Question

Question #5809d

Answer 1

`1.31 * 10^(-13)"m"`

Explanation:

So, you know that your neutron is moving at `1.00%` of the speed of light and that its has a mass of `1.67493 * 10^(-27)"kg"`.

The first important thing to notice here is that the neutron is moving relatively slow compared with the speed of light, so you won't have to take into account relativistic effects.

Now here's how you can think of what's asked of you here. You know that energy and mass are related through Einstein's equation

`color(blue)(E = m * c^2)" "`, where

`c` - the speed of light in vacuum, equal to `"299,792,458 m/s"`

The Planck - Einstein equation establishes a relationship between the energy of a photon and its frequency

`color(blue)(E = h * nu)" "`, where

`nu` - the frequency of the photon
`h` - Planck's constant, equal to `6.262 * 10^(-34)"J s"`

The idea here is that de Broglie believed that particles could also behave as waves, and as a consequence could be described by both of the above equations.

In essence, this implied that you have

`E_"particle" = E_"wave"`

`m * c^2 = h * nu`

Now, you know that frequency and wavelength have an inverse relationship

`color(blue)(c = lamda * nu)`

This means that you can write `nu = c/(lamda)` and plug it in the above equation to get

`m * c^color(red)(cancel(color(black)(2))) = h * color(red)(cancel(color(black)(c)))/(lamda) <=> m * c = h/(lamda)`

Now, particles cannot travel at the speed of light, so you can replace `c` with a velocity `v` to get

`lamda = h/(m * v) ->` the de Broglie wavelength equation

You'll often see this written as

`lamda = h/p`

Here `p = m * v` is the impulse of the particle.

You have the mass of the neutron and its speed, so plug in the values and solve for `lamda` - an important thing to note here is that

`"1 J" = 1("kg" * "m"^2)/"s"^2`

so Planck's constant can be written as

`h = 6.626 * 10^(-34)overbrace(("kg" * "m"^2)/"s"^color(red)(cancel(color(black)("2"))))^(color(blue)(="J")) * color(red)(cancel(color(black)("s"))) = 6.626 * 10^(_34)("kg" * "m"^2)/"s"`

`lamda = (6.626 * 10^(-34)(color(red)(cancel(color(black)("kg"))) * "m"^color(red)(cancel(color(black)(2))))/color(red)(cancel(color(black)("s"))))/(1.67493 * 10^(-27)color(red)(cancel(color(black)("kg"))) * "299,792,458" * 10^(-2)color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"))))`

`lamda = color(green)(1.31 * 10^(-13)"m") ->` rounded to three sig figs

So remember, it's very important to match the units you have for mass and speed with those used in for Planck's constant, kilograms and meters per second.