Position of Fermi Energy Level MCQ Quiz - Objective Question with Answer for Position of Fermi Energy Level - Download Free PDF

Explanation:

Fermi Level and Its Significance

Definition: The Fermi level is a concept in quantum mechanics that represents the energy level at which the probability of finding an electron is 50% at absolute zero temperature. It serves as a reference energy level, separating occupied energy states from unoccupied energy states in a material. The position of the Fermi level is crucial in determining the electrical and thermal properties of a material.

Working Principle: The Fermi level is not a fixed energy level but rather an indicator of the energy distribution of electrons within a material. In metals, the Fermi level lies within the conduction band, allowing free movement of electrons and making them good conductors. In semiconductors, the Fermi level lies between the valence and conduction bands, and its position can shift based on doping. In insulators, the Fermi level is typically far from the conduction band, making electron movement difficult.

Correct Option Analysis:

The correct answer is:

Option 2: Probability of occupancy of electrons or holes

This option accurately describes the Fermi level. The Fermi level is a measure of the probability that an electron or hole will occupy a given energy state at a specific temperature. This probability is governed by the Fermi-Dirac distribution function, which is given by:

F(E) = 1 / [1 + exp((E - Ef) / (k × T))]

Where:

• F(E): Probability of occupancy of an energy state with energy E
• Ef: Fermi level (energy)
• k: Boltzmann constant
• T: Absolute temperature

At absolute zero (T = 0 K), all energy states below the Fermi level are completely filled, and those above are empty. As the temperature increases, electrons gain thermal energy and can occupy higher energy states, leading to a distribution of electrons and holes around the Fermi level.

The position of the Fermi level in a material determines its conductivity and other electronic properties:

• In metals, the Fermi level resides within the conduction band, allowing for free electron movement and high conductivity.
• In semiconductors, the Fermi level lies in the band gap and shifts with doping. For n-type semiconductors, it moves closer to the conduction band, while for p-type, it moves closer to the valence band.
• In insulators, the Fermi level is far from the conduction band, making electron excitation difficult and reducing conductivity.

Importance in Semiconductors: In semiconductors, the Fermi level provides valuable insights into the type of doping (n-type or p-type) and the carrier concentration. It plays a crucial role in determining the behavior of devices like diodes, transistors, and solar cells.

Additional Information:

To further understand the analysis, let’s evaluate the other options:

Option 1: Doping of electrons

This option is incorrect because doping refers to the process of intentionally introducing impurities into a semiconductor to modify its electrical properties. While doping does affect the position of the Fermi level, the Fermi level itself is not a measure of doping. Instead, it represents the probability of electron or hole occupancy in energy states.

Option 3: Probability of occupancy of photons

This option is incorrect because the Fermi level specifically deals with the energy states of electrons (or holes) in a material. Photons, being quanta of light, do not have a Fermi level associated with them. Their behavior is described by other principles, such as Planck's law and the photon energy equation (E = h × ν).

Option 4: Probability of occupancy of wavelength

This option is incorrect because wavelength is a property of waves, including light and sound, and does not directly relate to the Fermi level. The Fermi level is concerned with electron or hole energy states, not with the occupancy of wavelengths.

Option 5: (No provided content for Option 5)

Since no information is provided for Option 5, we can conclude that it is not relevant to the question.

Conclusion:

The Fermi level is a fundamental concept in understanding the electronic properties of materials. It provides a probabilistic measure of electron and hole occupancy in energy states, which is critical for analyzing the behavior of conductors, semiconductors, and insulators. The correct answer, Option 2, highlights its role as a measure of the probability of occupancy of electrons or holes, distinguishing it from other unrelated concepts such as doping, photons, or wavelengths. By understanding the Fermi level, we can gain deeper insights into the operation of electronic devices and the design of advanced materials.

Given:

The Fermi energy of silver is \(E_F = 5.52 \, \text{eV}\) . We are asked to find the velocity of the conduction electron in silver.

Concept:

• The velocity of conduction electrons can be calculated using the relation between kinetic energy and velocity. The kinetic energy of the conduction electron is given by:
• \( E_F = \frac{1}{2} m v^2 \), where:
  • E_F is the Fermi energy,
  • m is the mass of the electron ( \(m = 9.11 \times 10^{-31} \, \text{kg}\) ),
  • v is the velocity of the electron.

The velocity can be found by rearranging the above equation:

\( v = \sqrt{\frac{2 E_F}{m}} \)

Calculation:

Given\( E_F = 5.52 \, \text{eV}\) and \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\) , we can convert the Fermi energy into joules:

\(⇒ E_F = 5.52 \times 1.602 \times 10^{-19} \, \text{J} = 8.84 \times 10^{-19} \, \text{J} \).

