Question

Young’s modulus for aluminum is $\mathbf{6}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{N}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. The density of aluminum is $\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$, and the mass of one mole $\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{atoms}\mathbf{)}$is 27g. If we model the interactions of neighbouring aluminum atoms as though they were connected by spring, determine the approximate spring constant of such a spring. Repeat this analysis for lead is: Young’s modulus for Lead $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{N}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$and the density of lead is $\mathbf{11}\mathbf{.}\mathbf{4}\mathbf{}{\mathbf{\text{g/cm}}}^{\mathbf{\text{3}}}$, and the mass of one mole is 207g. Make a note of these results, which we will use for various purposes later on. Note that aluminum is a rather stiff material, whereas lead is quite soft.

Short answer

The value of the approximate spring constant for aluminum is 15.8 N/m and that for lead is 5 N/m.

Step-by-step solution

Identification of given data

The given data is listed below.

• The Young’s modulus for aluminum is $Y_A=6.2\times {10}^{10}{\text{N/m}}^2$.
• The density of aluminum is ${\rho }_{Al}=2.7{\text{g/cm}}^3$.
• The mass of one mole of aluminum is $M_{Al}=27\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)=0.027\text{kg}$.
• Young modulus of Lead is $Y_{Pb}=1.6\times {10}^{10}{\text{N/m}}^2$.
• The density of lead is ${\rho }_{Pb}=11.4{\text{g/cm}}^3$.
• Molecular mass of Lead is $M_{Pb}=207\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)=0.207\text{kg}$.

Concept/ Definition of Young’s modulusYoung's modulus is defined as the proportion of longitudinal strain to longitudinal stress.

Young’s modulus is a number that is used extensively. It is determined by the material's properties rather than the amount of it in one location.

Determination of approximate spring constant of aluminum

The Young’s modulus of a material is given by

$Y=\frac{Fl}{Ae}$

Here F is the force, l is the length of the material, Ais the area of cross-section and e is the elongation.

The volume of aluminum atoms is given by

$V_{Al}=\frac{M_{Al}}{{\rho }_{Al}}\left(\frac{1\text{mol}}{{\text{N}}_A}\right)$

Here $M_{Al}$is the mass, ${\rho }_{Al}$is the density of aluminum and ${\mathrm{N}}_A$is Avogadro’s number.

Substitute all the values in the above.

$$\begin{gathered} V_{Al}=\frac{0.027\text{kg}}{2.7{\text{g/cm}}^3\left(\frac{1000{\text{kg/m}}^3}{1{\text{g/cm}}^3}\right)}\left(\frac{1\text{mol}}{6.023\times {10}^{23}\text{atom}}\right) \\ =1.7\times {10}^{-29}{\text{m}}^3 \end{gathered}$$

The spring constant for aluminum is given by

$$\begin{gathered} k_{Al}=Y_{Al}{d}_{Al} \\ =Y_{Al}{\left(V_{Al}\right)}^{1/3} \end{gathered}$$

Substitute all the values in the above.

$$\begin{gathered} k_{Al}=\left(6.2\times {10}^{10}{\text{N/m}}^2\right){\left(1.7\times {10}^{-29}{\text{m}}^3\right)}^{1/3} \\ =15.8\text{N/m} \end{gathered}$$

Thus, the value of the approximate spring constant of aluminum is $15.8N/m$.

Determination of approximate spring constant of such a spring for lead

The spring constant of lead is given by

$$\begin{gathered} k_{Pb}=Y_{Pb}{d}_{Pb} \\ =Y_{Pb}{\left(V_{Pb}\right)}^{1/3} \end{gathered}$$ …(i)

Here $Y_{Pb}$is the Young’s modulus of lead and $d_{Pb}$is the interatomic distance of lead.

The volume of lead is given by

$V_{Pb}=\frac{M_{Pb}}{{\rho }_{Pb}}\left(\frac{1\text{mol}}{{\text{N}}_A}\right)$

Substitute all the values in the above expression.

$$\begin{gathered} V_{Pb}=\frac{0.207\text{kg}}{11.4{\text{g/cm}}^3\left(\frac{{10}^3{\text{kg/m}}^3}{1{\text{g/cm}}^3}\right)}\left(\frac{1\text{mol}}{6.023\times {10}^{23}\text{atoms}}\right) \\ =3\times {10}^{-29}{\text{m}}^3 \end{gathered}$$

Substitute all the values in equation (i).

$$\begin{gathered} k_{Pb}=\left(1.6\times {10}^{10}{\text{N/m}}^2\right){\left(3\times {10}^{-29}{\text{m}}^3\right)}^{1/3} \\ =5\text{N/m} \end{gathered}$$

Thus, the spring constant for lead is 5 N/m.