Question

Question #f4c65

Answer 1

`1.72 * 10^(-2)"nm"`

Explanation:

The idea here is that particles that have mass have an associated wavelength called the de Broglie wavelength that depends on their momentum, `p`, as given by the de Broglie hypothesis.

More specifically, the de Broglie wavelength is equal to

`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"`
`p` - the momentum of the particle

In turn, the momentum depends on the particle's mass, `m`, and velocity, `v`

`color(blue)(ul(color(black)(p = m * v))) ->` describes the momentum of 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, `"J" * "s"`. As you know, `"1 J"` is equal to

`"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.