Question

Young’s modulus for aluminium is .The density of aluminium is ,and the mass of one mole is 27g. If we model the interactions of neighbouring aluminium 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 and the density of lead is , 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 aluminium is a rather stiff material, whereas lead is quite soft.

Short answer

The value of approximate spring constant for Aluminium is$15.8N/m$ and the spring constant for Lead is 5 N/m .

Step-by-step solution

Identification of given data

The given data can be listed below,

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

Concept/Significance of young modulus

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.

Determination of the approximate spring constant of Aluminium

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,

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

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

Substitute all the values in the above,

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

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,

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

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

Determination of the 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 modulus of Lead and$d_{Pb}$ is the interatomic distance of Lead.

The volume of Lead is given by,

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

Substitute all the values in the above expression.

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

Substitute all the values in the equation (i).

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

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