What is the edge length of the unit cell of a body-centred cubic crystal of an element whose atomic radius is 75 pm?
- 170 pm
- 175 pm
- 178 pm
- 173.2 pm
The Correct Option is D
Solution and Explanation
For a body-centred cubic (BCC) crystal, the relationship between the edge length $a$ and the atomic radius $r$ is:
$a = \frac{4r}{\sqrt{3}}$
Given: $r = 75 \, \text{pm}$
Substituting the value:
$a = \frac{4 \times 75}{\sqrt{3}} = \frac{300}{\sqrt{3}} \approx 173.2 \, \text{pm}$
Therefore, the edge length of the unit cell is 173.2 pm.
Correct option: (D) 173.2 pm
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Concepts Used:
Solid State
Solids are substances that are featured by a definite shape, volume, and high density. In the solid-state, the composed particles are arranged in several manners. Solid-state, in simple terms, means "no moving parts." Thus solid-state electronic devices are the ones inclusive of solid components that don’t change their position. Solid is a state of matter where the composed particles are arranged close to each other. The composed particles can be either atoms, molecules, or ions.
Types of Solids:
Based on the nature of the order that is present in the arrangement of their constituent particles solids can be divided into two types;
- Amorphous solids behave the same as super cool liquids due to the arrangement of constituent particles in short-range order. They are isotropic and have a broad melting point (range is about greater than 5°C).
- Crystalline solids have a fixed shape and the constituent particles are arranged in a long-range order.