A) \[{{A}_{4}}{{B}_{4}}{{O}_{7}}\]
B) \[{{A}_{8}}{{B}_{6}}{{O}_{7}}\]
C) \[{{A}_{8}}{{B}_{8}}{{O}_{7}}\]
D) \[{{A}_{6}}{{B}_{8}}{{O}_{6}}\]
Correct Answer: B
Solution :
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 \[\sqrt{2}\,mv'=(2m)\,v\]
You need to login to perform this action.
You will be redirected in
3 sec
