Introduction to X-ray Physics and Diffraction
Tianyang Huang, Ye Liu
1 Purpose
To study how voltage and current affect Kα, Kβ, and bremsstrahlung radiation
and to measure the lattice spacing of four different crystals through X-ray
diffraction.
2 Theory
This experiment mainly composed of two part: x-ray generation and x-ray
reflection.
The x-ray, a kind of electromagnetic wave, is generated by the rapid
deceleration of electrons, having direction perpendicular to electrons’
deceleration. The electrons are generated on the cathode and accelerated by
electric field to anode, where x-ray is generated. Thus the number of electron
being generated is depend on the magnitude of current while the energy (speed)
of the electron arrived at anode is depend on the magnitude of voltage. After the
electron arrived at atoms on anode, it may hit and bring an electron away from
the atoms. Then an outer layer electron of this atom will fall down to occupy the
position of the lost electron. This will generate x-ray. Because the electron can
only fall from an layer to another layer, the energy and wavelenth of these x-ray
are some fixed value. X-ray of layer n=2 to n=1 is λα = 0.7107A, of layer n=3
to n=1 is λβ = 0.7107A In addition to that, there is also another kind of way xray generated, the bremsstrahlung radiation, which has a continuous spectrum.
The x-ray will reflect at the surface of the target(crystal). This reflection may
occur on any layer of the atoms of the target crystal. The nearly by two layer have
a distance of d. This result to a 2𝑑𝑠𝑖𝑛(θ) distance difference of x-ray traveled.
Thus the only x-ray detected is the wavelength that have constructive interference.
This wavelength based on Brag formula 𝑛λ = 2𝑑𝑠𝑖𝑛(θ)
Procedure
1. Scan NaCl from lower limit 2.5 to the upper limit to 35.0 with current I = 1.00
mA, the measuring time per angular step ∆t = 1s, the tube voltage U = 35 kV,
and the angular step size ∆β = 0.1◦.
2. Change the β limits to now be 2.5◦ – 10.0◦. Scan for tube voltages of 35.0,
30.0, 25.0, 22.5, 20.0 and 15.0 kV.
3. return the tube to 35 kV, maintain the other previous settings, scan for
tube currents of 1.00, 0.80, 0.60, 0.40, 0.20 mA.
4. Change crystal in to LiF, KCl, and muscovite (mica), scan from 1.8◦ to 25◦.
5. select TARGET, then scan for the β limits to θT ±0.5 where θT is the Target
angle at the Brag peak
3 Week one
Use the Bragg formula, and the lattice spacing for NaCl, to obtain an
estimate for the wavelength/energy of each spectral line. Average these
estimates. How do your results compare with the known values?
k 1
6.4
1160
k 1
7.2
2318
k 2
12.9
245
k 2
14.6
525
k 3
19.6
56
k 3
22.2
116
 1  0.564  sin(7.2)  0.070688
 2  0.564  sin(14.6)  2  0.07108
 1  0.564  sin(22.2)  3  0.07103
 
0.070688  0.07108  0.07103
 0.07094
3
 1  0.564  sin(6.4)  0.06287
 2  0.564  sin(12.9)  2  0.06296
 1  0.564  sin(19.6)  3  0.06307
 
0.06287  0.06296  0.06307
 0.06206
3
Our result show that  is different 0.18% from known value, and  is about
0.41% differ. The values are closed to known values. And the energy is as follow:
E 
hc

