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A student wants to arrange 4 identical s...

A student wants to arrange 4 identical spheres (of radiusR) on a two dimensional floor as close as possible. Finally he could arrange them. Identify the type and dimensions of the figure obtained by joining their centers. Could he occupy the whole available space, if not, then what type of voids were generated. Calculate their number and radius of small ball that can be fitted in them.

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To solve the problem of arranging four identical spheres of radius \( R \) on a two-dimensional floor as close as possible, we can follow these steps: ### Step 1: Arrangement of Spheres To arrange four identical spheres as close as possible, we can position them in a square formation. The centers of the spheres will form the vertices of a square. ### Step 2: Identify the Figure When we join the centers of the four spheres, we obtain a square. The dimensions of the square can be determined as follows: - The distance between the centers of two adjacent spheres is equal to \( 2R \) (since the radius of each sphere is \( R \)). - Therefore, the side length of the square formed by the centers is \( 2R \). ### Step 3: Space Occupation Next, we need to determine if the spheres occupy the whole available space. Since the spheres are arranged in a square, they do not occupy the entire area of the square. There will be some unoccupied space. ### Step 4: Identify the Type of Voids The unoccupied space between the spheres can be identified as voids. In this arrangement, the voids formed are triangular voids. Specifically, there are two triangular voids formed at the corners of the square. ### Step 5: Calculate the Number of Voids In total, there are **two triangular voids** formed in this arrangement. ### Step 6: Calculate the Radius of Small Balls that can Fit in the Voids To calculate the radius of the small ball that can fit into the triangular voids, we can use the known ratio for triangular voids. The radius \( r \) of the small ball that can fit in a triangular void is given by: \[ r \approx 0.155R \] ### Summary of Results - **Type of Figure**: Square - **Dimensions of Figure**: Side length = \( 2R \) - **Type of Voids**: Triangular voids - **Number of Voids**: 2 - **Radius of Small Ball that can Fit in Voids**: \( r \approx 0.155R \)
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RESONANCE ENGLISH-SOLID STATE-Exercise -1
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  6. Metallic magensium has a hexagoanl close-packed structure and a densit...

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  7. Aluminium metal (atomic weight = 27 g) crystallises in the cubic syste...

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  8. Tom' cat arranges the glass balls, in a particular 3D array, that two ...

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  9. A student wants to arrange 4 identical spheres (of radiusR) on a two d...

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  10. Illustrate an octahedral void in the close packing of spheres. Why is ...

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  11. The number of tetrahedral and octahedral voids in a hexagonal close pa...

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  12. What is the number and closest distance between two octahedral voids a...

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  13. Suggest the probable structures of the unit cells of the following eac...

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  14. Why does ZnS not crystallize in the NaCl structure ?

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  15. Try to answer the following: (i) In Zinc Blende structure there is o...

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  16. KBr crystallizes in NaCl type of unit cell. K^(+) radius = 1.33Å, Br^(...

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  17. What do you understand by imperfections in ionic crystals? Name the ty...

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  18. How would you explain that (i) non-stoichimetric NaCl is yellow, (ii) ...

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  19. Account for the Silicon is an insulator but silicon doped with phospho...

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  20. How does the conductivity of a semiconductor change if its temperature...

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