91Ó°ÊÓ

The given data is listed below.

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

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.

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

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

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

Substitute all the values in the above.

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

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.

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

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

The spring constant of lead is given by

$$\begin{gathered} k_{Pb}=Y_{Pb}{d}_{Pb} \\ =Y_{Pb}{\left(V_{Pb}\right)}^{1/3} \end{gathered}$$ …(¾±)

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

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

Substitute all the values in the above expression.

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

Substitute all the values in equation (i).

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

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