What is the density of a solid'? Although the mass densities of solids vary greatly, the number densities (in $\mathbf{mol}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$) vary surprisingly little. The value. of course, hinges on the separation between the atoms-but where does a theoretical prediction start? The electron in hydrogen must not as a particle orbiting at a strict radius, as in the Bohr atom. but as a diffuse orbiting wave. Given a diffuse probability, identically repeated experiments dedicated to "finding" the electron would obtain a range of values - with a mean and standard deviation (uncertainty). Nevertheless, the allowed radii predicted by the Bohr model are very close to the true mean values.

(a) Assuming that atoms are packed into a solid typically jBohr radii apart, what would be the number of moles per cubic meter?

(b) Compare this with the typical mole density in a solid of${\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{mol}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$. What would be the value of j?

Step by step solution

01

Expression to determine the number of atoms and molecules and the typical mole of density.(a)

The expression to determine the number of atoms per meter cubes is given by,

$\frac{\mathbf{1}}{\mathbf{\left(}{\mathbf{ja}}_{\mathbf{0}}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{j}}^{\mathbf{3}}{\mathbf{a}}_{\mathbf{0}}^{\mathbf{3}}}$

The expression for the number of molecules per cubic meter is given by,

$\frac{\mathbf{1}}{{\displaystyle \frac{{\mathbf{\left(}{\mathbf{z}}_{\mathbf{0}}\mathbf{\right)}}^{\mathbf{3}}}{{\mathbf{N}}_{\mathbf{A}}}}}$

The expression for the typical mole density is given by,

$\frac{\mathbf{1}}{{\mathbf{j}}^{\mathbf{3}}}\mathbf{1}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{mol}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}\mathbf{=}\mathit{n}$

02

use the expression of atoms and molecules for calculation.

(a)

The number of atoms per meter cubes is calculated as,

$$\begin{gathered} \frac{1}{{\left(j_0\right)}^3}=\frac{1}{j^3(0.0529\mathrm{nm})} \\ =\frac{1}{j^3(0.0529\mathrm{nm})\left({\displaystyle \frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}}\right)} \\ =\frac{6.76\times {10}^{30}\mathrm{atom}/{\mathrm{m}}^3}{j^3} \end{gathered}$$

The number of molecules per cubic meter is calculated as,

$$\begin{gathered} \frac{{\displaystyle \frac{1}{{\left(j{u}_0\right)}^3}}}{N_A}=\frac{{\displaystyle \frac{6.67\times {10}^{30}\tan /{\mathrm{m}}^3}{j^3}}}{6.02\times {10}^{23}\mathrm{atom}/\mathrm{mol}} \\ =\frac{1}{j^3}1.12\times {10}^7\mathrm{mol}/{\mathrm{m}}^3 \end{gathered}$$

Therefore, the number of molecules per cubic meter is_{role="math" localid="1660023523746" $\frac{1}{j^3}1.12\times {10}^7\mathrm{mol}/{\mathrm{m}}^3$}.

03

Use the formula 1j31.12×107 mol/m3 for calculation.

(b)

The value of j is calculated as,

$$\begin{gathered} \frac{1}{j^3}1.12\times {10}^7\mathrm{mol}/{\mathrm{m}}^3=n \\ \frac{1}{j^3}1.12\times {10}^7\mathrm{mol}/{\mathrm{m}}^3={10}^5\mathrm{mol}/{\mathrm{m}}^3 \\ j=4.8 \end{gathered}$$

Therefore, the value of j is 4.8

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