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JEE MainChemistrySolid State

A solid metal crystallizes in a body-centered cubic (BCC) structure. The macroscopic density of the metal is and its molar mass is M . An interstitial guest atom fits perfectly at the edge center of the unit cell without causing any lattice distortion. If N_A is Avogadro's number, the correct algebraic expression for the radius r of the guest atom is :

Options

  1. A( 2 - 3 4 ) ( 4M N_A )^ 1/3
  2. B( 2 - 2 4 ) ( 2M N_A )^ 1/3
  3. C3 4 ( 2M N_A )^ 1/3
  4. D( 2 - 3 4 ) ( 2M N_A )^ 1/3

Correct answer

D. ( 2 - 3 4 ) ( 2M N_A )^ 1/3

Step-by-step solution

First, relate the edge length a of the unit cell to the macroscopic density . The density formula is: = Z M N_A a^3 For a body-centered cubic (BCC) lattice, the number of atoms per unit cell is Z = 2 . = 2M N_A a^3 a = ( 2M N_A )^ 1/3 Next, determine the radius r of the interstitial guest atom at the edge center. In a BCC lattice, the relationship between the host atom radius R and edge length a is 4R = 3 a . The empty space along the edge is a - 2R . The guest atom fits perfectly, so its diameter 2r equals this em

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