A hanging iron wire with diameter 0.08cm is initially 2.5m long. When a 52 kg is hung from it, the wire stretches an amount 1.27cm. A mole of iron has a mass of 56g, and its density is$\mathbf{7}\mathbf{.}\mathbf{87}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$ . (a) What is the length of an interatomic bond in iron (diameter of one atom)? (b) Find the approximate value of the effective spring stiffness of one interatomic bond in iron.

The given data can be listed below,

• The diameter of iron wire is,$d=0.08cm\left(\frac{1cm}{100cm}\right)=0.8\times {10}^{-3}cm$.
• The mass of object hanging on wire is,$m=52kg$
• The length of the wire is, $L=2.5m.$.
• The stretch in wire is,$∆L=1.27cm$
• The mass of one mole of iron is,$M=56g$
• Density of the iron is,$\rho =7.78g/c{m}^3$

The gradient of deformation versus applied force across a linear range is known as Young's modulus. The deformation graph, in other words, is a straight line. Deformation recovers when the force decreases within that range.

The mass of one atom is given by,

$m_a=\frac{M}{N_A}$

Here, M is the mass of one mole of iron and $N_A$is the Avogadro number.

Substitute value in the above,

$$\begin{gathered} m_a=\frac{56g}{6.02\times {10}^{-23}g/atom} \\ =9.3\times {10}^{-23}g/atom \end{gathered}$$

The length of interatomic bond is given by,

$d_a=\left(\frac{m_a}{\rho }\right)$

Here, $m_2$is the mass of one tom and$\rho$ is the density of iron atom.

Substitute all values in the above,

$$\begin{gathered} d_a={\left[\left(\frac{9.3\times {10}^{-23g/atom}}{7.78g/c{m}^3}\right)\right]}^{1/3} \\ =2.28\times {10}^{-10} \end{gathered}$$

Thus, the length of interatomic bond in iron is 2.27$\times {10}^{-10}m$.

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.

The young modules of iron is given by,

$Y=\frac{4mgL}{\left(∆L\right)\left(∆d^2\right)}$

Here, m is the mass of wire, g is the acceleration due to gravity, L is the length of the wire, is the stretch in wire and d is the diameter of wire.

Substitute all the values in the above,

$$\begin{gathered} Y=\frac{4\left(52\right)\left(9.8m/s^2\right)\left(2.5m\right)}{\left(1.27cm\right)\left({\displaystyle \frac{1m}{100cm}}\right)\mathrm{\pi }{\left(0.8\times {10}^{-3}\mathrm{m}\right)}^2} \\ =2.0\times {10}^{11}N/m^2 \end{gathered}$$

The stiffness of the interatomic force is given by,

$k=Y{d}_a$

Here, Y is the young modules and$d_a$ is the length of interatomic bond.

Substitute all the values in the above expression.

$$\begin{gathered} k=\left(2\times {10}^{11}N/m^2\right)\left(2.28\times {10}^{-10}m\right) \\ =45.6N/m \end{gathered}$$

Thus, the approximate value of effective spring stiffness of one interatomic bond isdata-custom-editor="chemistry" $45.6\mathrm{N}/\mathrm{m}$ .