157 pm
127 pm
Where:
Given that the edge length (a) is 361 pm, we can calculate the radius (r):
\(4r = \sqrt 2 \times 361\) pm
\(4r = 1.414 \times 361\) pm
\(4r = 510.454\) pm
\(r = \frac {510.454}{4}\)
\(r ≈ 127.6\) pm
The correct answer is (A}: \(127\) pm
To determine the radius of a copper (Cu) atom in a face-centered cubic (fcc) lattice, we begin by recognizing that in an fcc lattice, the atoms are positioned at each corner and the centers of each face of the cube. In such a lattice, the diagonal of the face of the cube is composed of four atomic radii, and this can be expressed by the equation:
\(\sqrt 2a=4r\)
Here, \(a\) is the edge length of the cube, and \(r\) is the atomic radius. Given that the unit cell edge length \(a\) is 361 pm, we can solve for \(r\) as follows:
\(r=\frac{√2a}{4}\)
Substituting the given edge length:
\(r=\frac{√2 \times 361}{4}\) pm
\(r=\frac{510.54 }{4}\) pm
\(r=127.635\) pm
Since the radius should be provided in whole numbers, we round off to 127 pm. Therefore, the radius of the Cu atom is 127 pm.
Identify the part of the sentence that contains a grammatical error:
Each of the boys have submitted their assignment on time.
Rearrange the following parts to form a meaningful and grammatically correct sentence:
P. a healthy diet and regular exercise
Q. are important habits
R. that help maintain good physical and mental health
S. especially in today's busy world