The second order Bragg diffraction of X-rays with `lambda=1.0 Å` from a set of parallel planes in a metal occurs at an angle `60^(@)`. The distance between the scattering planes in the crystals is

Options

A. `0.575Å`
B. `1.00Å`
C. `2.00Å`
D. `1.17Å`

Explanation

To solve the problem of finding the distance between the scattering planes in a crystal using Bragg's Law, we will follow these steps:

### Step 1: Write down the given information
- Wavelength of X-rays, \( \lambda = 1.0 \, \text{Å} \)
- Angle of diffraction, \( \theta = 60^\circ \)
- Order of diffraction, \( n = 2 \) (since it is second-order diffraction)

### Step 2: Recall Bragg's Law
Bragg's Law is given by the equation:
\[ n \lambda = 2d \sin \theta \]
Where:
- \( n \) is the order of diffraction
- \( \lambda \) is the wavelength of the X-rays
- \( d \) is the distance between the scattering planes
- \( \theta \) is the angle of diffraction

### Step 3: Rearrange the equation to solve for \( d \)
From Bragg's Law, we can rearrange the equation to find \( d \):
\[ d = \frac{n \lambda}{2 \sin \theta} \]

### Step 4: Substitute the known values into the equation
Now, substitute the values into the equation:
\[ d = \frac{2 \times 1.0 \, \text{Å}}{2 \sin(60^\circ)} \]

### Step 5: Calculate \( \sin(60^\circ) \)
We know that:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

### Step 6: Substitute \( \sin(60^\circ) \) into the equation
Substituting \( \sin(60^\circ) \) into the equation gives:
\[ d = \frac{2 \times 1.0 \, \text{Å}}{2 \times \frac{\sqrt{3}}{2}} = \frac{2 \, \text{Å}}{\sqrt{3}} \]

### Step 7: Simplify the expression
The \( 2 \) in the numerator and denominator cancels out:
\[ d = \frac{1 \, \text{Å}}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \, \text{Å} \]

### Step 8: Calculate the numerical value
To find the numerical value, we can approximate \( \sqrt{3} \approx 1.732 \):
\[ d \approx \frac{2}{1.732} \approx 1.1547 \, \text{Å} \]

### Step 9: Round off the value
Rounding off gives us:
\[ d \approx 1.17 \, \text{Å} \]

### Conclusion
The distance between the scattering planes in the crystal is approximately \( 1.17 \, \text{Å} \).