Manipal Medical Manipal Medical Solved Paper-2007

  • question_answer
    Copper has face-centred cubic (fee) lattice with interatomic spacing equal to\[2.54\overset{o}{\mathop{\text{A}}}\,\]. 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)  \[3.59\overset{o}{\mathop{\text{A}}}\,\]

    Correct Answer: D

    Solution :

     Interatomic spacing for fee lattice \[r={{\left[ {{\left( \frac{a}{2} \right)}^{2}}+{{\left( \frac{a}{2} \right)}^{2}}+{{(0)}^{2}} \right]}^{1/2}}=\frac{a}{\sqrt{2}}\] a being lattice constant. \[\therefore \] \[a=\sqrt{2}r=\sqrt{2}\times 2.54=3.59\overset{o}{\mathop{\text{A}}}\,\]


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