How Does Bragg's Law Determine the Angle of Gamma-Ray Deflection?

[MarkusNaslund19]

1. Question

A cubic crystal of interatomic spacing of 0.24 nm is used to select
gamma–rays of energy 100 keV from a radioactive source containing a continuum of energies. If the incident beam is normal to the crystal, at
what angle with respect to the incident beam do the 100 keV gamma–rays appear?

2. Equations and Variables

Bragg's Law: n(lambda) = 2dsinx

n is the order of intensity
lamba is the wavelength
d is the Bragg plane separation
x is the angle
E is energy, E=100keV=1.60x10^-14J

3. Attempt

Since the incident beam is normal to the crystal, d=0.24nm.

E=hf
E=h(c/lambda)
lambda=(hc)/E
=(6.626x10^-34)(3.00x10^8)/1.60x10^-14
=1.24x10^-11m

sinx=n(lambda)/2d
=[1(1.24x10^-11)]/[2(0.24x10^-9)]
= 0.0258

x = 1.48 degrees.

Just a bit unsure of myself. I guess it's the theory that is troubling me. I thought that the angle of incidence equals the angle of deflection. Thanks fellas.

 

[stunner5000pt]

i think you may be confusing the fact that angle of incidence must equal to the angle of REFLECTION

not deflection (refraction)

 

[m2003]

4. Response

Hello, thank you for your question. Based on your calculations, it appears that the 100 keV gamma-rays would appear at an angle of 1.48 degrees with respect to the incident beam. This is in accordance with Bragg's Law, which states that the angle of incidence is equal to the angle of deflection for constructive interference to occur. In this case, the Bragg plane separation, d, is equal to 0.24 nm, and the wavelength of the gamma-rays is 1.24x10^-11m. By plugging these values into the equation, we can see that the angle of deflection, or x, is indeed 1.48 degrees. This shows that Bragg X-Ray Scattering is a useful technique for selecting specific energies from a continuum of energies, as demonstrated in this scenario. I hope this helps clarify any confusion you may have had and thank you for your interest in scientific principles.