53. What fraction of the total volume of a cubic closest packed struc- ture is occupied by atoms? (Hint: V_(sphere)=(4)/(3)pi r^3 ) What fraction of the total volume of a simple cubic structure is occupied by atoms? Compare the answers. 54. Iron has a density of 7.86g/cm^3 and crystallizes in a body- centered cubic lattice. Show that only 68% of a body-centered lattice is actually occupied by atoms , and determine the atomic radius of iron. 55. Explain how doping silicon with either phosphorus or gal- lium increases the electrical conductivity over that of pure silicon. 56. Explain how a p-n junction makes an excellent rectifier.

53. In a cubic closest packed (ccp) structure, each sphere is in contact with 12 others, forming a face-centered cubic arrangement. The volume of each sphere is given by $V_{sphere}=\frac {4}{3}\pi r^{3}$. Since there are 4 spheres per unit cell in a ccp structure, the fraction of the total volume occupied by atoms is $\frac {4}{4+2\sqrt{2}}$, which is approximately 0.74 or 74%.<br /><br />In a simple cubic structure, each sphere is in contact with 6 others. The volume of each sphere is again given by $V_{sphere}=\frac {4}{3}\pi r^{3}$. Since there is only 1 sphere per unit cell in a simple cubic structure, the fraction of the total volume occupied by atoms is $\frac {1}{1+3\sqrt{2}}$, which is approximately 0.52 or 52%.<br /><br />Comparing the two, the fraction of the total volume occupied by atoms in a cubic closest packed structure is greater than that in a simple cubic structure.<br /><br />54. In a body-centered cubic (bcc) lattice, there are 2 atoms per unit cell. The volume of each sphere is given by $V_{sphere}=\frac {4}{3}\pi r^{3}$. The total volume of the unit cell is $a^{3}$, where $a$ is the lattice constant. The fraction of the total volume occupied by atoms is $\frac {2V_{sphere}}{a^{3}}$. Given that the density of iron is $7.86g/cm^{3}$, we can use the formula $\rho = \frac {m}{V}$ to find the atomic radius of iron.<br /><br />55. Doping silicon with phosphorus introduces extra electrons, making it an n-type semiconductor. Doping silicon with gallium introduces holes, making it a p-type semiconductor. The movement of electrons and holes under an electric field increases the electrical conductivity of the doped silicon compared to pure silicon.<br /><br />56. A p-n junction is formed by joining a p-type semiconductor to an n-type semiconductor. The electrons in the n-type semiconductor diffuse into the p-type semiconductor, filling the holes. This creates a depletion region devoid of charge carriers. When a p-type semiconductor is connected to a positive terminal and an n-type semiconductor is connected to a negative terminal, the depletion region widens, allowing current to flow in one direction only, making the p-n junction an excellent rectifier.