91Ó°ÊÓ

The given data can be listed below,

• The diameter of iron wire is,$d=0.08cm\left(\frac{1cm}{100cm}\right)=0.8×{10}^{-3}cm$.
• 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,$\mathrm{ÒÏ}=7.78g/c{m}^3$

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,

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

The length of interatomic bond is given by,

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

Here, $m_2$is the mass of one tom and$\mathrm{ÒÏ}$ is the density of iron atom.

Substitute all values in the above,

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

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

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.

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,

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

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.

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

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