We have that the edge length is mathematically given as
a=2.96929427×10^{−8}cm
Question Parameters:
the unit cell edge length for a 40 wt% Fe - 60 wt% V alloy.
The room-temperature density and atomic weight of Fe are 7.87 g/cm3 and 55.85 g/mol,
the room-temperature density and atomic weight of V are 6.10 g/cm3 and 50.94 g/mol, respectively.
Generally the equation for the Avg Density is mathematically given as
[tex]Pavg=\frac{100}{(Cv/pv)+(Cfe/pFe)}[/tex]
Therefore
[tex]Pavg=\frac{100}{(60/6.10)+(40/7.87)}[/tex]
Pavg=6.70301g/cm^3
Generally the equation for the atomic weight is mathematically given as
[tex]Aavg=\frac{100}{(Cv/Av)+(Cfe/AFe)}[/tex]
Therefore
Aavg=100/(60/50.94)+(40/55.85)
Pavg=52.796g/mol
Hence the unit cell is[tex]Vc=\frac{nA}{PNa}\\\\Therefore\\\\Vc=\frac{2*52.796}{6.7*6.02e23}[/tex]
Vc=nA/PNa}
Therefore
Vc=2*52.796/6.7*6.02e23
Vc=2.6179402*10^{−23}cm*3 unit cells
The unit of cell edge is
a=Vc^{1/3}
a=(2.6179402*10^{−23})^{1/3}
a=2.96929427×10^{−8}cm
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