Question #0a9a2

1 Answer
Nov 19, 2015

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"#.