Part A X rays with a wavelength of 0.12nm undergo first-order diffraction from a

ID: 2103380 • Letter: P

Question

Part A

X rays with a wavelength of 0.12nm undergo first-order diffraction from a crystal at a 73degrees angle of incidence.

What is the angle of second-order diffraction?

Part B

X rays with a wavelength of 0.19nm undergo first-order diffraction from a crystal at a 50degrees angle of incidence.

At what angle does first-order diffraction occur for x rays with a wavelength of 0.10nm ?

Part C

X rays diffract from a crystal in which the spacing between atomic planes is 0.187nm . The second-order diffraction occurs at 55.0 degrees.

What is the angle of the first-order diffraction?

Part D

The spacing between atomic planes in a crystal is 0.130nm .

If 15.0KeV x rays are diffracted by this crystal, what is the angle of first-order diffraction?

If 15.0KeV x rays are diffracted by this crystal, what is the angle of second-order diffraction?

Use "Bragg's law"

m*? = 2d*sin(?)

where

? is the wavelength,

d is the distance between crystal planes,

? is the angle of the diffracted wave (Theta).

and m is an integer known as the order of the diffracted beam.

d=0.12nm/2*sin(73)

sin(Theta 2)=2*0.12nm/2*d

X-rays have wavelengths much smaller then 0.19 mm (= 190 nm).

I don't know where your mistake is, but we can in any case say that

theta = arcsin(m*lambda/d) and thus that m*lambda1/d = sin(50 deg)

Then m*lambda2/d = sin(theta) = lambda2/lambda1*sin(50 deg) ==>

theta = arcsin(lambda2/lambda1*sin(50 deg)) = arcsin(10/19*sin(50 deg))

find