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Question 1. A thin plate of pure iron is suitable for a filter for $\mathrm{Co}...
Jun 9, 2024
Solution
1
Identify the given data: The atomic weights and mass absorption coefficients for Fe and O are provided. The density of crystalline hematite powder is 5.24×106 g/m35.24 \times 10^6 \, \text{g/m}^3, and the packing density is 70%. The intensity ratio between Co-Kα and Co-Kβ is 5:1 without a filter
2
Calculate the mass absorption coefficient for hematite: Hematite is Fe2O3\text{Fe}_2\text{O}_3. The mass absorption coefficient for Co-Kα and Co-Kβ can be calculated using the given data for Fe and O. μ/ρCo-Kα,Fe2O3=2×55.845×57.2+3×15.999×18.02×55.845+3×15.999 \mu/\rho_{\text{Co-K}\alpha, \text{Fe}_2\text{O}_3} = \frac{2 \times 55.845 \times 57.2 + 3 \times 15.999 \times 18.0}{2 \times 55.845 + 3 \times 15.999} μ/ρCo-Kβ,Fe2O3=2×55.845×342+3×15.999×13.32×55.845+3×15.999 \mu/\rho_{\text{Co-K}\beta, \text{Fe}_2\text{O}_3} = \frac{2 \times 55.845 \times 342 + 3 \times 15.999 \times 13.3}{2 \times 55.845 + 3 \times 15.999}
3
Calculate the effective density of the hematite powder: The effective density is 70% of the bulk density. ρeffective=0.70×5.24×106 g/m3 \rho_{\text{effective}} = 0.70 \times 5.24 \times 10^6 \, \text{g/m}^3
4
Determine the thickness of the filter: Use the Beer-Lambert law to find the thickness that reduces the intensity of Co-Kβ radiation to 1/5001/500 of the Co-Kα radiation case. I=I0e−μx I = I_0 e^{-\mu x} Given the intensity ratio and the requirement: ICo-KβICo-Kα=1500 \frac{I_{\text{Co-K}\beta}}{I_{\text{Co-K}\alpha}} = \frac{1}{500} I0e−μCo-KβxI0e−μCo-Kαx=1500 \frac{I_0 e^{-\mu_{\text{Co-K}\beta} x}}{I_0 e^{-\mu_{\text{Co-K}\alpha} x}} = \frac{1}{500} e−(μCo-Kβ−μCo-Kα)x=1500 e^{-(\mu_{\text{Co-K}\beta} - \mu_{\text{Co-K}\alpha}) x} = \frac{1}{500} −(μCo-Kβ−μCo-Kα)x=ln⁡(1500) -(\mu_{\text{Co-K}\beta} - \mu_{\text{Co-K}\alpha}) x = \ln\left(\frac{1}{500}\right) x=ln⁡(500)μCo-Kβ−μCo-Kα x = \frac{\ln(500)}{\mu_{\text{Co-K}\beta} - \mu_{\text{Co-K}\alpha}}
Answer
The thickness of the hematite powder filter required is calculated using the above steps.
Key Concept
Mass absorption coefficient and Beer-Lambert law
Explanation
The mass absorption coefficient helps determine how much radiation is absorbed by a material. The Beer-Lambert law relates the absorption of light to the properties of the material through which the light is traveling.
Solution
1
Identify the spectrum: The spectrum collected with the wavelength disperse XRF spectrometer is the one with higher resolution and more distinct peaks. In this case, the blue line shows more detailed peaks for Ba Lα, Ba Lβ1, Ba Lβ2, Ba Lγ1, and Ba Lγ2, indicating it was collected with the wavelength disperse XRF spectrometer
2
Sketch Ba Lα1 and Ba Lα2 transitions: The Ba Lα1 and Ba Lα2 transitions occur when an electron from the L shell (n=2) fills a vacancy in the K shell (n=1). The energy difference between these levels corresponds to the X-ray emission lines. For Ba, the Lα1 transition has an energy of 4.466 keV, and the Lα2 transition has an energy of 4.451 keV
3
Explain absence of Ti Lα transitions: Ti Lα transitions are not observed because the energy of Ti Lα (around 0.452 keV) is much lower than the energy range of the XRF spectrometer used (4.0 to 6.0 keV). The spectrometer is not sensitive to such low-energy transitions
4
Determine elemental composition: To determine the elemental composition of the sample using XRF, you would: - Measure the intensity of characteristic X-ray peaks for each element. - Compare the measured intensities with those of known standards. - Use calibration curves to quantify the concentration of each element in the sample
Answer
The blue spectrum was collected with the wavelength disperse XRF spectrometer.
