JEE Main & Advanced Chemistry The Solid State / ठोस प्रावस्था Sample Paper Topic Test - Solid State

  • question_answer
               
    DIRECTION: Read the passage given below and answer the questions that follows:
    In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms sandwiched in between them. A space-of this model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. There spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'.
    The empty space in this HCP unit cell is

    A) 74%

    B) 47.6%

    C) 32%

    D) 26%

    Correct Answer: D

    Solution :

    [d] In a hcp unit cell the space occupied is 74% calculated below                        
    \[\text{Packing fraction=}\frac{\text{Volume of the atoms in a unit cell}}{\text{Volume of a unit cell}}\]
                \[=\frac{6\times \frac{4}{3}\pi {{r}^{3}}}{24\sqrt{2}{{r}^{3}}}=\frac{\pi }{3\sqrt{2}}=\frac{22}{7}\times \frac{1}{3\sqrt{2}}=0.74\]
                or \[74%\]
                \[\therefore \]      Empty space is HCP unit cell
                            \[=(100-74)%\,~m=26%~\]
                i.e., the correct answer is option [d].


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