Question #c1b56
1 Answer
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,
The de Broglie wavelength depends on the momentum of the particle,
#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
Now, you know that the molecule has a speed of
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
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,
#"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.