Answer:
I think this answer is number B
Answer:
The distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2
Explanation:
From the image uploaded, a Face centered cubic structure (100) plane, there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell.
In terms of the atomic radius, R, we determine the distance between the centers of adjacent atoms.
Let this distance = AC
the two adjacent sides = AB and BC
AB = a = 2R
BC = a = 2R
Using Pythagoras theorem
AC² = AB² + BC²
AC² = a² + a²
AC² = 2a²
AC = √2a²
AC = a√2
But a = 2R
AC = 2R√2
Therefore, the distance between the centers of adjacent atoms for the FCC crystal structure along the [100] is 2R√2
Correct answer is option E. No dimensions
As we know formula Pressure (P) is
also,
- Dimensional formula of <em>Pressure is </em>

- Dimensional formula of <em>length is L </em>
- Dimensional formula of <em>mass is M</em>
- Dimensional formula of <em>velocity is </em>

So, as given W=
Dimensional formula of W =
since all terms get cancelled
Work is dimensionless i.e no dimensions
Learn more about dimensions here brainly.com/question/20351712
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