A hanging copper wire with diameter 1.4 mm $\left(1.4\times {10}^{-3}\right)\mathit{m}$ is initially 0.95m long. When a 36kg mass is hung from it, the wire stretches an amount 1.83mm, and when a 72kg mass is hung from it, the wire stretches an amount 3.66mm. A mole of copper has a mass of 63g, and its density is$\mathbf{9}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$ Find the approximate value of the effective spring stiffness of the interatomic force.

The given data can be listed below,

• The diameter of copper wire is$D=1.4\times {10}^{-3}m$ ,
• 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,$\rho =9g/c{m}^3$

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,

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

Here, d is the diameter of the wire.

Substitute all the values in the above,

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

The volume of a copper atom is given by,

$v=\frac{M}{\rho \times N_A}$

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

Substitute all the values in the above expression.

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

The interatomic distance between the atoms is given by,

$d=V^{1/3}$

Substitute value in the above,

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

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,

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

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

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

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

Substitute all the values in the above,

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

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.

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

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,

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

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