Question

Inter-planer spacing for a (034) plane in a simple cubic, whose lattice constant is \(4.5 \times 10^{-10}\) m, is:

• A. \(9 \times 10^{-10}\) m
• B. \(9 \times 10^{-11}\) m
• C. \(8 \times 10^{-10}\) m
• D. \(10^{-10}\) m

Hint:

Memorize the interplanar spacing formulas for different crystal systems. For cubic systems, the formula \(d = a/\sqrt{h^2+k^2+l^2}\) is universal and frequently tested. Also, be careful with unit conversions and scientific notation when selecting the final answer.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question requires the calculation of the interplanar spacing, \(d_{hkl}\), for a specific crystal plane in a simple cubic lattice using the given lattice constant.
Step 2: Key Formula or Approach:
For a cubic crystal system (simple, body-centered, or face-centered), the interplanar spacing \(d_{hkl}\) for a plane with Miller indices (hkl) is given by the formula:
\[ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \]
where 'a' is the lattice constant.
Step 3: Detailed Explanation:
We are given:


Lattice constant, \( a = 4.5 \times 10^{-10} \) m
Miller indices, (hkl) = (034)
First, calculate the sum of the squares of the Miller indices:
\[ h^2 + k^2 + l^2 = 0^2 + 3^2 + 4^2 = 0 + 9 + 16 = 25 \]
Next, find the square root of this sum:
\[ \sqrt{h^2 + k^2 + l^2} = \sqrt{25} = 5 \]
Finally, substitute the values into the formula:
\[ d_{034} = \frac{4.5 \times 10^{-10} \text{ m}}{5} = 0.9 \times 10^{-10} \text{ m} \]
To match the options, we can write this in scientific notation as:
\[ d_{034} = 9 \times 10^{-11} \text{ m} \]
Step 4: Final Answer:
The interplanar spacing for the (034) plane is \(9 \times 10^{-11}\) m.