Example 9.2 obtains a ratio of the number of particles expected in the n = 2state lo that in the ground state. Rather than the n = 2state, consider arbitrary n.

(a) Show that the ratio is $\frac{\mathbf{n}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{e}\mathbf{n}\mathbf{e}\mathbf{r}\mathbf{g}\mathbf{y}\mathbf{}{\mathbf{E}}_{\mathbf{n}}}{\mathbf{n}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{e}\mathbf{n}\mathbf{e}\mathbf{r}\mathbf{g}\mathbf{y}\mathbf{}{\mathbf{E}}_{\mathbf{1}}}\mathbf{=}{\mathit{n}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{6}\mathbf{c}\mathbf{V}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{n}}^{\mathbf{-}\mathbf{2}}\mathbf{)}\mathbf{/}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}$

Note that hydrogen atom energies are ${\mathit{E}}_{\mathbf{n}}\mathbf{=}\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{6}\mathit{e}\mathit{V}\mathbf{/}\mathit{s}{\mathit{t}}^{\mathbf{2}}$.

(b) What is the limit of this ratio as n becomes very large? Can it exceed 1? If so, under what condition(s)?

(c) In Example 9.2. we found that even at the temperature of the Sun's surface$\left(~6000K\right)$, the ratio for n = 2 is only 10^-8 . For what value of nwould the ratio be 0.01?

(d) Is it realistic that the number of atoms with high n could be greater than the number with low n ?

The expression for ratio of probability to find the particles is given by,

$\frac{\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{energy}\mathbf{}{\mathbf{E}}_{\mathbf{n}}}{\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{energy}\mathbf{}{\mathbf{E}}_{\mathbf{1}}}\mathbf{=}\frac{\mathbf{\left(}\mathbf{2}{\mathbf{n}}^{\mathbf{2}}\mathbf{\right)}\mathbf{N}\mathbf{\left(}{\mathbf{E}}_{\mathbf{n}}\mathbf{\right)}}{\mathbf{\left(}\mathbf{2}{\mathbf{n}}^{\mathbf{2}}\mathbf{\right)}\mathbf{N}\mathbf{\left(}{\mathbf{E}}_{\mathbf{1}}\mathbf{\right)}}$

a)

The expression for ratio of probability to find the particles is calculated as,

$\begin{array}{rcl}\frac{\mathrm{number}\mathrm{of}\mathrm{energy}{\mathrm{E}}_{\mathrm{n}}}{\mathrm{number}\mathrm{of}\mathrm{energy}{\mathrm{E}}_1}& =& \frac{\left(2{\mathrm{n}}^2\right)\mathrm{N}\left({\mathrm{E}}_{\mathrm{n}}\right)}{\left(2{\mathrm{n}}^2\right)\mathrm{N}\left({\mathrm{E}}_1\right)}\\ & =& \frac{\left(2{\mathrm{n}}^2\right)\left(NA{e}^{{\displaystyle \frac{-E}{k_{B}T}}}\right)}{\left(2{\left(1\right)}^2\right)\left(NA{e}^{\underline{-E_{B}T}}\right)}\\ & =& n^2{e}^{\frac{E_1-E_n}{k_{B}T}}\end{array}$

Further simplify the above,

$\begin{array}{rcl}\frac{\mathrm{number}\mathrm{of}\mathrm{energy}{\mathrm{E}}_{\mathrm{n}}}{\mathrm{number}\mathrm{of}\mathrm{energy}{\mathrm{E}}_1}& =& n^2{e}^{\frac{\left[\left({\displaystyle \frac{-13.6ev}{L^2}}\right)-\left(\frac{-13.6ev}{n^2}\right)\right]}{k_{B}T}}\\ & =& n^2{e}^{\frac{-13.6ev\left(1-n^{-2}\right)}{k_{B}T}}\end{array}$

b)

It is given:

The temperature is T = 6000K .

The ratio is 10^-8 .

The limit of the ratio is calculated as,

$\begin{array}{rcl}{e}^{-\frac{13.6eV}{k_{B}T}}& =& e^{\frac{13.6eV}{8.625\times {10}^{(-5eV/K)\left(3000K\right)}}}\\ & =& e^{-525.6}\\ & =& 5\times {10}^{-229}\end{array}$

So, for the reasonable values of n and T, the probability goes to 0 because exponential term is so small. The probability can be greater than 1 if temperature is very large.

c)

The value of is calculated as,

$\begin{array}{rcl}\frac{\mathrm{number}\mathrm{of}\mathrm{energy}{\mathrm{E}}_{\mathrm{n}}}{\mathrm{number}\mathrm{of}\mathrm{energy}{\mathrm{E}}_1}& =& n^2{e}^{\frac{-13.6ev\left(1-n^{-2}\right)}{k_{B}T}}0.01\\ & =& n^2{e}^{-26.28\left(1-n^{-2}\right)}n\\ & =& 50889.6\end{array}$

d)

It is not realistic to expect that number of atoms with high n could be greater than the number with low n because for this temperature should be greater than that on the surface of the Sun which is not possible.