Respuesta :
Answer:
The angle of diffraction are 67.75 deg and 53.57 deg.
Explanation:
Given:
Davisson and Germer experiment with nickel target for electrons bombarding.
Voltages : [tex]38\ eV[/tex] and [tex]54\ eV[/tex]
We have to find the angles that is [tex]\phi_3_8[/tex] and [tex]\phi_5_4[/tex] .
Concept:
- Davison Germer experiment is based on de Broglie hypothesis where it says matter has both wave and particle nature.
- When electrons get reflected from the surface of a metal target with an atomic spacing of [tex]D[/tex], they form diffraction patterns.
- The positions of diffraction maxima are given by [tex]Dsin(\phi) = n\lambda[/tex] .
- An atomic spacing is [tex]D = 0.215\ nm[/tex], when the principal maximum corresponds to n=1
- The wavelength is [tex]\lambda[/tex], and [tex]\lambda =\frac{1.227 \sqrt{V} } {\sqrt{V_o}}\ nm[/tex] .
Solution:
Finding the wavelength at [tex]V_o=38\ eV[/tex] .
⇒ [tex]\lambda_3_8 =\frac{1.227 \sqrt{V} } {\sqrt{38V}}\ nm[/tex]
⇒ [tex]\lambda_3_8 =0.199[/tex] nm
Plugging the values of wavelength.
⇒ [tex]sin(\phi)=\frac{\lambda}{D}[/tex]
⇒ [tex]\phi_3_8=sin^-1(\frac{0.199}{0.215} )[/tex]
⇒ [tex]\phi_3_8 =67.75[/tex] degrees.
Now
For for the electrons with energy [tex]54\ eV[/tex], [tex]V_0=54V[/tex] the wavelength is.
⇒ [tex]\lambda_5_4 =\frac{1.227 \sqrt{V} } {\sqrt{54V}} = 0.173[/tex] nm
And
⇒ [tex]\phi_5_4=sin^-1(\frac{0.173}{0.215} ) = 53.57[/tex] degrees.
So,
The angles of diffraction maxima are 67.75 deg and 53.57 deg.
