An electron beam of energy `10 KeV` is incident on metallic foil. If the interatomic distance is `0.55Å`. Find the angle of diffraction.

Answer

`lambda=D sin phi and lambda=(12.27)/(sqrt(V))Å so (12.27)/(sqrt(V))D sin`
`:. (12.27 " "10^(10))/(sqrt(10" "10^(3))) 0.55" "10^(10) sin `
`sin phi=(12.27)/(0.53" "1000)0.2231`
or `sin^(1)(0.2231) 12.89^(@)`

Explanation

`lambda=D sin phi` and `lambda=(12.27)/(sqrt(V))Å ` so `(12.27)/(sqrt(V))D sin`
`:. (12.27" "10^(10))/(sqrt(10" "10^(3)))0.55" "10^(10)sin`
`sin phi=(12.27)/(0.53" "100)0.2231` or `sin^(1)(0.2231), 12.89^(@)`