Gawaingnpang nkabuhayan,hamon at oportinidad

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

fredd [130]

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

3 0

3 years ago

What is the standard measurement system used by scientists around the world?

sleet_krkn [62]

Answer:

Hey mate....

Explanation:

This is ur answer....

<h3><em>The International System of Units (SI), commonly known as the metric system, is the international standard for measurement.</em></h3>

Hope it helps!

Brainliest pls!

Follow me! ◇

5 0

2 years ago

Read 2 more answers

Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 2.6 m long, its mass is 0.5

Irina-Kira [14]

Answer:

15.13 m/s

Explanation:

The wave speed of the stretched rope can be calculated using the following formula

$v = \sqrt{\frac{F_T}{\mu}}$

where $F_T = 44N$ is the tension on the rope and $\mu = m/L = 0.5 / 2.6 = 0.1923 kg/m$ is the density of the rope per unit length

$v = \sqrt{\frac{44}{0.1923}} = \sqrt{228.8} = 15.13 m/s$

6 0

3 years ago

A diver shines a flashlight upward from beneath the water at a 35.2° angle to the vertical. at what angle does the light leave t

AnnyKZ [126]

We can solve the problem by using Snell's law:
$n_i \sin \theta_i = n_r \sin \theta_r$
where
$n_i$ is the refractive index of the first medium (in this case, air, so $n_i = 1.00$ )
$n_r$ is the refractive index of the second medium (in this case, water, so $n_r = 1.33$ )
$\theta _i$ is the angle of incidence of the light, with respect to the vertical, so $\theta_i = 35.2^{\circ}$
$\theta_r$ is the angle of refraction of the light inside the water, with respect to the vertical

Re-arranging the equation and using the data of the problem, we can find the the angle of refraction of the light inside the water:
$\theta_r = \arcsin ( \frac{n_i}{n_r} \sin \theta_i )=\arcsin ( \frac{1.00}{1.33} \sin 35.2^{\circ} )=25.7^{\circ}$

3 0

3 years ago

A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a right turn and travels 1.33 km east before maki

charle [14.2K]

Answer:

$AD\approx1.7582\ km$