Q. 7.66 - An Introduction to Thermal Physics

Question

Consider a collection of 10,000 atoms of rubidium- 87 , confined inside a box of volume ${\left({10}^{-5}m\right)}^3$.

(a) Calculate ${\epsilon }_0$, the energy of the ground state. (Express your answer in both joules and electron-volts.)

(b) Calculate the condensation temperature, and compare $k{T}_c to {\epsilon }_0$.

(c) Suppose that $T= 0.9T_c$ How many atoms are in the ground state? How close is the chemical potential to the ground-state energy? How many atoms are in each of the (threefold-degenerate) first excited states?

(d) Repeat parts (b) and (c) for the case of ${10}^6$ atoms, confined to the same volume. Discuss the conditions under which the number of atoms in the ground state will be much greater than the number in the first excited state.

Step-by-Step Solution

Answer

(a). The energy of the Rubidium in the ground state is $7.1\times {10}^{-14}$.

(b). Condensation temperature is $8.6\times {10}^{-8}K$.

(c). The number of particles in excited state is less than that of the number of particles in ground state.

(d). The number of particles in ground state is greater than that of the number of particles in excited state. 

Step 1. Given information

The energy levels of a particle confined to an infinite well with walls located at $x=0 and x=L$ will be,

$E_n=n^2\frac{h^2}{8m{L}^2}$

Plank's Constant, $h=6.626\times {10}^{-34}J.s$

Mass of particle, $m= 14.4\times {10}^{-26}kg$

Length of box,    $L={10}^{-5}m$

and $n$ is the positive integer $(n=1,2,3,...)$

Step 2. (a) Considering a free particle inside a box with lengths L 1 ,   L 2 ,   a n d   L 3 along the axis x, y, and z respectively

Therefore the energy levels of particle confined to this box are,

$E_{n_x,n_y,n_z}=E_{n_x}+E_{n_y}+E_{n_z}$

              

                $=n_x^2\frac{h^2}{8m{L}_x^2}+n_y^2\frac{h^2}{8m{L}_y^2}+n_z^2\frac{h^2}{8m{L}_z^2}$

            

               $=\frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2}\right)$

here the box is taken as cube, so $L_x=L_y=L_z=L$. Thus the equation become

  

             $E_{n_x,n_y,n_t}=\frac{h^2}{8m{L}^2}\left(n_x^2+n_y^2+n_z^2\right)$          ( Equation-1)

Step 3. Now finding the value of ε 0 ,  For the ground state,  n x , n y , n z = ( 1 , 1 , 1 ) . 

Therefore the equation-1 become

${\epsilon }_0=\frac{h^2}{8m}\left(\frac{(1{)}^2}{L^2}+\frac{(1{)}^2}{(L{)}^2}+\frac{(1{)}^2}{(L{)}^2}\right)$

 

    $=\frac{h^2}{8m}\left(\frac{3}{L^2}\right)$

 ${\epsilon }_0=\frac{3h^2}{8m{L}^2}$             (Equation-2)

Step 4. Substituting the value of h ,   L ,   m in Equation-2

we get,

${\epsilon }_0=\frac{3{\left(6.626\times {10}^{-34} \mathrm{J}·\mathrm{s}\right)}^2}{8\left(14.4\times {10}^{-26} \mathrm{kg}\right){\left({10}^{-5}\right)}^2}$

    $=1.14\times {10}^{-32} \mathrm{J}$

Converting the values from $J to eV$.

${\epsilon }_0=1.14\times {10}^{-32} \mathrm{J}\left(\frac{1.0\mathrm{eV}}{1.602\times {10}^{-49} \mathrm{J}}\right)$

    $=7.1\times {10}^{-14}\mathrm{eV}$

The energy of the Rubidium in the ground = $7.1\times {10}^{-14}eV$.