 2.800  1015 J
E 
hc

 3.155 1015 J
For the initial scan from 2.5◦ to 35.0◦ make three plots, each on separate
figures. One for count rate vs. angle, another for count rate vs. nλ, and finally
one more for count rate vs. E/n.
Fig 1. Counts rate vs angle theta.
Fig 2. Counts rate vs nλ
Fig 3. Counts rate vs E/n
For the six scans varying the tube voltages, make one plot which shows all
six scans on the same figure of count rate vs. E/n.
Fig 4. Counts rate vs E/n with different voltage
For the five scans varying the tube currents, make one plot which shows all
five scans on the same figure of count rate vs. E/n.
Fig 5. Counts rate vs E/n with different Current
Determine the threshold of X-ray emission for each tube voltage, and make
a scatterplot (with six points) of the threshold of X-ray emission vs. tube
voltage. What relationship do you observe? Is this what you expect?
Fig 6. Threshold vs V scatterplot
The higher the voltage is, the higher the threshold energy become which exactly
agree with what we expect.
Do you note any threshold of X-ray emission in your plot which varies over
the tube currents? Why or why not?
Threshold don’t vary when current vary, since wave has quantized energy.
Change of voltage changes energy of individual x-ray emission. But change of
current only effect to satisfy of x-ray.
For the five scans varying the tube currents, make a plot of I at the Kα, n =
1 peak vs. i
Fig 7. I at Kα vs i
Plot I/i vs. i, and perform a linear least squares fit.
Fig 8. I/i vs i
Our model:
f ( x)  m * x  b
m  454.7
b  2497
2
and the relationship after substitute the value to the formula I / i  c  Dt c , the
m in the fitting model is Dt and b is equal to c. Thus c=2497, Dt=7.30e-5.
Use the formula I′ = I/(1−IDt) to correct all of your data for this dead-time.
Re- plot the graphs you made. Do the corrections make much of a difference?
Fig 9,10. Replot of Count rate vs E/n with corrected data.
Based on the graph and the data points shows that there is no much difference
between.
Using your (corrected) plots which shows the six scans varying the tube
voltages, you will note that not all of the scans have distinct spectral lines.
Determine which tube voltages created Kα and Kβ peaks.
Only at V=35,30,25,22.5 create Kα and Kβ peaks.
For the tube voltages which do have Kα and Kβ peaks, make a scatterplot
with two different series of data on it: I′b vs. the tube voltage and I′s vs. the
tube voltage.
Fig 11. Scatter plot of I’b vs voltage.
Repeat the above two steps for the five scans varying the tube currents.
Fig 12. Scatter plot of I’b vs current
Question: Why don’t all tube voltages create Kα and Kβ peaks? Do all tube
currents? Explain.
That’s because the voltage 20 and 15 can’t accelerate electrons to enough energy
to create x-ray at the peak wavelength. There is alwaus spectral emission for
different current because there is enough energy for each single electron.
Now observe a power-law relationship like I′/i = A(V −VT )^α where
A,VT , and α are as of yet unknown. Using your scans of count rate vs. tube
voltage, make a plot of I′/i at the Kα peak vs. tube voltage. Only use scans
which have a Kα peak. Does your plot look roughly like a power-law? Fit our
plot to determine A,VT , and α. What do you get for A, VT , and α? Do you
have any expectations for what VT should be?
Fig 13. I/i vs V
The plot looks roughly like a a power-law, but the data point is not enough since
there is only few peaks.
Fig 14. ln(I/i) vs ln(V)
Fig 15. Linear fit of ln(I/i) vs ln(V)
After the linear fit, we find α=0.5526, VT=22.48, A=534.7. We are expecting
VT to be higher than 20V and lower than 22.5V
Week 2:
Plot nλ/2 vs. sin θ.
Fig 15. nλ/2 vs sin θ for NaCl
Our fit model and data for NaCl:
f ( x)  m * x  b
m  0.355
b  0.00078
Fig 16. nλ/2 vs sin θ for KCl
Our fit model and data for KCl:
f ( x)  m * x  b
m  0.3171
b  0.004459
Fig 17. nλ/2 vs sin θ for Mica
Our fit model and data for Mica:
f ( x)  m * x  b
m  0.1908
b  0.02077
Fig 18. nλ/2 vs sin θ for LiF
Our fit model and data for LiF:
f ( x)  m * x  b
m  0.4372
b  0.001986
The slope will then correspond to the lattice spacing where m is the lattice spacing,
for NaCl is 0.355, for KCl is 0.3171, for Mica is 0.1908, for LiF is 0.4372. And
the intercept will reflect misalignment of the crystal. If the crystal is misaligned,
there will be a shift in the θ value. As a result, all calculation will have a bigger
uncertainty.
Are low angle or high angle measurements best? Why or why not?
The low angle measurements are best. That’s because the values are significant
at low angle. The peaks are more accurately located.
Plot both of the rocking curves and estimate the full width at half-maximum.