(a) What is the separation in energy between the lowest two energy levels for a container 20 cmon a side containing argon atoms? Assume, for simplicity, that the argon atoms are trapped in a one-dimensional well20cmwide. The molar mass of argon is$\mathbf{39}\mathbf{.}\mathbf{9}\mathit{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{.}$

(b) At 300k, to the nearest power of ten, what is the ratio of the thermal energy of the atoms to this energy separation?

(c) At what temperature does the thermal energy equal the energy separation?

Only a few distinct states of an electron in an infinite potential well are possible. The energy connected to these quantum states if the well is one-dimensional and has length Lare

${\mathit{E}}_{\mathbf{n}}\mathbf{=}\left(\frac{h^2}{8m{L}^2}\right){\mathit{n}}^{\mathbf{2}}$ …… (1)

Where, h is plank’s constant, and n is quantum number, m is mass of atom, and is L length.

We have, given that L = 20 cm

The molar mass of argon is$39.9g/mol$.

$T=300K$

Plank constant is given by$h=6.626\times {10}^{-34}J.s$

Boltzmann constant is given by$k_b=1.38\times {10}^{-23}J/K$

(a)

From equation (1) we have,

$E_n=\left(\frac{h^2}{8m{L}^2}\right)n^2$

Where, his plank’s constant, and n is quantum number, m is mass of atom, and Lis length.

Now, for findingenergy between the lowest two energy levels for a container 20 cm on a side containing argon atoms, we have to find

localid="1661852775995" $$\begin{gathered} ∆E_{21}=E_2-E_1 \\ ∆E_{21}=\left(\frac{h^2}{8m{L}^2}\right)\left(2^2\right)-\left(\frac{h^2}{8m{L}^2}\right)\left(1^2\right) \\ ∆E_{21}=\left(\frac{h^2}{8m{L}^2}\right)\left(2^2-1^2\right) \\ ∆E_{21}=\frac{3h^2}{8m{L}^2}..............\left(2\right) \end{gathered}$$

Now, we have given that the molar mass of argon is 39.9 g/mol .

So, mass of argon is given by

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

Where, M is molar mass of argon and$N_A$is Avogadro number.

So, mass of Argon is given by,

$$\begin{gathered} m=\frac{0.0399kg/mol}{6.022\times {10}^{23}1/mol} \\ =6.626\times {10}^{-26}kg \end{gathered}$$

Now, by substituting all the numerical values in equation (2) we get,

$$\begin{gathered} ∆E_{21}=\frac{3{\left(6.626\times {10}^{-34}J.s\right)}^2}{8\left(6.626\times {10}^{-26}kg\right){\left(0.20m\right)}^2} \\ =6.21\times {10}^{-41}J \end{gathered}$$

Now, we know that$1eV=1.6\times {10}^{-19}J$

So, the value of$∆E_{21}$becomes,

$∆E_{21}=3.88\times {10}^{-22}eV$

Hence, the separation in energy between the lowest two energy levels is

$∆E_{21}=3.88\times {10}^{-22}eV$

(b)

Here thermal energy is same as average kinetic energy and it is given by,

$k=\frac{3}{2}{k}_{b}T$……. (3)

Where K is thermal energy,$k_b$ is Boltzmann constant, and T is temperature.

Now, substituting the values in equation (3) we get,

$$\begin{gathered} k=\frac{3}{2}\left(1.38\times {10}^{-23}J/K\right)\left(300K\right) \\ =6.21\times {10}^{-21}J \end{gathered}$$

Now, we know that$1eV=1.6\times {10}^{-19}J$

So, the value of becomes,

$K=3.88\times {10}^{-2}eV$

Now, the ratio of the thermal energy of the atoms to this energy separation is given by

$$\begin{gathered} \frac{K}{∆E_{21}}=\frac{3.88\times {10}^{-2}eV}{3.88\times {10}^{-22}eV} \\ ={10}^{20} \end{gathered}$$

Hence, the ratio of the thermal energy of the atoms to this energy separation is${10}^{20}$ .

(c)

Now, by step 3 we have, energy separation is$∆E_{21}=3.88\times {10}^{-22}eV$

Also, by step 4 we have, thermal energy is given by,$K=\frac{3}{2}{k}_{b}T$

Now, let energy separation and thermal energy are equal.

So, we get the equation,

$$\begin{gathered} ∆E_{21}=K \\ ∆E_{21}=\frac{3}{2}{k}_{b}T \\ T=\frac{2∆E_{21}}{3K_b} \end{gathered}$$

Now, by substituting the numerical values in above equation we get,

Here we have$k_b=1.38\times {10}^{-23}J/K$

Now, we know that$1eV=1.6\times {10}^{-19}J$

So, the value of$k_b$ becomes

$$\begin{gathered} k_b=\frac{2\left(3.88\times {10}^{-22}eV\right)}{3\left(8.617\times {10}^{-5}eV/K\right)} \\ =3.00\times {10}^{-18}K \end{gathered}$$

Hence, the temperature at which the thermal energy equal the energy separation is$3.00\times {10}^{-18}K$ .