2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    A solid has a 'bcc' structure. If the distance of nearest approach between two atoms is 1.73A,the edge length of the cell is   JEE Main Online Paper (Held On 07 May 2012)

    A) 314.20pm           

    B)                        1.41pm

    C)                        200pm 

    D)                        216pm

    Correct Answer: C

    Solution :

                    For bcc structure\[\text{d=}\frac{\sqrt{3}a}{2}\]where d = distance between two atoms a = edge length \[\text{1}\text{.73=}\frac{\sqrt{3}}{2}a\]\[\text{a=}\frac{2\times 1.73}{\sqrt{3}}=2{\AA}=200pm\]                


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