(a) This peak in the intensity of scattered electrons is due to constructive interference between electron waves scattered by different surface atoms. Vaa = 54 V u = 50" 15° 30" 45" 60° 75° 90" If the scattered waves are in phase, there is a peak in the intensity of scattered electrons. (b) Incident waves in phase 50 d Atoms on surface of crystal k-
(a) This peak in the intensity of scattered electrons is due to constructive interference between electron waves scattered by different surface atoms. Vaa = 54 V u = 50" 15° 30" 45" 60° 75° 90" If the scattered waves are in phase, there is a peak in the intensity of scattered electrons. (b) Incident waves in phase 50 d Atoms on surface of crystal k-
Related questions
Question
In an electron-diffraction experiment using an accelerating voltage of 54 V, an intensity maximum occurs for u = 50° (see Fig. ). X-ray diffraction indicates that the atomic spacing in the target is d = 2.18 * 10-10 m = 0.218 nm. The electrons have negligible kinetic energy before being accelerated. Find the electron wavelength.
Transcribed Image Text:(a)
This peak in the intensity of scattered
electrons is due to constructive
interference between electron waves
scattered by different surface atoms.
Vaa
= 54 V
u = 50"
15° 30"
45"
60°
75° 90"
If the scattered waves are in phase,
there is a peak in the intensity of
scattered electrons.
(b)
Incident waves
in phase
50
d
Atoms on surface of crystal
k-
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 2 images
