energy & copper & probability

[logistic_guy]

Calculate the energy which has \(\displaystyle 90\) percent occupancy probability for copper at \(\displaystyle \bold{(a)} \ T = 300 \ \text{K}; \bold{(b)} \ T = 1200 \ \text{K}\).

 

[khansaheb]

Calculate the energy which has \(\displaystyle 90\) percent occupancy probability for copper at \(\displaystyle \bold{(a)} \ T = 300 \ \text{K}; \bold{(b)} \ T = 1200 \ \text{K}\).

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Please share your work/thoughts about this problem

 

[logistic_guy]

Calculate the energy which has \(\displaystyle 90\) percent occupancy probability for copper at \(\displaystyle \bold{(a)} \ T = 300 \ \text{K}; \bold{(b)} \ T = 1200 \ \text{K}\).

This problem can be solved by the Fermi-Dirac probability function:

\(\displaystyle f(E) = \frac{1}{e^{(E - E_F)/kT} + 1}\)

The goal is to find \(\displaystyle E\), but unfortunately the Fermi energy, \(\displaystyle E_F\), for copper is not given.



It is up to us. We can find it from scratch or we can just look it up!

 

[logistic_guy]

It is up to us. We can find it from scratch or we can just look it up!

We choose the first option. Now suppose we were given the density of states \(\displaystyle g(E)\):

\(\displaystyle g(E) = \frac{8\sqrt{2}\pi m^{\frac{3}{2}}}{h^3}\sqrt{E}\)

There is a relation between the number of conduction electrons per unit volume, \(\displaystyle \frac{N}{V}\), the density of states, \(\displaystyle g(E)\), and the Fermi energy, \(\displaystyle E_F\). That is:

\(\displaystyle \frac{N}{V} = \int_{0}^{E_F} g(E) \ dE = \int_{0}^{E_F} \frac{8\sqrt{2}\pi m^{\frac{3}{2}}}{h^3}\sqrt{E} \ \ dE\)

 

[logistic_guy]

\(\displaystyle \frac{N}{V} = \int_{0}^{E_F} g(E) \ dE = \int_{0}^{E_F} \frac{8\sqrt{2}\pi m^{\frac{3}{2}}}{h^3}\sqrt{E} \ \ dE\)

Solving this integral yields:

\(\displaystyle \frac{N}{V} = \frac{8\sqrt{2}\pi m^{\frac{3}{2}}}{h^3}\frac{2}{3}E^{\frac{3}{2}}_F\)

 

[logistic_guy]

\(\displaystyle \frac{N}{V} = \frac{8\sqrt{2}\pi m^{\frac{3}{2}}}{h^3}\frac{2}{3}E^{\frac{3}{2}}_F\)

If we solve for \(\displaystyle E_{F}\), we get:

\(\displaystyle E_{F} = \frac{h^2}{8m}\left(\frac{3N}{\pi V}\right)^{\frac{2}{3}}\)

 

[logistic_guy]

\(\displaystyle E_{F} = \frac{h^2}{8m}\left(\frac{3N}{\pi V}\right)^{\frac{2}{3}}\)

\(\displaystyle h \rightarrow \) Planck's constant
\(\displaystyle m \rightarrow \) mass of one electron

But we still don't know the value of:

\(\displaystyle \frac{N}{V} \rightarrow\) the number of conduction electrons per unit volume

From Chemistry we know that the mass density of copper is:

\(\displaystyle \rho = 8900 \ \text{kg/}\text{m}^3 \)

And we also know that the atomic mass of copper is:

\(\displaystyle m_{\text{C}_{\text{u}}} = 63.5 \ \text{g/}\text{mole} = 0.0635 \ \text{kg/}\text{mole}\)

All is remained is \(\displaystyle N \rightarrow\) Avogadro number. That is:

\(\displaystyle N = 6.022 \times 10^{23} \ \text{electrons/mole}\)

Then,

\(\displaystyle \frac{N}{V} = \frac{N}{m_{\text{C}_{\text{u}}}/\rho} = \frac{6.022 \times 10^{23}}{0.0635}8900 = 8.44 \times 10^{28} \ \text{electrons/}\text{m}^3\)

 

[logistic_guy]

We are ready to find the Fermi energy.

\(\displaystyle E_{F} = \frac{h^2}{8m}\left(\frac{3N}{\pi V}\right)^{\frac{2}{3}} = \frac{(6.626 \times 10^{-34})^2}{8(9.109 \times 10^{-31})}\left(\frac{3(8.44 \times 10^{28})}{\pi}\right)^{\frac{2}{3}} = 1.124 \times 10^{-18} \ \text{J}\)

Convert it to \(\displaystyle \text{eV}\).

\(\displaystyle E_{F} = 1.124 \times 10^{-18} \ \text{J} = 1.124 \times 10^{-18} \ \text{J} \frac{\text{eV}}{1.602 \times 10^{-19} \ \text{J}} = 7 \ \text{eV}\)

 

[logistic_guy]

Now we are ready to use the Fermi-Dirac probability function:

\(\displaystyle f(E) = \frac{1}{e^{(E - E_F)/kT} + 1}\)

Before we plug in numbers, let us convert \(\displaystyle kT\) to \(\displaystyle \text{eV}\). \(\displaystyle \left(k \right.\) is just the \(\displaystyle \left. \text{Boltzmann Constant} \right)\)

\(\displaystyle kT = 1.38 \times 10^{-23} \ \frac{\text{J}}{\text{K}} \times 300 \ \text{K} \times \frac{\text{eV}}{1.602 \times 10^{-19} \ \text{J}} = 0.025843 \ \text{eV}\)

Plug in numbers.

Calculate the energy

\(\displaystyle 0.9 = \frac{1}{e^{(E - 7)/0.025843} + 1}\)

This gives:

\(\displaystyle E = \textcolor{blue}{6.94 \ \text{eV}}\)