Metallic gold crystallizes in a face-centered cubic lattice, with one au atom per lattice point. if the edge length of the unit
1 answer:
The metallic radius (r) of au is 144.2pm
A face-centered cubic unit cell structure is made up of atoms arranged in a cube with six additional whole atoms placed in the centre of each cube face and a fraction of an atom at each of the cube's four corners. Eight additional unit cells share the atoms at the cube's corner.
Atoms in an FCC crystal are in contact along the unit cell's diagonal. 4r = √2 a, where r is the radius of an atom and a is the edge length of a unit cell.
Putting the values in the above expression,we get
r = √2×408 / 4
r =
r =
r = 144.2 pm
Therefore, The metallic radius (r) of au is 144.2pm
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Answer:
1.08 m/s
Explanation:
This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.
Time taken to fall 9.5 m
vertical acceleration = a = 9.8 m/s^2.
vertical velocity = 0, (since there is only horizontal component for velocity, )
distance traveled s = 9.5 m.
Substituting these values in the equation
⇒ t= 1.392 sec
Velocity needed
We know the time taken (1.392 s) to travel 1.5 m,
So velocity = 1.5 m / 1.392 s = 1.08 m/s
hence velocity of the diver must be at least 1.08 m/s