Question #c1b56

1 Answer
Nov 13, 2016

Answer:

#9.74 * 10^(-2)"nm"#

Explanation:

The idea here is that matter can also behave like a wave, as described by the de Broglie hypothesis.

In your case, a molecule of oxygen, #"O"_2#, will exhibit wave-like behavior and have a wavelength called the de Broglie wavelength associated with it.

The de Broglie wavelength depends on the momentum of the particle, #p#, which in turn depends on the mass of the particle, #m#, and its velocity, #v#. You will thus have

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

and

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

Here

#lamda# - the wavelength of the molecule
#h# - Planck's constant, equal to #6.626 * 10^(-34)"J s"#

Now, you know that the molecule has a speed of #"128 m s"^(-1)#, which can be used here instead of velocity.

In order to find the mass of a single molecule of oxygen, use the molar mass of oxygen gas and Avogadro's constant.

Oxygen gas has a molar mass of approximately #"32.0 g mol"^(-1)#, which means that #1# mole of oxygen molecules has a mass of #"32.0 g"#.

This means that a single molecule of oxygen gas will have a mass of

#1 color(red)(cancel(color(black)("molecule O"_2))) * overbrace((1color(red)(cancel(color(black)("mole O"_2))))/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules O"_2)))))^(color(purple)("Avogadro's constant")) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2))))#

# = 5.314 * 10^(-23)"g" = 5.314 * 10^(-26)"kg"#

This means that the momentum of the molecule will be

#p = 5.314 * 10^(-26)"kg" * "128 m s"^(-1)#

#p = 6.802 * 10^(-24)"kg m s"^(-1)#

Now, you need to mindful of units here. Notice that Planck's constant is given in joules per second, #"J" * "s"#. As you know, #"1 J"# is equal to

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

This means that Planck's constant can also be written 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)#

You can now say that the de Broglie wavelength associated with this molecule is

#lamda = (6.626 * 10^(-34) color(red)(cancel(color(black)("kg"))) "m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-1)))))/(6.802 * 10^(-24)color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1))))) = 9.74 * 10^(-11)"m"#

Expressed in nanometers, the answer will be

#9.74 * 10^(-11)color(red)(cancel(color(black)("m"))) * (10^9 "nm")/(1color(red)(cancel(color(black)("m")))) = color(darkgreen)(ul(color(black)(9.74 * 10^(-2)"nm")))#

The answer is rounded to three sig figs.