Answer :
Answer:
λ=1.7pm
This electron would not be useful in a Davisson-Germer type scattering experiment.
θ=0.45⁰
Explanation:
a) We need to use the relativistic for of the kinetic energy:
[tex]K_{E}=mc^{2}-m_{0}c^{2}[/tex]. (1)
Here m₀ is the mass rest and c the speed of light in vacuum.
We can write this equation in terms of the linear momentum (p) using this expression: [tex]p^{2}c^{2}=-m_{0}^{2}c^{4}+(mc^{2})^{2}[/tex] (2)
if we solve this equation for p and put into the first equation we will have the KE in terms of p.
From (2) we have: [tex]mc^{2}=\sqrt{p^{2}c^{2}+(mc^{2})^{2}}[/tex] (3)
Let's substitute mc² in (1).
[tex]K_{E}=\sqrt{p^{2}c^{2}+(m_{0}c^{2})^{2}}-m_{0}c^{2}[/tex] (4)
Now, let's solve this for p:
[tex]p=\frac{\sqrt{K_{E}(K_{E}+2m_{0}c^{2})} }{c}[/tex]
And finally, using the De Broglie wavelength equation[tex]\lambda=\frac{h}{p}=\frac{hc}{\sqrt{K_{E}(K_{E}+2m_{0}c^{2})}} =1.7*10^{-12}[/tex].
[tex]m_{0}c^{2}=0.511*10^{6}eV[/tex]
[tex]h=4.136*10^{-15} eVs[/tex]
[tex]c=3*10^{8}m/s[/tex]
b) To determine the angle of the first order diffraction, we use the Bragg equation:
[tex]n\lambda=dsin(\theta).[/tex]
Now , in the Davisson and Germer, they use a nickel target, so the inter atomic distance for this particular element is around 0.215 nm. We take n=1 for the first order
[tex]sin(\theta)=\frac{\lambda}{d}=7.8*10^{-3}[/tex] and [tex]\theta=0.45^{o}[/tex].
If we see, this angle is to small to implement the experiment, so it would not be useful in a Davisson-Germer type scattering experiment.