Good Luck! 1. A crystal, made of atoms of radius \'R\', has a face-centered cubi

Question

Good Luck! 1. A crystal, made of atoms of radius 'R', has a face-centered cubic (FCC) structure at temperatures above 1 to 400 constant of the HCP unit cell at 000 K. The lattice constant of the FCC unit cell at 1001 K is"z" On cooling , this crystal transforms to hexagonal-closed packed (HCP) structure. The lattice 401 K being a' and cell height'c';c/a ratio being 1.633.O cooling to 300 K, it transforms to simple cubic structure. Its lattice constant is "p' at 299 K. G) What is the change in volume between 1001 K and 401 K? (i) What is the volume change when the crystal transforms between 401 K and 299 K? (iii) what is the overall volume change between 1001 K to simple cubic at 299 K? (iv) If the atomic weight of the atom is Q, what is the density at 1001 K, and at 299 K? 2. The attractive force between two atoms as a function of distance R is given by the expression: att The repulsive force is given by the expression: rep Both constants A and B are >0. Calculate the () equilibrium radius, Ro, and (ii) equilibrium force, Fo. Prove that this is a stable equilibrium. 3. On page 333 of the text book, the Fe-FesC phase diagram is shown. Just below the eutectoid temperature (i) What is the composition (in wt%C) of F&C; phase? (ii) What is the composition (in wt%C) of the ferrite phase? (iii) If the overall alloy has 0.2 wt%C, then what is the wt% of ferrite and the FeC phases? (iv) If the overall alloy has 0.2wt%C, what is the wt% of the perlite phase present? (v) If the overall alloy has 0.4wt%C, what is the wt% of ferrite phase that is present only in the pearlite phase? (vi) In the Fe-FesC phase diagram, state all the reactions between distinct phases by writing cach reaction, its temperature, and the composition at which it occurs. Be careful! Look carefully! Think!

Explanation / Answer

lattice contant is defined as the lengths of the sides of the unit cell such as a,b,c .but for a cubic unit cell all values are equal so it has only one lattice contant .

here atomic radius is given as R

at T=1000k it forms FCC unit cell and its lattice contant is =z

for FCC 20.5z=4R

z=2×20.5R

volume of cube=z3=(2×20.5R)3=22.62R3 -(1)

at 400k unitcell has HCP structure where a is the side and h is the height of the hexagon

c÷a=1.633(given)

for HCP a=2R

then c=1.633×2R=3.266R

volume of HCP unit cell is given by 6×30.5a2c÷4

put a=2R and c=3.266R on solving be get

volume of HCP=25.79R3 -(2)

at 300k it has simple cubic unit cell

for simple cubic a=2R

volume =a3=(2R)3=8R3 -(3)

(1) change in volume between 1000k to 400k

from eq(1)and eq(2)

25.79R3-22.62R3=3.1625R3

(2)change in volume between 401k to 299k

from eq(2) and eq(3)

25.79R3-8R3=17.79R3

(3)overall volume change from 1001k to 299k

from eq(1) and eq(3)

22.62R3-8R3=14.62R3

atomic weight=Q

than atomic mass=Q÷g (g=gravitational canstant)

than mass of one atom =Q÷(g×n)

where n=6.022×1023 called avagadro's number

than for density at 1000k it has FCC unit cell structure for FCC no. of atoms in an unit cell=4

than mass of unit cell=4Q÷(n×g)

density =mass÷volume

density=4Q÷22.62R3ng = .1768Q÷ngR3

now to find density at 299k structure is simple cubic

total no of atoms in a simple cubic unit cell= 1

mass of unit cell=Q÷ng

density= mass÷volume

density=Q÷8R3ng=.125Q÷ngR3  

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