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Question: Silver metal is at\[{961^ \circ }C\]. At the melting point, what fraction of the conduction electron...

Silver metal is at961C{961^ \circ }C. At the melting point, what fraction of the conduction electrons are in states with energies greater than the Fermi energy of5.5eV5.5eV?

Explanation

Solution

In order to solve this question, we are going to firstly take the temperature and the fermi energy of the silver as given and taking the atomic mass of silver, we put all of these values into the formula for the fermi energy and comparing the two sides we will find the electrons per unit volume.

Formula used: The fermi energy formula for the metals is given by
Ef=h22m(3N8πV)23{E_f} = \dfrac{{{h^2}}}{{2m}}{\left( {\dfrac{{3N}}{{8\pi V}}} \right)^{\dfrac{2}{3}}}
Where, Ef{E_f}is the fermi energy, mmis the mass of the metal, hhis the planck’s constant, NNis the number of electrons present in the metal and VVis the volume.

Complete step-by-step solution:
In the question, we are given that the silver metal is at the temperature equal toT=961CT = {961^ \circ }C,
The fermi energy of the metal is given as:
Ef=5.5eV{E_f} = 5.5eV
The atomic mass of silver is equal to
m=107.8gm = 107.8g
The fermi energy formula for the metals is given by
Ef=h22m(3N8πV)23{E_f} = \dfrac{{{h^2}}}{{2m}}{\left( {\dfrac{{3N}}{{8\pi V}}} \right)^{\dfrac{2}{3}}}
Rearranging the terms in order to get the fraction of electrons per unit volume, we get
NV=8π(2mEf)323h3\dfrac{N}{V} = \dfrac{{8\pi {{\left( {2m{E_f}} \right)}^{\dfrac{3}{2}}}}}{{3{h^3}}}
Here, hhis the Planck’s constant.

NV=8×3.14(2×103×107.8×5.5×1.6×1019)323(6.626×1034)3=2.6127×3.16×105×1024×25.12864.8423×10102 NV=0.2496×1073 NV=2.4×1072  \dfrac{N}{V} = \dfrac{{8 \times 3.14{{\left( {2 \times {{10}^{ - 3}} \times 107.8 \times 5.5 \times 1.6 \times {{10}^{ - 19}}} \right)}^{\dfrac{3}{2}}}}}{{3{{\left( {6.626 \times {{10}^{ - 34}}} \right)}^3}}} = \dfrac{{2.6127 \times 3.16 \times {{10}^{ - 5}} \times {{10}^{ - 24}} \times 25.12}}{{864.8423 \times {{10}^{ - 102}}}} \\\ \Rightarrow \dfrac{N}{V} = 0.2496 \times {10^{73}} \\\ \Rightarrow \dfrac{N}{V} = 2.4 \times {10^{72}} \\\

Thus, the fraction of conduction electrons is 2.4×1072m32.4 \times {10^{72}}{m^{ - 3}}

Note: It is important to note that Fermi energy is a concept in quantum mechanics usually referring to the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero temperature. The probability of electrons at this energy is half as that of the total.