Step 5. (b) To find the Condensate temperature, T c we have k B T c = 0 . 527 h 2 2 π m N V 2 3

Here,

 $N/V$= number of atoms per unit volume,

 $k_B$ = Boltzmann constant = $8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}$

$V=L^3$= Volume of box

Substituting the value of $V=L^3$ in above equation,

$k_B{T}_C=0.527\left(\frac{h^2}{2\pi m^2}\right){\left(\frac{N}{L^3}\right)}^{\frac{2}{3}}$

         $=\frac{0.527}{2\pi }\left(\frac{h^2}{m{L}^2}\right)(N{)}^{\frac{2}{3}}$

Step 6. Substituting the value of 8 3 ε 0  for  h 2 m L 2

we get,

$k_B{T}_c=\frac{0.527}{2\pi }\left(\frac{8}{3}{\epsilon }_0\right)(N{)}^{\frac{2}{3}}$

$\frac{k_B{T}_c}{{\epsilon }_0}=(0.224)(N{)}^{\frac{2}{3}}$

substituting $N=10000$

$\frac{k_B{T}_c}{{\epsilon }_0}=(0.224)(10,000{)}^{\frac{2}{3}}$

          $=104$

$T_c=\frac{\left(104\right){\epsilon }_0}{k_B}$

Substituting the value of ${\epsilon }_0, k_B$

$T_c=\frac{\left(104\right)\left(7.1\times {10}^{-14}\mathrm{eV}\right)}{8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}}$

    $=8.6\times {10}^{-8} \mathrm{K}$

Therefore, Condensation temperature $= 8.6\times {10}^{-8}K$

Step 7. (c) To find the number of atoms in ground state, we have

the relation between temperature and condensate temperature,

$T=0.9T_c$

$\frac{T}{T_c}=0.9$

The formula for number of atoms in ground state is,

$N_0=\left[1-{\left(\frac{T}{T_e}\right)}^{\frac{3}{2}}\right]N$

$N_0=\left[1-(0.9{)}^{\frac{3}{2}}\right]N$

     $=(0.146) N$

As $N=10000$,

$N_0=1460$

Therefore number of atoms in ground state$=1460$

Step 8. To find the number of atoms in excited state, we have

Formula of chemical potential, $\mu ={\epsilon }_0-\frac{k_{B}T}{N_0}$

                                                   ${\epsilon }_0-\mu =\frac{k_{B}T}{N_0}$

Substituting $T= 0.9T_c$,

${\epsilon }_0-\mu =\frac{k_B\left(0.9T_c\right)}{N_0}$

         $=\frac{0.9k_B{T}_c}{N_0}$

Puting the value of $k_B, T_c, and N_0$,

${\epsilon }_0-\mu =\frac{0.9\left(8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(8.6\times {10}^{-8} \mathrm{K}\right)}{1460}$

${\epsilon }_0-\mu =4.6\times {10}^{-15}\mathrm{eV}$

         $=(0.065)\left(7.1\times {10}^{-14}\mathrm{eV}\right)$

         $=(0.065){\epsilon }_0$

$\mu ={\epsilon }_0-(0.065){\epsilon }_0$

The formula for the energy of first excited state is,

${\epsilon }_1=\frac{h^2}{8m{L}^2}\left(2^2+1^2+1^2\right)$

   $=\frac{6h^2}{8m{L}^2}$

    $=2\left(\frac{3h^2}{8m{L}^2}\right)$

Substituting $\frac{3h^2}{8m{L}^2}\text{ = }{\epsilon }_0$

${\epsilon }_1=2{\epsilon }_0$

The formula for number of atoms in excited state is 

$N_1=\frac{1}{e^{\left({\epsilon }_1-\mu \right)/k_{B}T}-1}$

Puting the value of ${\epsilon }_1, \mu , k_B, T$, we get

$N_1=\frac{1}{e^{\left(2{\epsilon }_0-\left[{\epsilon }_0-(0.065){\epsilon }_0\right]/k_B\left[0.9T_c\right]\right.}-1}$

