- #1
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.
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.
