# Question c1b56

Nov 13, 2016

$9.74 \cdot {10}^{- 2} \text{nm}$

#### 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, ${\text{O}}_{2}$, 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

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

and

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

Here

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

Now, you know that the molecule has a speed of ${\text{128 m s}}^{- 1}$, 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 ${\text{32.0 g mol}}^{- 1}$, which means that $1$ mole of oxygen molecules has a mass of $\text{32.0 g}$.

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))))

$= 5.314 \cdot {10}^{- 23} \text{g" = 5.314 * 10^(-26)"kg}$

This means that the momentum of the molecule will be

$p = 5.314 \cdot {10}^{- 26} {\text{kg" * "128 m s}}^{- 1}$

$p = 6.802 \cdot {10}^{- 24} {\text{kg m s}}^{- 1}$

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

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

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")))

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

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

$9.74 \cdot {10}^{- 11} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{m"))) * (10^9 "nm")/(1color(red)(cancel(color(black)("m")))) = color(darkgreen)(ul(color(black)(9.74 * 10^(-2)"nm}}}}$

The answer is rounded to three sig figs.