    $=\frac{1}{e^{\left({\epsilon }_0+(0.065){\epsilon }_0\right)/(0.9)k_B{T}_c}-1}$

    $=\frac{1}{e^{(1.065){ϵ}_0/(0.9)k_B{T}_c}-1}$

Substituting the values of ${\epsilon }_0, k_B, T_c$, we get

$N_1=\frac{1}{e^{(1.065)\left(7.1\times {10}^{-14}\mathrm{eV}\right)(0.9)\left(8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(8.6\times {10}^{-8} \mathrm{K}\right)}-1}$

     $=87$

Thus, in one excited state we have $87$ particles and for three excited state we have $3\left(87\right)=261$

Since, $261<1460$

Therefore, The number of particles in excited state is less than that of the number of particles in ground state. 

Step 9. (d) The new value of N = 10 6 ,

Substituting the value of $N={10}^6$ in equation $\frac{k_B{T}_c}{{\epsilon }_0}=(0.224)(N{)}^{\frac{2}{3}}$

                                                                            $\frac{k_B{T}_C}{{\epsilon }_0}=(0.224){\left({10}^6\right)}^{\frac{2}{3}}$

                                                                                       $=2240$

$T_c=\frac{\left(2240\right){\epsilon }_0}{k_B}$

Puting the value of ${\epsilon }_0, k_B$ 

$T_c=\frac{\left(2240\right)\left(7.1\times {10}^{-14}\mathrm{eV}\right)}{8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}}$

    $=1.85\times {10}^{-6} \mathrm{K}$

therefore the condensate temperature = $=1.85\times {10}^{-6} \mathrm{K}$$=1.85\times {10}^{-6} \mathrm{K}$

Step 10. For the number of particles in ground state

we have,

$N_0=(0.146)N$

Substituting the value of $N={10}^6$

$N_0=(0.146)\left({10}^6\right)$

     $=1.46\times {10}^5$

Therefore the number of particles in ground state $=1.46\times {10}^5$

Step 11. To find the number of particles in excited state

Substituting the new value of $N_0, T_c, k_B$ in the equation ${\epsilon }_0-\mu =\frac{0.9k_B{T}_c}{N_0}$

${\epsilon }_0-\mu =\frac{0.9\left(8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(1.85\times {10}^{-6} \mathrm{K}\right)}{1.46\times {10}^5}$

$=9.8\times {10}^{-16}\mathrm{eV}$

$=(0.0138)\left(7.1\times {10}^{-14}\mathrm{eV}\right)$

$\approx (0.014){\epsilon }_0$

$\mu ={\epsilon }_0-(0.014){\epsilon }_0$

 Formula for energy of first excited state = ${\epsilon }_1=2\left(\frac{3h^2}{8m{L}^2}\right)$

substituting  $\frac{3h^2}{8m{L}^2}\text{ = }{\epsilon }_0$

${\epsilon }_1=2{\epsilon }_0$

 Formula for the number of atoms in the excited state = $N_1=\frac{1}{e^{\left({ϵ}_1-\mu \right)/k_{B}T}-1}$

Substituting the value of ${\epsilon }_1, \mu , k_B, T$

$N_1=\frac{1}{e^{\left(2{\epsilon }_0-\left[c_0-(0.014){\epsilon }_0\right]/k_B\left[0.9T_c\right]\right.}-1}$

      $=\frac{1}{e^{(1.014){ϵ}_0/(0.9)k_B{T}_c}-1}$

Putting the values of ${\epsilon }_0, k_B, T_c$

$N_1=\frac{1}{e^{(1.014)\left(7.1\times {10}^{-14}\mathrm{ev}\right)/(0.9)\left(8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)\left(1.85\times {10}^{-6} \mathrm{K}\right)}-1}$

     $\approx 1990$

Thus, in one excited state we have $1990$ particles and for the three excited states we have $3\left(1990\right)=5970$

Since, $5970<1.46\times {10}^5$

Therefore, the number of particles in ground state is greater than that of the number of particles in excited state.