# Question #f4c65

##### 1 Answer

#### Explanation:

The idea here is that particles that have **mass** have an associated wavelength called the **de Broglie wavelength** that depends on their **momentum**, *de Broglie hypothesis.*

More specifically, the de Broglie wavelength is equal to

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

Here

In turn, the momentum depends on the particle's *mass*, *velocity*,

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

The problem provides you with the *speed*, which you can use instead of velocity in your calculations, and the mass of the oxygen molecule, which basically means that it provides you with its **momentum**

#p = 5.31 * 10^(-26)"kg" * "725 m s"^(-1)#

#p = 3.850 * 10^(-23)"kg m s"^(-1)#

Now, notice that Planck's constant is given in *joules per second*,

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

which means that you can also express Planck's constant 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)#

Plug in your value to find the de Broglie wavelength of the oxygen molecule

#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)))))/(3.850 * 10^(-23)color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1))))) = 1.72 * 10^(-11)"m"#

Wavelengths of this magnitude are usually expressed in *nanometers*

#1.72 * 10^(-11) color(red)(cancel(color(black)("m"))) * (10^9"nm")/(1color(red)(cancel(color(black)("m")))) = color(darkgreen)(ul(color(black)(1.72 * 10^(-2) "nm")))#

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