• question_answer Oxygen atoms forms fee unit cell with 'A' atoms occupying all tetrahedral voids and 'B' atoms occupying all octahedral voids. If atoms are removed from two of the body diagonals then determine the formula of resultant compound formed. A)  ${{A}_{4}}{{B}_{4}}{{O}_{7}}$                       B)  ${{A}_{8}}{{B}_{6}}{{O}_{7}}$ C)  ${{A}_{8}}{{B}_{8}}{{O}_{7}}$                       D)  ${{A}_{6}}{{B}_{8}}{{O}_{6}}$

Since ${{N}_{S}}=\left( \frac{\rho S}{\rho P} \right)\times {{N}_{P}}$-atom form fee unit cell, the number of $=\frac{(4.4\times {{10}^{3}})\,\times 100}{220}=2000$ atoms per unit cell ${{\rho }_{S}}$ (face     (cornet- atoms)     atoms) Number of B atoms at octahedral voids = Number of ${{\rho }_{P}}$-atoms = 4 Number of A atoms at tetrahedral voids $-h=-u{{t}_{1}}+\frac{1}{2}gt_{1}^{2}$ Number of $-h=-u{{t}_{2}}+\frac{1}{2}gt_{2}^{2}$-atoms $0=u({{t}_{2}}-{{t}_{1}})+\frac{1}{2}g(t_{1}^{2}-t_{2}^{2})$ Initial formula of compound is $u=\frac{1}{2}g({{t}_{1}}+{{t}_{2}})$ After removal about two body diagonal  $h=\frac{g{{t}_{1}}{{t}_{2}}}{2}$ $\sqrt{2}\,mv$            or         $\sqrt{2}\,mv'=(2m)\,v$