Answer:
The percentage volume change is -3.0%
Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn
Value are defined as;
Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3
Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3
Density of alpha-Mn = 7.47g/cm3
Density of beta-Mn = 7.26g/cm3
Atomic weight of Mn = 54.938g/mol
Atomic radius of Mn = 0.112nm
STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;
Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000
(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31
Therefore the number of Atom in alpha-Mn is 58.31 atom per cell
STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn
Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000
(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14
Therefore the number of Atom in beta-Mn is 20.14 atom per cell
STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn
V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000
For Alpha-Mn:
[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%
For Beta-Mn:
[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%
STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA
change = final state - initial state
Therefore;
Change = 111.766 - 114.998 = -3.23%
Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume
