The radius of an indium atom is 0.163 nm. If indium crystallizes in a face-centered unit cell, what is the length of an edge of the unit cell

The edge length of the unit cell is 0.461 nm

Explanation:

We are given:

Atomic radius of iridium = 0.163 nm

To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:

Putting values in above equation, we get:

Hence, the edge length of the unit cell is 0.461 nm

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

indicate valleys, hills, steepness and gentleness of slopes.

explanation:

contour lines on a map indicate elevation, or depth. numeric notations on contour lines refer to the level of elevation, or height about sea level.