1) (From Textbook Problem 3.82 ) The first three diffraction peaks of a metal powder are 20 = 44.4, 64.6, and 81.70 using Cuk -radiation. Is this a bec or an fcc metal? 2. (From Textbook Problem 4.2) The Al-Mg system shows incomplete solid solution . Which of the Hume -Rothery rules are violated ? Note that r = 0.143 nm, r = 0.160 nm, and electronegativityes can be found in Fig. 2.21 (Al: 1.5; Mg: 1.2 ).

Answer is as follows : 

1) As edge length of unit cell is approximately the same for all three diffraction peaks, the metal has a bcc            structure. 

Step-by-step explanation

Step: 1 The difference in path lengths between the adjacent x—ray beams is integral number (n) of radiation wavelength (1). at = 2dsin6 ...... (1) Here, d is the spacing between adjacent crystal planes. 9, angle of scattering. The radiation wavelength ( ,1.) for CuKa, radiation is 0.1542 nm and for powder samples n = 1. The angles (2 9) of ?rst three diffraction peaks of a metal powder are 44.6", 64.6", and 813°- Solve equation (1) for d: 2. d= , 25m6

Step: 2 Substitute the values and calculate dj, d2, and d3 for three diffraction peaks: 0.1542 nm d,
= 44.4 2sin 2 = 0.204 nm 0.1542 nm d, = 64.6 2sin 2 = 0.144 nm 0.1542 nm d; = 81.7 2sin 2 = 0.1 18
nm

Step: 3 Reflection rules of X-ray diffraction for the BCC, FCC metal structures: Crystal Diffraction occurs when structure (h + k + / = even): 110 BCC 200 211 (h, k, I unmixed - all even numbers or all odd numbers): FCC 111 200 220 The spacing between adjacent hkl planes is given by, a d nki V12 + * 2 + 12

Step: 4 Assume metal powder to be BCC, calculate edge length of unit cell (a): a=d xvh2+k2+ 12 = 0.204 nmxv12 + 12 + 02 = 0.288 nm a=d2 x Vh2 + k2 + 12 = 0.144 nmxv22 + 02 + 02 = 0.288 nm a =d3 x Vh2+ k2+12 = 0.118 nmxv22 + 12 + 12 = 0.289 nm As edge length of unit cell is approximately the same for all three diffraction peaks, the metal has a bec structure.

Ans:- The Hume- Rothery rules helps to confirm if two elements can form a substitution solid solution or not According to these rules two elements form a substitution solid solution if . They have similar atomic radii . They have similar electronegativity . They have same crystal structure . They have similar valency . - The Al- My system shows incomplete solid solution because voilates the following Hume-Rothery rules- 1. They have large electronegativity difference . Al has 1. 5 electronegativity and My has 1.2 electronegativity. 2. They have different crystal structure. Al has FCC crystal and My has HCP crystal. 3. They have different valency . Al has valency of 3 and My has 2.