91Ó°ÊÓ

The given data can be listed below,

• The diameter of copper wire is$D=1.4×{10}^{-3}m$ ,
• The length of the copper wire is,$L=0.95m$.
• The mass of first hanging object is,$m_1=36kg$
• The mass of second hanging object is, $m_2=72kg$.
• The length of wire stretched due to first mass is, $l_1=1.83mm$.
• The length of wire stretched due to second mass is,$l_2=3.66mm$ .
• The mass of one mole copper is, $M=63g$
• The density of copper is,$\mathrm{ÒÏ}=9g/c{m}^3$

The capacity of a spring to withstand a force is called spring stiffness. The force required to compress or stretch the spring increases as the spring stiffens.

The cross-sectional area of the wire is given by,

$A=\mathrm{Ï€}{\left(\frac{\mathrm{d}}{2}\right)}^2$

Here, d is the diameter of the wire.

Substitute all the values in the above,

$$\begin{gathered} A=3.14×\left(\frac{1.4×{10}^{-3}}{2}\right) \\ =1.538×{10}^{-3}{m}^2 \\ =1.54×{10}^{-6}{m}^2 \end{gathered}$$

The volume of a copper atom is given by,

$v=\frac{M}{\mathrm{ÒÏ}×N_A}$

Here, M is the mass of one mole of copper, is the density of copper and is the Avogadro number whose value is $6.023×{10}^{23}$.

Substitute all the values in the above expression.

$$\begin{gathered} V=\frac{63g}{9g/c{m}^3×6.023×{10}^{23}atom} \\ =1.17×{10}^{-23}c{m}^3/atom \end{gathered}$$

The interatomic distance between the atoms is given by,

$d=V^{1/3}$

Substitute value in the above,

$$\begin{gathered} d_a=\sqrt{1.17×{10}^{-23}c{m}^3{\left(\frac{1m}{100cm}\right)}^3} \\ =2.26×{10}^{-10}m \end{gathered}$$

The stiffness of the wire due to first mass is given by,

$k=\frac{m_{1}g}{l_1}$

Here,$m_1$is the mass of first hanging mass and $l_1$is the length stretched due to mass.

Substitute all the values in the above,

$$\begin{gathered} k_1=\frac{\left(36kg\right)\left(9.8m/s^2\right)}{1.83×{10}^{-3}m} \\ =192.78×{10}^{3}N/m \end{gathered}$$

The number of side-by-side atoms in chain is given by,

$$\begin{gathered} N_{chain}=\frac{A_{wire}}{A_{atom}} \\ =\frac{\mathrm{Ï€}{\left(\mathrm{d}/2\right)}^2}{\left(d_a\right)} \end{gathered}$$

Here, d is the diameter of wire and is the interatomic distance of atoms.

Substitute all the values in the above,

$$\begin{gathered} N_{chain}=\frac{\mathrm{π}{\left(1.4×{10}^{-3}\mathrm{m}/2\right)}^2}{{\left(2.26×{10}^{-10}m\right)}^2} \\ =30.14×{10}^{12}chains \end{gathered}$$

The number of bonds in an atomic chain of wire is given by,

$N_{bond}=\frac{L}{d_a}$

Here, Lis the length of copper wire.

Substitute values in the above equation.

$$\begin{gathered} N_{bond}=\frac{3m}{2.26×{10}^{-10}m} \\ =4.20×{10}^{9}bonds \end{gathered}$$

The stiffness of a single interatomic bond is very smaller than the stiffness of whole copper wire it can be given by,

$k_{aj}=\frac{N_{bond}{k}_1}{N_{chain}}$

Here, $N_{bond}$ is the number of bonds in atomic chain, $k_1$is the stiffness of first mass and $N_{chain}$ is the number of chains.

Substitute all the values in the above expression,

$$\begin{gathered} K_{aj}=\frac{\left(4.20×{10}^9\right)\left(192.78×{10}^{3}N/m\right)}{30.14×{10}^{12}} \\ =27N/m \end{gathered}$$

Thus, the approximate value of stiffness of interatomic force is$27N/m$ .