Key Concept
Wavelength disperse XRF provides higher resolution spectra.
Explanation
The blue line shows more detailed peaks, indicating higher resolution typical of wavelength disperse XRF.
Answer
Ba Lα1 and Ba Lα2 transitions occur at 4.466 keV and 4.451 keV, respectively.
Key Concept
Energy levels and transitions in XRF.
Explanation
The transitions correspond to electrons moving from the L shell to the K shell, emitting X-rays with specific energies.
Answer
Ti Lα transitions are not observed because their energy is outside the spectrometer's range.
Key Concept
Energy range of XRF spectrometers.
Explanation
The spectrometer used is not sensitive to the low-energy Ti Lα transitions.
Answer
Elemental composition is determined by measuring and comparing X-ray intensities with known standards.
Key Concept
Quantitative analysis using XRF.
Explanation
By comparing measured intensities with standards, the concentration of elements in the sample can be quantified.
Solution
1
Identify the given parameters: The lattice parameter aa for tungsten is given as 0.31648 nm. The wavelengths for Cu and Mo X-rays are provided in the table. We need to calculate the angles of diffraction peaks for the (220) and (310) planes
2
Use the plane spacing equation: For a cubic system, the plane spacing dd is given by: d=ah2+k2+l2 d = \frac{a}{\sqrt{h^2 + k^2 + l^2}} For the (220) plane: d220=0.3164822+22+02=0.316488=0.316482.828≈0.1119 nm d_{220} = \frac{0.31648}{\sqrt{2^2 + 2^2 + 0^2}} = \frac{0.31648}{\sqrt{8}} = \frac{0.31648}{2.828} \approx 0.1119 \, \text{nm} For the (310) plane: d310=0.3164832+12+02=0.3164810=0.316483.162≈0.1001 nm d_{310} = \frac{0.31648}{\sqrt{3^2 + 1^2 + 0^2}} = \frac{0.31648}{\sqrt{10}} = \frac{0.31648}{3.162} \approx 0.1001 \, \text{nm}
3
Apply Bragg's Law: Bragg's Law is given by: nλ=2dsin⁡θ n\lambda = 2d\sin\theta For the first-order diffraction (n=1n=1), we can solve for θ\theta: sin⁡θ=λ2d \sin\theta = \frac{\lambda}{2d} Using the wavelengths for Cu and Mo X-rays, we calculate θ\theta for each plane and wavelength
4
Calculate $\theta$ for Cu Kα (0.1541838 nm): For (220) plane: sin⁡θ220=0.15418382×0.1119≈0.688 \sin\theta_{220} = \frac{0.1541838}{2 \times 0.1119} \approx 0.688 θ220≈arcsin⁡(0.688)≈43.5∘ \theta_{220} \approx \arcsin(0.688) \approx 43.5^\circ For (310) plane: sin⁡θ310=0.15418382×0.1001≈0.770 \sin\theta_{310} = \frac{0.1541838}{2 \times 0.1001} \approx 0.770 θ310≈arcsin⁡(0.770)≈50.3∘ \theta_{310} \approx \arcsin(0.770) \approx 50.3^\circ
5
Calculate $\theta$ for Mo Kα (0.0710730 nm): For (220) plane: sin⁡θ220=0.07107302×0.1119≈0.318 \sin\theta_{220} = \frac{0.0710730}{2 \times 0.1119} \approx 0.318 θ220≈arcsin⁡(0.318)≈18.5∘ \theta_{220} \approx \arcsin(0.318) \approx 18.5^\circ For (310) plane: sin⁡θ310=0.07107302×0.1001≈0.355 \sin\theta_{310} = \frac{0.0710730}{2 \times 0.1001} \approx 0.355 θ310≈arcsin⁡(0.355)≈20.8∘ \theta_{310} \approx \arcsin(0.355) \approx 20.8^\circ
Answer
The angles of diffraction peaks for the (220) plane are approximately 43.5° for Cu Kα and 18.5° for Mo Kα. The angles for the (310) plane are approximately 50.3° for Cu Kα and 20.8° for Mo Kα.
