Manipal Medical Manipal Medical Solved Paper-2013

  • question_answer
    Copper has facecentered cubic (fee) lattice with interatomic spacing equal to\[2.54\text{ }\overset{o}{\mathop{\text{A}}}\,\], the value of lattice constant for this lattice is

    A)  \[1.27\overset{o}{\mathop{\text{A}}}\,\]             

    B)  \[5.08\overset{o}{\mathop{\text{A}}}\,\]

    C)  \[2.54\overset{o}{\mathop{\text{A}}}\,\]                

    D)  \[3.57\overset{o}{\mathop{\text{A}}}\,\]

    Correct Answer: D

    Solution :

     Interactive spacing for fee lattice \[r={{\left[ {{\left( \frac{q}{2} \right)}^{2}}+{{\left( \frac{q}{2} \right)}^{2}}+{{(0)}^{2}} \right]}^{1/2}}\] \[=\frac{q}{\sqrt{2}}\] Being lattice contact \[q=\sqrt{2}r=\sqrt{2}\times 2.84\] \[=3.5\overset{o}{\mathop{\text{A}}}\,\]


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