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 2šš šš(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 šλ = 2šš šš(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 āt = 1s, the tube voltage U = 35 kV, and the angular step size āβ = 0.1ā¦. 2. Change the β limits to now be 2.5ā¦ – 10.0ā¦. 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 1.8ā¦ to 25ā¦. 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? kļ¢ 1 6.4 1160 kļ” 1 7.2 2318 kļ¢ 2 12.9 245 kļ” 2 14.6 525 kļ¢ 3 19.6 56 kļ” 3 22.2 116 ļ¬ļ” 1 ļ½ 0.564 ļ“ sin(7.2) ļ½ 0.070688 ļ¬ļ” 2 ļ½ 0.564 ļ“ sin(14.6) ļø 2 ļ½ 0.07108 ļ¬ļ” 1 ļ½ 0.564 ļ“ sin(22.2) ļø 3 ļ½ 0.07103 ļ¬ļ” ļ½ 0.070688 ļ« 0.07108 ļ« 0.07103 ļ½ 0.07094 3 ļ¬ļ¢ 1 ļ½ 0.564 ļ“ sin(6.4) ļ½ 0.06287 ļ¬ļ¢ 2 ļ½ 0.564 ļ“ sin(12.9) ļø 2 ļ½ 0.06296 ļ¬ļ¢ 1 ļ½ 0.564 ļ“ sin(19.6) ļø 3 ļ½ 0.06307 ļ¬ļ¢ ļ½ 0.06287 ļ« 0.06296 ļ« 0.06307 ļ½ 0.06206 3 Our result show that ļ¬ļ” is different 0.18% from known value, and ļ¬ļ¢ is about 0.41% differ. The values are closed to known values. And the energy is as follow: Eļ” ļ½ hc ļ¬ļ” ļ½ 2.800 ļ“ 10ļ15 J Eļ¢ ļ½ hc ļ¬ļ¢ ļ½ 3.155 ļ“10ļ15 J For the initial scan from 2.5ā¦ to 35.0ā¦ 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) ļ½ m * x ļ« b m ļ½ ļ454.7 b ļ½ 2497 2 and the relationship after substitute the value to the formula I / i ļ½ c ļ 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) ļ½ m * x ļ« b m ļ½ 0.355 b ļ½ ļ0.00078 Fig 16. nλ/2 vs sin θ for KCl Our fit model and data for KCl: f ( x) ļ½ m * x ļ« b m ļ½ 0.3171 b ļ½ ļ0.004459 Fig 17. nλ/2 vs sin θ for Mica Our fit model and data for Mica: f ( x) ļ½ m * x ļ« b m ļ½ 0.1908 b ļ½ ļ0.02077 Fig 18. nλ/2 vs sin θ for LiF Our fit model and data for LiF: f ( x) ļ½ m * x ļ« b m ļ½ 0.4372 b ļ½ ļ0.001986 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.