Question

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temperature the crystal structure for this alloy is bcc

Answer 1

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$, we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$, we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting  $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

Answer 2

The edge of a unit cell, by convention, always connects equivalent locations.

The unit cell edge length for an 85 wt% fe -15wt% v alloy is 0.289 nm.

What are alloys?

An alloy is a mixture of chemical elements, mostly metals, and non-metals, but it must include at least one metal.

Now

Given, the wt% of Fe is  85 and the wt% of V is 15.

To calculate the edge length of the unit cell having a bcc structure.

By formula of density

$\bold{\rho_a_v_e = \dfrac{Z \times M_a_v_e}{a_3 \times N_a} }$

To calculate the edge length

$\bold{a =( \dfrac{Z \times M_a_v_e}{P_a_v_e \times N_a}) ^\frac{1}{3} }$

To calculate $M_a_v_e$

$\bold{M_a_v_e = \dfrac{100}{\dfrac{wt\% (Fe)}{M_F_e} +\dfrac{wt\% (V)}{M_V} } }$

To calculate the $\bold{\rho_a_v_e}$

$\bold{\rho_a_v_e = \dfrac{100}{\dfrac{wt\% (Fe)}{\rho_F_e} +\dfrac{wt\% (V)}{\rho_V} } }$

Now,

Mass of Fe is 50.85 g/mol

Mass of V is 50.941 g/mol

The specific density of both elements:

Fe is 7.874 g

V is  6.10 g

By putting the $M_a_v_e$ and $\bold{\rho_a_v_e}$ formula in edge length formula

$a=[ \dfrac{\bold{ Z( \dfrac{100}{\dfrac{wt\% (Fe)}{M_V} +\dfrac{wt\% (V)}{M_V} }) }}{\bold { N_A( \dfrac{100}{\dfrac{wt\% (Fe)}{\rho_F_e} +\dfrac{wt\% (V)}{\rho_V} }) }}]\dfrac{1}{3}$

$a=[ \dfrac{\bold{ 2 atoms / unit\;cell ( \dfrac{100}{\dfrac{85\% }{55.85\;g/mol } +\dfrac{15\% }{50.941\'g/mol} }) }}{\bold { 6.023\times 10^2^3\;a/mol( \dfrac{100}{\dfrac{85\%}{7.874 g } +\dfrac{15\%}{ 6.10 g } }) }}]\dfrac{1}{3}$

$\bold{a =2.89 \times 10^-^8\; cm = 0.289\;cm}$

Thus, the unit cell edge length is 0.289 cm.

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