Ask question

Ask question

All categories

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?

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

#SPJ4

You might be interested in

Given that 1 pound is equal to 4.45 newton’s what is the weight of a 500N child in pounds?

Bond [772]

Answer:

112.36 pounds

Explanation:

Since 1 pound = 4.45 Newtons, a 500N child in pounds = 500÷4.45 = 112.36 pounds (approximately).

4 0

2 years ago

Protons are the smallest particle of an element that retains the original Characteristics of the element. True or False

dlinn [17]

Answer:

False; that’s an atom

Explanation:

5 0

3 years ago

Read 2 more answers

What energy comes from swinging a bat

hjlf

Answer:

Kinetic Energy

Explanation:

The potential energy is being converted into kinetic energy. The hitter has struck the ball transferring the kinetic energy from the swinging bat to the ball.

8 0

3 years ago

You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr

Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

$v_r = 27 - 10 = 17 m/s$

relative deceleration of car

$a_r = -7 m/s^2$

now the distance before they stop with respect to each other is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 17^2 = 2 *(-7)*d$

$d = 20.6 m$

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

$v_f^2 - v_i^2 = 2 a s$

$0^2 - 27^2 = 2 * (-7)* s$

$s = 52.1 m$

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

$v_f - v_i = at$

$0 - 27 = (-7) * t$

$t = 3.86 s$

now the distance covered by truck in same time

$d = 3.86 * 10 = 38.6 m$

now after the car will stop its distance from the truck is

$D = 25 + 38.6 - 52.1 = 11.5 m$

<em>so the distance between them is 11.5 m</em>

6 0

3 years ago

You have a double slit experiment, with the distance between the two slits to be 0.025 cm. A screern is 120 cm behind the double

dimulka [17.4K]

Answer:

The wavelength of the light is 633 nm.

Explanation:

Given that,

Distance between the two slits d= 0.025 cm

Distance between the screen and slits D = 120 cm

Distance between the slits y= 1.52 cm

We need to calculate the angle

Using formula of double slit

$\tan\theta=\dfrac{y}{D}$

Where, y = Distance between the slits

D = Distance between the screen and slits

Put the value into the formula

$\tan\theta=\dfrac{1.52}{120}$

$\theta=\tan^{-1}\dfrac{1.52}{120}$

$\theta=0.725$

We need to calculate the wavelength

Using formula of wavelength

$d\sin\theta=n\lambda$

Put the value into the formula

$0.025\times\sin0.725=5\times\lambda$

$\lambda=\dfrac{0.025\times10^{-2}\times\sin0.725}{5}$

$\lambda=6.326\times10^{-7}\ m$

$\lambda=633\ nm$

Hence, The wavelength of the light is 633 nm.

4 0

2 years ago