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A topaz crystal has an interplanar spacing \((d)\) of \(1.36 \AA^{\circ}(1 \AA=\) \(\left.1 \times 10^{-10} \mathrm{~m}\right) .\) Calculate the wavelength of the \(\mathrm{X}\) ray that should be used if \(\theta=15.0^{\circ}\) (assume \(n=1\) ).

Short Answer

Expert verified
The wavelength of the X-ray that should be used is approximately \(7.20 \times 10^{-11}\) m.

Step by step solution

01

Convert angstroms to meters

We are given the interplanar spacing d in units of angstroms (Å), but we need to convert this to meters for our calculation. Recall that 1 Å = 1 × 10⁻¹⁰ m. \[ d = 1.36 \ Å × \frac{1 \times 10^{-10} \ m}{1 \ Å} = 1.36 \times 10^{-10} \ m \]
02

Convert degrees to radians

We are given the angle θ in degrees, but we need to convert this to radians for our calculation. Recall that 1° = π/180 radians. \[ \theta = 15.0 \ ° × \frac{\pi \ rad}{180 \ °} = \frac{15\pi}{180} \ \text{radians} \]
03

Apply Bragg's Law

Now that we have the interplanar spacing in meters and the angle in radians, we can plug these values into Bragg's Law to solve for the wavelength λ. Recall that Bragg's Law is: \[ n\lambda = 2d \sin{\theta} \] Plugging in the values (n = 1, d = 1.36 × 10⁻¹⁰ m, θ = 15.0°): \[ \lambda = \frac{2 \times (1.36 \times 10^{-10} \ m) \times \sin{\frac{15\pi}{180}}}{1} \]
04

Calculate the wavelength λ

Now, we can calculate the wavelength λ: \[ \lambda \approx 7.20 \times 10^{-11} \ \text{m} \] Therefore, the wavelength of the X-ray that should be used is approximately \(7.20 \times 10^{-11}\) m.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Interplanar Spacing
Interplanar spacing is a crucial concept in crystallography. It refers to the distance between adjacent planes in a crystal structure, which is measured using X-ray diffraction. This spacing, denoted by the symbol \(d\), helps determine the wavelength needed for X-rays in order to explore the crystal's properties. A common measurement unit for interplanar spacing is the angstrom (\(\AA\)), where \(1 \AA = 1 \times 10^{-10} \text{ meters}\). Converting this to meters is essential before applying formulas like Bragg's Law in scientific calculations.

To convert interplanar spacing from angstroms to meters, you simply multiply the given value in angstroms by \(1 \times 10^{-10} \). This conversion is straightforward and prepares your measurements for topology analysis or other calculations that require SI units. Understanding this basic conversion is vital for any student dealing with crystallography or related fields.
Wavelength Calculation
The calculation of the wavelength of X-rays used in examining crystal structures is crucial for understanding material properties. One of the primary tools for this is Bragg's Law, which can be represented by \( n\lambda = 2d \sin\theta \). This law shows the relationship between the wavelength \(\lambda\), the interplanar spacing \(d\), and the angle of incidence \(\theta\). Here, \(n\) is the order of reflection, commonly set to 1 in basic problems to simplify calculations.

To find the wavelength, rearrange Bragg's Law to solve for \(\lambda\):
\[ \lambda = \frac{2d \sin\theta}{n} \]
With the interplanar spacing already in meters and the angle in radians, inserting these values into the equation allows you to find the wavelength. This calculation is pivotal for determining the appropriate X-ray wavelength that penetrates a crystal and reflects properly, ensuring clear readings and discoveries related to the structure of the material.
Angle Conversion
Converting angles from degrees to radians is a necessary step when performing calculations involving trigonometric functions and scientific formulas like Bragg's Law. The reason for this conversion is that these functions typically require the input angle in radians instead of degrees.

To convert degrees to radians, use the conversion factor \(\frac{\pi \text{ radians}}{180 \text{ degrees}}\). For example, if your angle is \(15.0^\circ\), the conversion to radians involves multiplying:
\[ \theta = 15.0 \times \frac{\pi}{180} \]
This conversion simplifies the mathematical equations you'll perform, making calculations more aligned with the standards in physics and engineering.

Understanding the necessity and method of converting angle measurements ensures that when applying formulas or using tools like calculators, the values are correctly interpreted. This step is essential for accuracy in discussing wave interactions with crystals or analyzing other problems involving wave mechanics.

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Most popular questions from this chapter

What is the formula for the compound that crystallizes with a cubic closest packed array of sulfur ions, and that contains zinc ions in \(\frac{1}{8}\) of the tetrahedral holes and aluminum ions in \(\frac{1}{2}\) of the octahedral holes?

MnO has either the \(\mathrm{NaCl}\) type structure or the \(\mathrm{CsCl}\) type structure (see Exercise 67). The edge length of the \(\mathrm{MnO}\) unit cell is \(4.47 \times 10^{-8} \mathrm{~cm}\) and the density of \(\mathrm{MnO}\) is \(5.28 \mathrm{~g} / \mathrm{cm}^{3}\) a. Does \(\mathrm{Mn} \mathrm{O}\) crystallize in the \(\mathrm{NaCl}\) or the \(\mathrm{CsCl}\) type structure? b. Assuming that the ionic radius of oxygen is \(140 . \mathrm{pm}\), estimate the ionic radius of manganese.

Barium has a body-centered cubic structure. If the atomic radius of barium is \(222 \mathrm{pm}\), calculate the density of solid barium.

What quantity of energy does it take to convert \(0.500 \mathrm{~kg}\) ice at \(-20 .{ }^{\circ} \mathrm{C}\) to steam at \(250 .{ }^{\circ} \mathrm{C} ?\) Specific heat capacities: ice, \(2.03 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} ;\) liquid, \(4.2 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} ;\) steam, \(2.0 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} ; \Delta H_{\mathrm{vap}}=\) \(40.7 \mathrm{~kJ} / \mathrm{mol} ; \Delta H_{\mathrm{fus}}=6.02 \mathrm{~kJ} / \mathrm{mol}\)

Rubidium chloride has the sodium chloride structure at normal pressures but assumes the cesium chloride structure at high pressures. (See Exercise 67.) What ratio of densities is expected for these two forms? Does this change in structure make sense on the basis of simple models? The ionic radius is \(148 \mathrm{pm}\) for \(\mathrm{Rb}^{+}\) and 181 pm for \(\mathrm{Cl}^{-}\).

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