Question

You hang a heavy ball with a mass of 14 kg from a gold wire 2.5 m long that is 2 mm in diameter. You measure the stretch of the wire, and find that the wire stretched 0.00139 m. (a) Calculate Young’s modulus for the wire. (b) The atomic mass of gold is 197 g/mole, and the density of gold is. Calculate the interatomic spring stiffness for gold.

Short answer

a)$7.85\times {10}^{10}N/m$

b)$20.096N/m$

Step-by-step solution

Identification of the given data 

The given data can be listed below as-

• The mass of the object is = 14 kg.
• The atomic weight of the object is =197 g/mol.
• The density of the object is $\rho =19.3g/c{m}^3$.
• The length of the object is L=2.5m.
• The radius of the object is$r=1mm=1\times {10}^{3}m$.
• The change in the length of the object is$∆L=0.00139m$

Significance of the Young’s modulus

The ratio of the stress and the strain of an object gives the value of the Young’s modulus.

The equation of the Young’s modulus gives the Young’s modulus and the interatomic spring stiffness.

Step 3:Calculate the Young’s Modulus for the wire

the equation of the Number of moles in the wire is,

$n=\frac{m}{mw}$

Here, n is the number of moles, mw is the atomic weight

Substituting the values in the above equation,

$$\begin{gathered} n=\frac{14\times {10}^{3}g}{197g/mole} \\ =71.659mol \end{gathered}$$

The equation of the number of atoms is expressed as,

$N=n\times N_A$

Here, N is the number of Atoms, n is the number of moles and is the Avogadro's number

Substituting the values in the above equation,

$$\begin{gathered} N=71.06mol\times 6.023\times {10}^{23} \\ =4.27\times {10}^{25}mol\left[N_A=6.023\times {10}^{23}\right] \end{gathered}$$

The equation of the interatomic bond length can be expressed as-

$d={\left(\frac{ml\rho }{N}\right)}^{\frac{1}{3}}$

Here, d is the interatomic bond length and$\rho$is the density

Substituting the values in the above equation,

$$\begin{gathered} d=\left(\frac{14\times {10}^{3}g}{4.27\times {10}^{25}mol\times 19.3g/c{m}^3}\right) \\ =2.56\times {10}^{-8}cm \end{gathered}$$

or

$=2.56\times {10}^{-10}m$

The equation of the Young’s modulus of the wire is,

$$\begin{gathered} Y=\frac{Stress}{Strain} \\ =\frac{F}{A}\times \frac{L}{∆L} \end{gathered}$$

Y is the Young’s modulus,$∆l$ = change in length, F= force applied load (load), L = original length and A= Cross sectional area

Substituting the values in the above equation,

$$\begin{gathered} Y=\frac{14kg\times 9.8m/s^2}{3.14\times {\left({10}^{-3}m\right)}^2}\times \frac{2.5}{1.39\times {10}^{-3}m} \\ =7.85\times {10}^{10}N/m^2 \end{gathered}$$

Thus, the Young’s modulus for the wire is$=7.85\times {10}^{10}N/m^2$.

Calculate the Interatomic spring stiffness for gold

b)

The equation of the interatomic spring stiffness is,

$k_s=Yd$

Here, $k_s$is the interatomic spring stiffness, Y and d are the Young’s modulus and the density respectively.

Substituting the values in the above equation,

$k=7.85\times {10}^{10}N/m^2\times 2.56\times {10}^{-10}m$

$=20.096N/m$

Thus, the interatomic spring stiffness for gold is$20.096N/m$.