91Ó°ÊÓ

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

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

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

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

Length of box, $L={10}^{-5}m$

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

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

Therefore the equation-1 become

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

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

we get,

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

$=1.14×{10}^{-32}\mathrm{J}$

Converting the values from $JtoeV$.

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

$=7.1×{10}^{-14}\mathrm{eV}$

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

Here,

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

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

$V=L^3$= Volume of box

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

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

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

we get,

$k_B{T}_c=\frac{0.527}{2\mathrm{Ï€}}\left(\frac{8}{3}{\mathrm{Î\mu }}_0\right)(N{)}^{\frac{2}{3}}$

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

substituting $N=10000$

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

$=104$

$T_c=\frac{\left(104\right){\mathrm{Î\mu }}_0}{k_B}$

Substituting the value of ${\mathrm{Î\mu }}_0,k_B$

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

$=8.6×{10}^{-8}\mathrm{K}$

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

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$=1460$

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

${\mathrm{Î\mu }}_0-\mathrm{μ}=\frac{k_{B}T}{N_0}$

Substituting $T=0.9T_c$,

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

$=\frac{0.9k_B{T}_c}{N_0}$

Puting the value of $k_B,T_c,and{N}_0$,

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

${\mathrm{Î\mu }}_0-\mathrm{μ}=4.6×{10}^{-15}\mathrm{eV}$

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

$=(0.065){\mathrm{Î\mu }}_0$

$\mathrm{μ}={\mathrm{Î\mu }}_0-(0.065){\mathrm{Î\mu }}_0$

The formula for the energy of first excited state is,

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

$=\frac{6h^2}{8m{L}^2}$

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

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

${\mathrm{Î\mu }}_1=2{\mathrm{Î\mu }}_0$

The formula for number of atoms in excited state is

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

Puting the value of ${\mathrm{Î\mu }}_1,\mathrm{μ},k_B,T$, we get

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

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

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

Substituting the values of ${\mathrm{Î\mu }}_0,k_B,T_c$, we get

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

$=87$

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.

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

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

$=2240$

$T_c=\frac{\left(2240\right){\mathrm{Î\mu }}_0}{k_B}$

Puting the value of ${\mathrm{Î\mu }}_0,k_B$

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

$=1.85×{10}^{-6}\mathrm{K}$

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

we have,

$N_0=(0.146)N$

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

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

$=1.46×{10}^5$

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

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

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

$=9.8×{10}^{-16}\mathrm{eV}$

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

$≈(0.014){\mathrm{Î\mu }}_0$

$\mathrm{μ}={\mathrm{Î\mu }}_0-(0.014){\mathrm{Î\mu }}_0$

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

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

${\mathrm{Î\mu }}_1=2{\mathrm{Î\mu }}_0$

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

Substituting the value of ${\mathrm{Î\mu }}_1,\mathrm{μ},k_B,T$

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

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

Putting the values of ${\mathrm{Î\mu }}_0,k_B,T_c$

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

$≈1990$

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

Since, $5970<1.46×{10}^5$

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