Question

Question #0a9a2

Answer 1

Here's what I got.

Explanation:

The idea here is that you need to use two very famous equations, the Einstein equation, which relates mass and energy

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

`c` - the speed of light in vacuum, approximately equal to `3 * 10^8"m s"^(-1)`

and the Planck - Einstein equation, which describes the relationship between energy and frequency for photons

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

`nu` - the frequency of the photon
`h` - Planck's constant, equal to `6.262 * 10^(-34)"kg m"^2 "s"^(-1)`

Now, de Broglie believed that particles can also behave as waves. As a result, he thought that both of these equations could be used to describe a particle and its associated de Broglie wavelength.

`E_"particle" = E_"wave"`

`m * c^2 = h * nu " " " " color(purple)((1))`

As you know, frequency and wavelength have an inverse relationship that depends on the speed of light

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

This means that you can use this relationship to find an equation that connects wavelength and mass.

`c = lamda * nu implies nu = c/(lamda)`

Plug this into equation `color(purple)((1))` to get

`m * c^color(red)(cancel(color(black)(2))) = h * color(red)(cancel(color(black)(c)))/(lamda) implies lamda = h/(m * c) " " " " color(purple)((2))`

For particles with mass, you can replace the speed of light with their specific velocity, `v`. Since your atom is traveling at `1%` the speed of light, you can say that its velocity will be

`v = 1/100 * c = 10^(-2) * c`

Rearrange equation `color(purple)((2))` to solve for `m`, the mass of the atom

`lamda = h/(10^(-2) * c * m)`

`m = h/(lamda * 10^(-2) * c)`

Plug in your values - do not forget to convert the wavelength from picometers to meters - to get

`m = (6.626 * 10^(-34)"kg" color(red)(cancel(color(black)("m"^2))) color(red)(cancel(color(black)("s"^(-1)))))/(1.46 * 10^(-3) * 10^(-12)color(red)(cancel(color(black)("m"))) * 10^(-2) * 3 * 10^8color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))`

`m = 1.513 * 10^(-25)"kg"`

To find the identity of the atom, use the known unified atomic mass unit, `u`, which is equal to `1/12"th"` of the mass of a single carbon-12 atom in the neutral state

`"1 u" = 1.661 * 10^(-27)"kg"`

In your case, you have

`1.513 * 10^(-25)color(red)(cancel(color(black)("kg"))) * "1 u"/(1.661 * 10^(-27)color(red)(cancel(color(black)("kg")))) = "91.1 u"`

The closest match to this value is zirconium, `"Zr"`, which has an atomic number equal to `40` and a relative atomic mass of `"91.224 u"`.