If the edge length of an NaH unit cell is 488pm, what is the length (in picometers) of an Na−H bond? Na−H has face-centered cubic unit cell. Please help
244pm
Hlo..here is the answer ..in face centred cubic unit cell Na-H Na is present on each corners of the cube and H will be present at edge centre ...Hence the formula for this will a/2 then we have 488/2 then the answer will be 244 I hope you understand..good luck �
To find the length of an Na−H bond in a face-centered cubic (FCC) unit cell, we need to consider the diagonal distance across the unit cell.
In an FCC unit cell, the atoms are located at the corners and the center of each face of the cube. This creates a face-centered arrangement. Each Na−H bond can be visualized as connecting a sodium (Na) atom to a hydrogen (H) atom.
The diagonal distance across the FCC unit cell can be found using the Pythagorean theorem. One edge of the unit cell is given as 488 pm. Let's denote this as "a". The diagonal distance, denoted as "d", can be calculated as follows:
d^2 = a^2 + a^2 + a^2
d^2 = 3a^2
d = sqrt(3a^2)
Substituting the given value for "a":
d = sqrt(3 * 488^2)
Calculating this:
d ≈ sqrt(356352)
d ≈ 597.02 pm
Therefore, the length of an Na−H bond is approximately 597 picometers (pm).
To find the length of an Na-H bond in a face-centered cubic (FCC) unit cell, we can start by visualizing the arrangement of atoms in the unit cell.
In an FCC unit cell, there are atoms at each corner of the cube and at the center of each face. This makes a total of 8 corner atoms and 6 face-centered atoms per unit cell. Each Na atom is surrounded by 4 H atoms, forming a tetrahedral arrangement, and each H atom is surrounded by 4 Na atoms.
Now, let's consider the distance between the centers of two neighboring Na atoms. Since there is one Na atom at each corner and one Na atom at the center of each face, the distance from the center of one Na atom to the center of a neighboring Na atom is half the length of the unit cell edge. This can be calculated as follows:
Na-Na distance = 0.5 * edge length
Substituting the given edge length of 488 pm (picometers):
Na-Na distance = 0.5 * 488 pm = 244 pm
Now, let's calculate the length of an Na-H bond. Since the Na atoms form a tetrahedral arrangement, the Na-H bonds can be visualized as radii extending from the Na atoms' centers to the outermost H atoms. In this arrangement, the Na-H distance is equal to the distance from the center of a Na atom to the center of a neighboring H atom.
Since each Na atom is surrounded by 4 H atoms, the Na-H distance is 1/4th of the Na-Na distance:
Na-H distance = (1/4) * Na-Na distance
= (1/4) * 244 pm
= 61 pm
Therefore, the length of an Na-H bond in an NaH FCC unit cell is 61 picometers.