For a certain metal the stiffness of the interatomic bond and the mass of one atom are such that the spacing of the quantum oscillator energy levels is $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}$J. A nanoparticle of this metal consisting of atoms has a total thermal energy of role="math" localid="1657867970311" $\mathbf{18}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}$J. (a) What is the entropy of this nanoparticle? (b) The temperature of the nanoparticle is 87 K. Next we add$\mathbf{18}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}$ J to the nanoparticle. By how much does the entropy increase?

The given data can be listed below as-

• The energy levels of the spacing of the quantum oscillator is$1.5\times {10}^{-23}\mathrm{J}$ .
• The atoms inside the nanoparticle is 10.
• The thermal energy of the nanoparticle isrole="math" localid="1657872735111" $18\times {10}^{-23}\mathrm{J}$.
• The value of the Boltzmann constant is$1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$ .

The entropy is the energy that is available for doing a certain amount of work.

The concept of entropy gives the entropy of the nanoparticle and the increase of the entropy.

a)

The equation of the energy quanta of a system can be expressed as,

$\mathrm{q}=\frac{\mathrm{thermal}\mathrm{energy}}{\mathrm{quantum}\mathrm{energy}\mathrm{levels}}$

Here, q is the energy quanta of the system

Substituting the values in the above equation,

$$\begin{gathered} \mathrm{q}=\frac{18\times {10}^{-23}\mathrm{J}}{1.5\times {10}^{-23}\mathrm{J}} \\ =12\mathrm{J} \end{gathered}$$

The equation of the oscillator is expressed as,

$\mathrm{N}=3\mathrm{a}$

Here, N is the number of oscillatorsand a is the atoms.

Substituting the values in the above equation,

$$\begin{gathered} N=3\times 10 \\ =30 \end{gathered}$$

The equation of thenumber of the ways to arrange the energy quanta of a system is,

$\Omega =\frac{(q+N-1)!}{q!(N-1)!}$

Here,$\Omega$ is the number of ways, q is the energy quanta and N is the oscillators

Substituting the values in the above equation,

$$\begin{gathered} \Omega \hspace{0.33em}=\frac{(12+30-1)!}{12!(30-1)!} \\ =7.9\times {10}^9 \end{gathered}$$

the equation of the entropy of the nanoparticle can be expressed as,

$S=k\ln \Omega$

Here, S is the entropy of the nanoparticle, k is the Boltzmann’s constant and $\Omega$is the number of ways to arrange energy for the nanoparticle.

Substituting the vales in the above equation,

$$\begin{gathered} \mathrm{S}=\mathrm{K}\mathrm{ln\Omega } \\ =1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\ln (7.9\times {10}^9)\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} \\ =3.15\times {10}^{-22}\mathrm{J}/\mathrm{K} \end{gathered}$$

Thus, the entropy of the nanoparticle is role="math" localid="1657868298073" $3.15\times {10}^{-22}\mathrm{J}/\mathrm{K}$.

b)

The given temperature of nanoparticle = 87 K

The given extra energy added to the nanoparticle $18\times {10}^{-23}\mathrm{J}$

The equation of the slope at the temperature is expressed as-

$s=\frac{1}{T}$

Here, s is the slope and T is the temperature

Substituting the values in the above equation,

$$\begin{gathered} \mathrm{s}=\frac{1}{87\mathrm{K}} \\ =0.0115{\mathrm{K}}^{-1} \end{gathered}$$

With the help of the extra energy, the entropy goes up.

The equation change in the entropy is expressed as-

$\mathrm{\Delta E}=\mathrm{sE}$

Here,$∆E$is the change in the entropy and E is the added energy

Substituting the values in the above equation,

$$\begin{gathered} ∆E=0.0115K^{-1}\times 18\times {10}^{-23}\mathrm{J} \\ =2.1\times {10}^{-24}\mathrm{J}/\mathrm{K} \end{gathered}$$

Thus, the entropy has increased by about $2.1\times {10}^{-24}\mathrm{J}/\mathrm{K}$.