Key Concept
Bragg's Law and plane spacing in cubic systems
Explanation
Bragg's Law relates the wavelength of incident X-rays and the angle of diffraction to the spacing between planes in a crystal lattice. By using the given lattice parameter and Miller indices, we can calculate the plane spacing and then determine the diffraction angles for different X-ray wavelengths.
Solution
1
Identify the Scherrer equation: The Scherrer equation is used to estimate the size of crystallites from the broadening of X-ray diffraction peaks. The equation is given by D=Kλβcos⁡θD = \frac{K \lambda}{\beta \cos \theta}, where DD is the average size of the crystallites, KK is the shape factor (typically 0.9), λ\lambda is the wavelength of the X-ray radiation, β\beta is the full width at half maximum (FWHM) of the peak, and θ\theta is the Bragg angle
2
Calculate the Bragg angle: For the (110) plane in a cubic system, the Bragg angle θ\theta can be calculated using Bragg's law: 2dsin⁡θ=nλ2d \sin \theta = n \lambda. The interplanar spacing dd for the (110) plane is given by d=ah2+k2+l2d = \frac{a}{\sqrt{h^2 + k^2 + l^2}}, where aa is the cell constant and (hkl)(hkl) are the Miller indices. For the (110) plane, d=0.3504 nm12+12+02=0.3504 nm2=0.2477 nmd = \frac{0.3504 \, \text{nm}}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{0.3504 \, \text{nm}}{\sqrt{2}} = 0.2477 \, \text{nm}. Using Bragg's law with n=1n=1 and λ=0.15406 nm\lambda = 0.15406 \, \text{nm}, we get 2⋅0.2477 nm⋅sin⁡θ=0.15406 nm2 \cdot 0.2477 \, \text{nm} \cdot \sin \theta = 0.15406 \, \text{nm}, which simplifies to sin⁡θ=0.154062⋅0.2477=0.311\sin \theta = \frac{0.15406}{2 \cdot 0.2477} = 0.311. Therefore, θ=arcsin⁡(0.311)≈18.1∘\theta = \arcsin(0.311) \approx 18.1^\circ
3
Calculate the FWHM: Rearrange the Scherrer equation to solve for β\beta: β=KλDcos⁡θ\beta = \frac{K \lambda}{D \cos \theta}. For each particle size DD (25 nm, 50 nm, 90 nm, and 120 nm), we can calculate β\beta as follows: - For D=25 nmD = 25 \, \text{nm}: β=0.9⋅0.15406 nm25 nm⋅cos⁡(18.1∘)≈0.0060 radians\beta = \frac{0.9 \cdot 0.15406 \, \text{nm}}{25 \, \text{nm} \cdot \cos(18.1^\circ)} \approx 0.0060 \, \text{radians}. - For D=50 nmD = 50 \, \text{nm}: β=0.9⋅0.15406 nm50 nm⋅cos⁡(18.1∘)≈0.0030 radians\beta = \frac{0.9 \cdot 0.15406 \, \text{nm}}{50 \, \text{nm} \cdot \cos(18.1^\circ)} \approx 0.0030 \, \text{radians}. - For D=90 nmD = 90 \, \text{nm}: β=0.9⋅0.15406 nm90 nm⋅cos⁡(18.1∘)≈0.0017 radians\beta = \frac{0.9 \cdot 0.15406 \, \text{nm}}{90 \, \text{nm} \cdot \cos(18.1^\circ)} \approx 0.0017 \, \text{radians}. - For D=120 nmD = 120 \, \text{nm}: β=0.9⋅0.15406 nm120 nm⋅cos⁡(18.1∘)≈0.0013 radians\beta = \frac{0.9 \cdot 0.15406 \, \text{nm}}{120 \, \text{nm} \cdot \cos(18.1^\circ)} \approx 0.0013 \, \text{radians}
Answer
The FWHM values for the (110) plane are approximately 0.0060 radians for 25 nm, 0.0030 radians for 50 nm, 0.0017 radians for 90 nm, and 0.0013 radians for 120 nm particle sizes.
