91Ó°ÊÓ

The given data can be listed below,

• The young modulus of Aluminium is,$Y_A=6.2×{10}^{10}N/m^2$
• The density of Aluminium is,${\mathrm{ÒÏ}}_{Al}=2.7g/c{m}^3$
• The mass of one mole of aluminium is,$M_{Al}=27g\left(\frac{1kg}{1000kg}\right)=0.027kg$
• Young modulus of Lead is,$Y_{Pb}=1.6×{10}^{10}N/m^2$
• The density of lead is,${\mathrm{ÒÏ}}_{Pb}=11.4g/c{m}^3$
• Molecular mass of Lead is,$M_{Pb}=207g\left(\frac{1kg}{1000g}\right)=0.207kg$

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

The young modulus of a material is given by,

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

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

The volume of aluminium atoms is given by,

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

Here, M_Al is the mass,${\mathrm{ÒÏ}}_{Al}$ is the density of Aluminium and$N_A$ is the Avogadro number.

Substitute all the values in the above,

$$\begin{gathered} V=\frac{0.027}{2.7g/c{m}^3\left({\displaystyle \frac{1000kg/m^3}{1g/c{m}^3}}\right)}\left(\frac{1mol}{3.023×{10}^{23}atom}\right) \\ =1.7×{10}^{-29}{m}^3 \end{gathered}$$

The spring constant for Aluminum is given by,

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

Substitute values in above,

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

Thus, the value of approximate spring constant of Aluminium 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 modulus of Lead and$d_{Pb}$ is the interatomic distance of Lead.

The volume of Lead is given by,

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

Substitute all the values in the above expression.

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

Substitute all the values in the equation (i).

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

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