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

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SUMMARY

Bragg's Law is utilized to determine the angle of gamma-ray deflection in a cubic crystal with an interatomic spacing of 0.24 nm, specifically for gamma-rays of energy 100 keV. The calculations reveal that the angle of deflection is 1.48 degrees, derived from the equation n(lambda) = 2dsinx, where lambda is calculated to be 1.24x10^-11 m. The discussion clarifies that the angle of incidence equals the angle of reflection, not deflection, addressing common misconceptions about these terms.

PREREQUISITES
  • Understanding of Bragg's Law
  • Knowledge of gamma-ray energy calculations
  • Familiarity with wavelength and frequency relationships
  • Basic principles of crystallography
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  • Study the derivation and applications of Bragg's Law in X-ray diffraction
  • Learn about gamma-ray spectroscopy techniques
  • Explore the relationship between energy, wavelength, and frequency in quantum mechanics
  • Investigate the properties of cubic crystals and their impact on diffraction patterns
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Physicists, materials scientists, and anyone involved in gamma-ray analysis or crystallography will benefit from this discussion.

MarkusNaslund19
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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.
 
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i think you may be confusing the fact that angle of incidence must equal to the angle of REFLECTION

not deflection (refraction)
 

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