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Calculate the number of atoms per unit cell of a metal having lattice parameter 2.9A0 and density 7.87 gm/cm3. Atomic weight of metal is 55.85, Avagadro number is 6.023 x 1023 /gm-mole.

written 3.5 years ago by teamques10 ★ 68k

modified 2.9 years ago by pedsangini276 • 4.8k

engineering physics

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written 3.5 years ago by teamques10 ★ 68k

Given:

$a=2.9A^\circ =2.9\times 10^{-8} cm$

$ρ=7.87$gm/cm3

$A =55.85$

$N=6.023\times 10^{23}$/gm-mole

Formula:

$a^3 ρ=\dfrac{nA}{N}$

Solution:

$a^3 ρ=\dfrac{nA}{N}$

$∴n=Na^3 \dfrac{ρ}{A}$

$n= \dfrac{6.023\times 10^{23}\times {(2.9\times 10^{-8})}^3\times 7.87}{55.85}$

$n=2$

Number of atoms per unit cell is 2.

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