Question

A fcc lattice of a metal M and a bcc lattice of metal N contain same number of 2.25×10^22 unit cells.if density of M is twice than that of N the ratio between the number of atoms per unit cell is​

Answer 1

Given that,

A fcc lattice of a metal M and a bcc lattice of metal N contain same number of $2.25\times10^{22}\ unit\ cells$

If the density of M is twice of N

$d_{M}=2d_{N}$

We need to calculate the ratio between the number of atoms per unit cell

Using formula of number of atoms per unit cell

For M and N,

$\dfrac{Z_{M}}{Z_{N}}=\dfrac{\dfrac{a^3d_{M}}{m_{M}}}{\dfrac{a^3d_{N}}{m_{N}}}$

Put the value into the formula

$\dfrac{Z_{M}}{Z_{N}}=\dfrac{\dfrac{(\dfrac{4r}{\sqrt{2}})^3\times2d_{N}}{m}}{\dfrac{(\dfrac{4r}{\sqrt{3}})^3\times d_{N}}{m}}$

$\dfrac{Z_{M}}{Z_{N}}=\dfrac{3\sqrt{3}}{\sqrt{2}}$

Hence, The ratio between the number of atoms per unit cell is $\dfrac{3\sqrt{3}}{\sqrt{2}}$.