<u>Answer:</u>
<u>For a:</u> The edge length of the unit cell is 314 pm
<u>For b:</u> The radius of the molybdenum atom is 135.9 pm
<u>Explanation:</u>
To calculate the edge length for given density of metal, we use the equation:
![\rho=\frac{Z\times M}{N_{A}\times a^{3}}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7BZ%5Ctimes%20M%7D%7BN_%7BA%7D%5Ctimes%20a%5E%7B3%7D%7D)
where,
= density = ![10.28g/cm^3](https://tex.z-dn.net/?f=10.28g%2Fcm%5E3)
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal (molybdenum) = 95.94 g/mol
= Avogadro's number = ![6.022\times 10^{23}](https://tex.z-dn.net/?f=6.022%5Ctimes%2010%5E%7B23%7D)
a = edge length of unit cell =?
Putting values in above equation, we get:
![10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm](https://tex.z-dn.net/?f=10.28%3D%5Cfrac%7B2%5Ctimes%2095.94%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%20%28a%29%5E3%7D%5C%5C%5C%5Ca%5E3%3D%5Cfrac%7B2%5Ctimes%2095.94%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%2010.28%7D%3D3.099%5Ctimes%2010%5E%7B-23%7D%5C%5C%5C%5Ca%3D%5Csqrt%5B3%5D%7B3.099%5Ctimes%2010%5E%7B-23%7D%7D%3D3.14%5Ctimes%2010%5E%7B-8%7Dcm%3D314pm)
Conversion factor used:
Hence, the edge length of the unit cell is 314 pm
To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:
![R=\frac{\sqrt{3}a}{4}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B%5Csqrt%7B3%7Da%7D%7B4%7D)
where,
R = radius of the lattice = ?
a = edge length = 314 pm
Putting values in above equation, we get:
![R=\frac{\sqrt{3}\times 314}{4}=135.9pm](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B%5Csqrt%7B3%7D%5Ctimes%20314%7D%7B4%7D%3D135.9pm)
Hence, the radius of the molybdenum atom is 135.9 pm