Given that the Fermi energy of gold is 5.54 eV, the number density of electrons is ________ $\times 10^{28}$ $m^{-3}$ (upto one decimal place) (Mass of electron = $9.11 \times 10^{-31}$ kg; $h = 6.626 \times 10^{-34}$J. s; 1 eV = $1.6 \times 10^{-19}$J)
Calculating Electron Number Density from Fermi Energy
This solution explains how to calculate the electron number density ($n$) using the given Fermi energy ($E_F$) of gold.
1. Formula for Fermi Energy
The relationship between Fermi energy ($E_F$) and electron number density ($n$) for a free electron gas is given by:
$ E_F = \frac{h^2}{2m_e} \left( \frac{3n}{\pi} \right)^{2/3} $
Where:
- $E_F$ is the Fermi energy
- $h$ is Planck's constant
- $m_e$ is the mass of an electron
- $n$ is the electron number density
2. Convert Fermi Energy to Joules
The Fermi energy is given in electron volts (eV). Convert it to Joules (J) for calculation:
$ E_F = 5.54 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}) = 8.864 \times 10^{-19} \text{ J} $
3. Rearrange Formula for Number Density ($n$)
Rearrange the Fermi energy formula to solve for $n$:
$ \left( \frac{2m_e E_F}{h^2} \right) = \left( \frac{3n}{\pi} \right)^{2/3} $
$ \left( \frac{2m_e E_F}{h^2} \right)^{3/2} = \frac{3n}{\pi} $
$ n = \frac{\pi}{3} \left( \frac{2m_e E_F}{h^2} \right)^{3/2} $
4. Substitute Values and Calculate
Substitute the given values and constants:
- $E_F = 8.864 \times 10^{-19}$ J
- $m_e = 9.11 \times 10^{-31}$ kg
- $h = 6.626 \times 10^{-34}$ J.s
First, calculate the term inside the parenthesis:
$ \frac{2m_e E_F}{h^2} = \frac{2 \times (9.11 \times 10^{-31} \text{ kg}) \times (8.864 \times 10^{-19} \text{ J})}{(6.626 \times 10^{-34} \text{ J.s})^2} $
$ \frac{2m_e E_F}{h^2} = \frac{1.6163 \times 10^{-48}}{4.3904 \times 10^{-67}} \approx 3.6814 \times 10^{18} \text{ m}^{-2} $
Now, raise this to the power of $3/2$:
$ \left( 3.6814 \times 10^{18} \right)^{3/2} = (3.6814)^{1.5} \times (10^{18})^{1.5} \approx 7.0435 \times 10^{27} \text{ m}^{-3} $
Finally, calculate $n$:
$ n = \frac{\pi}{3} \times (7.0435 \times 10^{27} \text{ m}^{-3}) $
$ n \approx \frac{3.14159}{3} \times 7.0435 \times 10^{27} \text{ m}^{-3} \approx 7.377 \times 10^{27} \text{ m}^{-3} $
5. Final Result in Required Format
Express the result in the format $\_\_\_\_\_ \times 10^{28}$ $m^{-3}$, rounded to one decimal place:
$ n \approx 0.7377 \times 10^{28} \text{ m}^{-3} $
Rounding to one decimal place gives $0.7$.
Important Questions from Free Electron Theory Fermi Energy Velocity
- Copper has an electron number density of $8.3 \times 10^{28} \text{ m}^{-3}$. Its Fermi energy in eV (rounded off to one decimal place) is _____
($\hbar = 1.06 \times 10^{-34} \text{ J.s}$, mass of electron $\text{m}_e = 9.10 \times 10^{-31} \text{ kg}$, charge of electron $= 1.60 \times 10^{-19} \text{ C}$) - Consider one mole of a monovalent metal at absolute zero temperature, obeying the free electron model. Its Fermi energy is $E_F$. The energy corresponding to the filling of $\frac{N_A}{2}$ electrons, where $N_A$ is the Avogadro number, is $2^n E_F$. The value of $n$ is
Crystal structures of two metals A and B are two-dimensional square lattices with same lattice constant $a$. Electrons in metals behave as free electrons. The Fermi surfaces corresponding to A and B are shown by solid circles in figures.
The electron concentrations in A and B are $n_A$ and $n_B$, respectively. The value of $(\frac{n_B}{n_A})$ isIf $X$ is the dimensionality of a free electron gas, the energy ($E$) dependence of density of states is given by $E^{½X-Y}$, where $Y$ is ________.
- Potassium metal has electron concentration of $1.4 \times 10^{28}m^{-3}$ and the corresponding density of states at Fermi level is $6.2 \times 10^{46}$ Joule$^{-1} m^{-3}$. If the Pauli paramagnetic susceptibility of Potassium is $n \times 10^{-k}$ in standard scientific form, then the value of $k$ (an integer) is __________ (Magnetic moment of electron is $9.3 \times 10^{-24}$ Joule T$^{-1}$; permeability of free space is $4\pi \times 10^{-7}$ T m A$^{-1}$)
