explanation:

For starters, you are right, that is the equation that you must use in order to find the velocity of the atom.

You know that the de Broglie wavelength is given by

$l a m {\mathrm{da}}_{\text{matter}} = \frac{h}{m \cdot v}$

Here

• $l a m {\mathrm{da}}_{\text{matter}}$ is its de Broglie wavelength
• $h$ is Planck's constant, equal to $6.626 \cdot {10}^{- 34} \text{J s}$
• $m$ is the mass of the particle
• $v$ is its velocity

Rearrange the equation to solve for $v$

$v = \frac{h}{m \cdot l a m {\mathrm{da}}_{\text{matter}}}$

Now, you know that argon has a molar mass of ${\text{39.948 g mol}}^{- 1}$. This tells you that $1$ mole of argon, which is equivalent to $6.022 \cdot {10}^{23}$ atoms of argon, has a mass of $\text{39.948 g}$.

In order to find the mass of a single atom of argon, you need to use the definition of the mole, which is given by Avogadro's constant, as a conversion factor.

You will end up with

1 color(red)(cancel(color(black)("atom Ar"))) * overbrace("39.948 g"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Ar")))))^(color(blue)("Avogadro's constant")) = 6.634 * 10^(-23)color(white)(.)"g"

Now all you have to do is to plug in your values and find the velocity of the atom--do not forget to convert the wavelength from nanometers to meters!

$l a m {\mathrm{da}}_{\text{matter" = 5.2 color(red)(cancel(color(black)("pm"))) * "1 m"/(10^12color(red)(cancel(color(black)("pm")))) = 5.2 * 10^(-12)color(white)(.)"m}}$

Also, keep in mind that you have

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

so the units you have for Planck's constant can be converted to

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

At this point, you should notice that you need to convert the mass of the atom from grams to kilograms, so

m = 6.634 * 10^(-23)color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = 6.634 * 10^(-26)color(white)(.)"kg"#

You can thus say that the velocity of the atom is equal to

$v = \left(6.626 \cdot {10}^{- 34} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg"))) "m"^color(red)(cancel(color(black)(2))) "s"^(-1))/(6.634 * 10^(-26)color(red)(cancel(color(black)("kg"))) * 5.2 * 10^(-12)color(red)(cancel(color(black)("m}}}}\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{v = 1.9 \cdot {10}^{3} \textcolor{w h i t e}{.} {\text{m s}}^{- 1}}}}$

The answer is rounded to two sig figs.