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dmitriy555 [2]
2 years ago
8

Metallic gold crystallizes in a face-centered cubic lattice, with one au atom per lattice point. if the edge length of the unit

cell is found to be 408 pm, what is the metallic radius of au in pm?
Physics
1 answer:
vaieri [72.5K]2 years ago
6 0

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 = \frac{1.414*408}{4}

r = \frac{576.9}{4}

r = 144.2 pm

Therefore, The metallic radius (r) of au is 144.2pm

Learn more about metallic radius (r) here;

brainly.com/question/14885097

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Answer:

Answer:

6.68 x 10^16 m/s^2

Explanation:

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charge of electron, q = 1.6 x 10^-19 C

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Let a be the acceleration of the electron.

The force due to electric field on electron is

F = q E

where q be the charge of electron and E be the electric field

F = 1.6 x 10^-19 x 3.8 x 10^5

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According to Newton's second law

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6.08 x 10^-14 = 9.1 x 10^-31 x a

a = 6.68 x 10^16 m/s^2

Explanation:

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Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th
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The tank has a volume of \dfrac\pi3R^2H, where H=6\,\rm m is its height and R=\dfrac d2=2\,\rm m is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

\dfrac26=\dfrac rh\implies r=\dfrac h3

The volume of water in the tank at any given time is

V=\dfrac\pi3r^2h

and can be expressed as a function of the water level alone:

V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3

Implicity differentiating both sides with respect to time t gives

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}

We're told the water level rises at a rate of \dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min} at the time when the water level is h=2\,\mathrm m=200\,\mathrm{cm}, so the net change in the volume of water \dfrac{\mathrm dV}{\mathrm dt} can be computed:

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})

We're told the water is leaking out at a rate of 10,500\,\frac{\mathrm{cm}^3}{\rm min}, so we find the rate at which it's being pumped in to be

\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}

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4 0
3 years ago
If you increase the distance by double the work will _____________
VMariaS [17]

Answer: the work will also increase by double

Explanation:

This is because they are directly proportional in the formula w=f x d

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