2- Assuming that the oscillating ionic compound is cesium chloride, if you know the atomic weight of chlorine is 35.5 gm/mol and the atomic weight of cesium is 133 gm/mol, and assuming that Hooke's constant is y=4N/m²? NA = 6.02 x 1023 mol-¹ 1. Calculate the maximum frequency of the optical branch? 2. Calculate the frequency of the optical and acoustic branch at the boundary? 3. Determine the range of the forbidden frequency region?

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2- Assuming that the oscillating ionic compound is cesium chloride, if you know
the atomic weight of chlorine is 35.5 gm/mol and the atomic weight of cesium
is 133 gm/mol, and assuming that Hooke's constant is y=4N/m²?
NA = 6.02 x 1023 mol-¹
1. Calculate the maximum frequency of the optical branch?
2. Calculate the frequency of the optical and acoustic branch at the boundary?
3. Determine the range of the forbidden frequency region?
4. Determine the Debye Temperature?
5. If you know that the lattice constant a = 3.14 Å, draw the first Brillion zone
curve with its optical and acoustic branches.
6. If you know that the two atoms have the same mass equal to the mass of the
smaller atom, then repeat the previous calculations and redraw the
differentiation curve?
Transcribed Image Text:2- Assuming that the oscillating ionic compound is cesium chloride, if you know the atomic weight of chlorine is 35.5 gm/mol and the atomic weight of cesium is 133 gm/mol, and assuming that Hooke's constant is y=4N/m²? NA = 6.02 x 1023 mol-¹ 1. Calculate the maximum frequency of the optical branch? 2. Calculate the frequency of the optical and acoustic branch at the boundary? 3. Determine the range of the forbidden frequency region? 4. Determine the Debye Temperature? 5. If you know that the lattice constant a = 3.14 Å, draw the first Brillion zone curve with its optical and acoustic branches. 6. If you know that the two atoms have the same mass equal to the mass of the smaller atom, then repeat the previous calculations and redraw the differentiation curve?
Expert Solution
Part 1)

1. m1=35.5gm/mol m2=133g/molMaximum frequency of optical branch areω+ = 2γ1m1+1m2ω+ = 2×4×135.5+1133=0.53432. Optical branch at boundary ω+ = 2γ1m1 =2×4×135.5=0.4747Acoustical branch at boundary ω+ = 2γ1m2 =2×4×1133=0.245

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