JEE Main & Advanced Physics Semiconducting Devices Question Bank Solids and Crystals

  • question_answer
    The nearest distance between two atoms in case of a bcc lattice is equal to                                         [J & K CET 2004]

    A)            \[a\frac{\sqrt{2}}{3}\]      

    B)            \[a\frac{\sqrt{3}}{2}\]

    C)            \[q\sqrt{3}\]                        

    D)            \[\frac{a}{\sqrt{2}}\]

    Correct Answer: B

    Solution :

               The nearest distance between two atoms in a bcc lattice = 2 (atomic radius) \[=2\times \left( \frac{\sqrt{3}\,a}{4} \right)=\frac{\sqrt{3}a}{2}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner