Question #0a9a2
1 Answer
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
and the Planck - Einstein equation, which describes the relationship between energy and frequency for photons
#color(blue)(E = h * nu)" "# , where
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
#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 = 1/100 * c = 10^(-2) * c#
Rearrange equation
#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,
#"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,