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yawa3891 [41]
3 years ago
7

Determine the number of atoms per unit cell in a (a) face-centered cubic, (b) body- centered cubic, and (c) diamond lattice.

Physics
1 answer:
fredd [130]3 years ago
3 0

Answer:

a) 4

b) 2

c) 8

Explanation:

In a cubic lattice, each atom of the vertex is shared among other 8 unit cells, so the atoms on the vertex contribute to 1/8 to a given unit cell.

a) Ina face-centered cubic we have 8 atoms on the vertex and one in each face, which is shared with another unit cell, so it contribute to 1/2

Therefore, the total atoms are:

8*(1/8) + 6*(1/2) = 4

b) In the body centered cubic structure, the centered atom is not shared with another cell, therefore it contribute to 1 to the given cell:

The number of atoms per unit cel is:

8*(1/8) + 1 = 2

c) The diamond lattice is similar to the face-centered cubic lattice but it contains two identical atoms per lattice point.

Therefore it must contain twice atoms than the face-centered cubic lattice:

that is, it has 8 atoms per unit cell

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3 years ago
Canola oil is less dense than water, so it floats on water, but its index of refraction is 1.47, higher than that of water. When
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therefore critical angle c= 69.79°

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which is higher than that of water

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to calculate critical angle of light going from the oil into water

we know that

sinc= \frac{n_{water}}{n_{canola}}

now putting values we get

sinc= \frac{1.33}{1.47}

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8 0
3 years ago
A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu
seropon [69]

Answer:

\theta=50\ revolution

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, \omega_i=600\ rpm=62.83\ rad/s

Finally it shuts off, \omega_f=0

Time taken, t = 10 s

Let \alpha is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

\alpha =\dfrac{\omega_f-\omega_i}{t}

\alpha =\dfrac{0-62.83}{10}

\alpha =-6.283\ rad/s^2

Let \theta is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }

\theta=\dfrac{-62.83^2}{2\times -6.283}

\theta=314.15\ radian

\theta=49.99\ revolution

or

\theta=50\ revolution

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0
3 years ago
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