
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
507k+ views
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