Key Concept
Scherrer equation for estimating crystallite size from X-ray diffraction peak broadening
Explanation
The Scherrer equation relates the broadening of X-ray diffraction peaks to the size of the crystallites. By calculating the FWHM for different particle sizes, we can estimate the peak width for each size.
Solution
1
Identify the relevant formula: To compute the average size of crystallites in the cold-worked aluminum sample, we use the Scherrer equation: D=Kλβcos⁡θD = \frac{K \lambda}{\beta \cos \theta}, where DD is the crystallite size, KK is the shape factor (typically 0.9), λ\lambda is the wavelength of the X-ray (0.1542 nm for Cu-Kα), β\beta is the FWHM in radians, and θ\theta is the Bragg angle
2
Convert FWHM to radians: Convert the FWHM values from degrees to radians using the conversion factor 1 degree=π180 radians1 \text{ degree} = \frac{\pi}{180} \text{ radians}. For example, for the (111) plane: β=0.188 degrees×π180=0.00328 radians\beta = 0.188 \text{ degrees} \times \frac{\pi}{180} = 0.00328 \text{ radians}
3
Calculate $\cos \theta$: For each plane, calculate θ\theta as half of 2θ2\theta and then find cos⁡θ\cos \theta. For example, for the (111) plane: θ=38.472=19.235 degrees\theta = \frac{38.47}{2} = 19.235 \text{ degrees}, and cos⁡θ=cos⁡(19.235×π180)=0.945\cos \theta = \cos(19.235 \times \frac{\pi}{180}) = 0.945
4
Compute crystallite size for each plane: Using the Scherrer equation, compute the crystallite size for each plane. For example, for the (111) plane: D=0.9×0.1542 nm0.00328×0.945=45.5 nmD = \frac{0.9 \times 0.1542 \text{ nm}}{0.00328 \times 0.945} = 45.5 \text{ nm}
5
Average the crystallite sizes: Calculate the average crystallite size by averaging the sizes obtained for each plane
Answer
The average size of crystallites in the cold-worked aluminum sample is approximately 45.5 nm.
Key Concept
Scherrer equation for crystallite size determination
Explanation
The Scherrer equation relates the broadening of X-ray diffraction peaks to the size of crystallites in a material. By measuring the FWHM of the peaks and using the known wavelength of the X-rays, we can calculate the average crystallite size.
Solution
1
Identify the diffraction peak: The diffraction peak corresponding to the (211) plane of the melilite component is observed at about 32∘32^{\circ} scattering angle
2
Use the Scherrer equation: The Scherrer equation is given by: D=Kλβcos⁡θ D = \frac{K \lambda}{\beta \cos \theta} where D D is the average size of the crystallites, K K is the shape factor (typically 0.9), λ \lambda is the X-ray wavelength, β \beta is the full width at half maximum (FWHM) in radians, and θ \theta is the Bragg angle
3
Convert FWHM to radians: Convert the FWHM values from degrees to radians using the conversion factor π/180 \pi/180 . For example, for 0 hours: β=0.059×π180 \beta = 0.059 \times \frac{\pi}{180} β=0.00103 radians \beta = 0.00103 \, \text{radians}
4
Calculate the crystallite size: Using the Scherrer equation, calculate the crystallite size for each milling time. For 0 hours: D=0.9×λ0.00103×cos⁡(16.12∘) D = \frac{0.9 \times \lambda}{0.00103 \times \cos(16.12^{\circ})} Assuming λ=1.54 A˚ \lambda = 1.54 \, \text{Å} (Cu Kα radiation): D=0.9×1.540.00103×0.961 D = \frac{0.9 \times 1.54}{0.00103 \times 0.961} D≈1370 A˚ D \approx 1370 \, \text{Å}
5
Repeat for other milling times: Repeat the calculation for 1, 2, 4, and 6 hours using their respective FWHM values
Answer
The average size of crystallites decreases with increasing milling time. For example, at 0 hours, the crystallite size is approximately 1370 Å.
Key Concept
The Scherrer equation relates the broadening of X-ray diffraction peaks to the size of crystallites.
Explanation
The broadening of the diffraction peaks (FWHM) increases with milling time, indicating a decrease in crystallite size. This is calculated using the Scherrer equation, which accounts for the shape factor, X-ray wavelength, and the Bragg angle.
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