
Silver metal is at\[{961^ \circ }C\]. At the melting point, what fraction of the conduction electrons are in states with energies greater than the Fermi energy of\[5.5eV\]?
Answer
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Hint: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
\[{E_f} = \dfrac{{{h^2}}}{{2m}}{\left( {\dfrac{{3N}}{{8\pi V}}} \right)^{\dfrac{2}{3}}}\]
Where, \[{E_f}\]is the fermi energy, \[m\]is the mass of the metal, \[h\]is the planck’s constant, \[N\]is the number of electrons present in the metal and \[V\]is the volume.
Complete step-by-step solution:
In the question, we are given that the silver metal is at the temperature equal to\[T = {961^ \circ }C\],
The fermi energy of the metal is given as:
\[{E_f} = 5.5eV\]
The atomic mass of silver is equal to
\[m = 107.8g\]
The fermi energy formula for the metals is given by
\[{E_f} = \dfrac{{{h^2}}}{{2m}}{\left( {\dfrac{{3N}}{{8\pi V}}} \right)^{\dfrac{2}{3}}}\]
Rearranging the terms in order to get the fraction of electrons per unit volume, we get
\[\dfrac{N}{V} = \dfrac{{8\pi {{\left( {2m{E_f}} \right)}^{\dfrac{3}{2}}}}}{{3{h^3}}}\]
Here, \[h\]is the Planck’s constant.
\[
\dfrac{N}{V} = \dfrac{{8 \times 3.14{{\left( {2 \times {{10}^{ - 3}} \times 107.8 \times 5.5 \times 1.6 \times {{10}^{ - 19}}} \right)}^{\dfrac{3}{2}}}}}{{3{{\left( {6.626 \times {{10}^{ - 34}}} \right)}^3}}} = \dfrac{{2.6127 \times 3.16 \times {{10}^{ - 5}} \times {{10}^{ - 24}} \times 25.12}}{{864.8423 \times {{10}^{ - 102}}}} \\
\Rightarrow \dfrac{N}{V} = 0.2496 \times {10^{73}} \\
\Rightarrow \dfrac{N}{V} = 2.4 \times {10^{72}} \\
\]
Thus, the fraction of conduction electrons is \[2.4 \times {10^{72}}{m^{ - 3}}\]
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.
Formula used: The fermi energy formula for the metals is given by
\[{E_f} = \dfrac{{{h^2}}}{{2m}}{\left( {\dfrac{{3N}}{{8\pi V}}} \right)^{\dfrac{2}{3}}}\]
Where, \[{E_f}\]is the fermi energy, \[m\]is the mass of the metal, \[h\]is the planck’s constant, \[N\]is the number of electrons present in the metal and \[V\]is the volume.
Complete step-by-step solution:
In the question, we are given that the silver metal is at the temperature equal to\[T = {961^ \circ }C\],
The fermi energy of the metal is given as:
\[{E_f} = 5.5eV\]
The atomic mass of silver is equal to
\[m = 107.8g\]
The fermi energy formula for the metals is given by
\[{E_f} = \dfrac{{{h^2}}}{{2m}}{\left( {\dfrac{{3N}}{{8\pi V}}} \right)^{\dfrac{2}{3}}}\]
Rearranging the terms in order to get the fraction of electrons per unit volume, we get
\[\dfrac{N}{V} = \dfrac{{8\pi {{\left( {2m{E_f}} \right)}^{\dfrac{3}{2}}}}}{{3{h^3}}}\]
Here, \[h\]is the Planck’s constant.
\[
\dfrac{N}{V} = \dfrac{{8 \times 3.14{{\left( {2 \times {{10}^{ - 3}} \times 107.8 \times 5.5 \times 1.6 \times {{10}^{ - 19}}} \right)}^{\dfrac{3}{2}}}}}{{3{{\left( {6.626 \times {{10}^{ - 34}}} \right)}^3}}} = \dfrac{{2.6127 \times 3.16 \times {{10}^{ - 5}} \times {{10}^{ - 24}} \times 25.12}}{{864.8423 \times {{10}^{ - 102}}}} \\
\Rightarrow \dfrac{N}{V} = 0.2496 \times {10^{73}} \\
\Rightarrow \dfrac{N}{V} = 2.4 \times {10^{72}} \\
\]
Thus, the fraction of conduction electrons is \[2.4 \times {10^{72}}{m^{ - 3}}\]
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.
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