Answer: The number of iron atoms in one unit cell are 2
Explanation:
To calculate the number of iron atoms in 1 unit cell, we use the equation to calculate the density of metal, which is:
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density = [tex]7.87g/cm^3[/tex]
Z = number of atom in unit cell = ?
M = atomic mass of metal = 55.85 g/mol
[tex]N_{A}[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]
a = edge length of unit cell = [tex]287pm=287\times 10^{-10}cm[/tex] (Conversion factor: [tex]1cm=10^{10}pm[/tex] )
Putting values in above equation, we get:
[tex]7.87=\frac{Z\times 55.85}{6.022\times 10^{23}\times (287\times 10^{-10})^3}\\\\Z=\frac{7.87\times 6.022\times 10^{23}\times (287\times 10^{-10})^3}{55.85}=2[/tex]
Hence, the number of iron atoms in one unit cell are 2
According to the question,
We know,
→ [tex]\rho = \frac{Z\times M}{N_A\times a^3}[/tex]
By putting the values,
[tex]7.87 = \frac{Z\times 55.85}{6.022\times 10^{23}\times (287\times 10^{-10})^3}[/tex]
[tex]Z = \frac{7.87\times 6.022\times 10^{23}\times (287\times 10^{-10})^3}{55.85}[/tex]
[tex]= 2[/tex]
Thus the solution above is right.
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