# Question #28c8a

##### 1 Answer

#### Answer:

#### 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

#lamda_ "matter" = h/(m * v) #

Here

#lamda_ "matter"# is its de Broglie wavelength#h# isPlanck's constant, equal to#6.626 * 10^(-34)"J s"# #m# is the mass of the particle#v# is its velocity

Rearrange the equation to solve for

#v = h/(m * lamda_"matter")#

Now, you know that argon has a **molar mass** of **mole** of argon, which is equivalent to **atoms** of argon, has a mass of

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*!

#lamda_"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

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

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

#"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 = (6.626 * 10^(-34) color(red)(cancel(color(black)("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"))))#

#color(darkgreen)(ul(color(black)(v = 1.9 * 10^3color(white)(.)"m s"^(-1))))#

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