Now, using the equation for velocity:

\(⇒ v = \sqrt{\frac{2 \times 8.84 \times 10^{-19}}{9.11 \times 10^{-31}}} \\ ⇒ v = \sqrt{1.94 \times 10^{12}} \\ ⇒ v = 1.39 \times 10^6 \, \text{m/s} .\)

∴ The velocity of conduction electrons in silver is \(1.39 \times 10^6 \, \text{m/s}\) .

The correct option is 3

Concept:

\({\phi _{fp}} = {E_i} - {E_f}\)

\( = \frac{{KT}}{q}\ln \left( {\frac{{{N_A}}}{{{n_i}}}} \right)\)

\( = 25 \times {10^{ - 3}} \times \ln \left( {\frac{{4 \times {{10}^{17}}}}{{1.5 \times {{10}^{10}}}}} \right)\)

= 0.427 eV

As the impurity is acceptor type the Fermi level goes down by 0.427 eV than the intrinsic level.

The correct answer is option 1):(close to the conduction band)

Concept:

Fermi Level:

• The Fermi-level in an intrinsic semiconductor is nearly midway between the conductive and valence band.
• Fermi level is the highest energy state occupied by electrons in a material at absolute zero temperature

The Fermi for an n-type semiconductor lies closer to the conduction band as shown:

• Similarly, the Fermi level for a p-type lies near the valence band.
• As the temperature increases above zero degrees, the extrinsic carriers in the conduction band and the valence band increases.
• Since the intrinsic concentration also depends on temperature, ni also increases.
• But for small values of temperature, the extrinsic concentration dominates in comparison to the intrinsic concentration
• As the temperature continues to increase, the semiconductor starts to loose its extrinsic property and becomes intrinsic, as ni becomes comparable to the extrinsic concentration.

• The bandgap of intrinsic semiconductor material is in order of 0.7 – 1 eV at room temperature
• At absolute zero temperature, the thermal energy is (KT = 25 meV) is not enough to excite electrons from the valence band to the conduction band
• Hence there are no electrons in the conduction band and correspondingly no holes in the valence band
• Thus, an intrinsic semiconductor is an insulator at absolute zero temeprature.

• The bandgap of intrinsic semiconductor material is in order of 0.7 – 1 eV at room temperature
• At absolute zero temperature, the thermal energy is (KT = 25 meV) is not enough to excite electrons from the valence band to the conduction band
• Hence there are no electrons in the conduction band and correspondingly no holes in the valence band
• Thus, an intrinsic semiconductor is an insulator at absolute zero temeprature.

The correct answer is option 1):(close to the conduction band)

Concept:

Fermi Level:

• The Fermi-level in an intrinsic semiconductor is nearly midway between the conductive and valence band.
• Fermi level is the highest energy state occupied by electrons in a material at absolute zero temperature

The Fermi for an n-type semiconductor lies closer to the conduction band as shown:

• Similarly, the Fermi level for a p-type lies near the valence band.
• As the temperature increases above zero degrees, the extrinsic carriers in the conduction band and the valence band increases.
• Since the intrinsic concentration also depends on temperature, ni also increases.
• But for small values of temperature, the extrinsic concentration dominates in comparison to the intrinsic concentration
• As the temperature continues to increase, the semiconductor starts to loose its extrinsic property and becomes intrinsic, as ni becomes comparable to the extrinsic concentration.

Concept:

For a given position of Fermi-level, the concentration of electrons and holes are given by:

\({n_0} = {N_c}{e^{ - \frac{{\left( {{E_c} - {E_F}} \right)}}{{kT}}}}\)

\({p_0} = {N_v}{e^{ - \frac{{\left( {{E_F} - {E_V}} \right)}}{{kT}}}}\)

N­c and Nv are the effective density of states.

We can now calculate the intrinsic Fermi level position.

For the intrinsic semiconductor, the electron and hole concentrations are equal.

∴ we can write:

\({N_c}\;{e^{ - \left( {\frac{{{E_c} - {E_{Fi}}}}{{kT}}} \right)}} = {N_v}\;{e^{ - \left( {\frac{{{E_{Fi}} - {E_v}}}{{kT}}} \right)}}\)

Taking the natural log of both sides of the above equation, we get:

\({E_{{F_i}}} = \frac{1}{2}\left( {{E_c} + {E_v}} \right) + \frac{1}{2}kT\;In\;\left( {\frac{{{N_v}}}{{{N_c}}}} \right)\)

Since, \(\frac{1}{2}\left( {{E_c} + {E_v}} \right) = {E_{midgap}}\)

\({E_{{F_i}}} = {E_{midgap}} + \frac{1}{2}kT\;In\left( {\frac{{{N_v}}}{{{N_c}}}} \right)\)

\({E_{Fi}} - {E_{midgap}} = \frac{1}{2}kT\;In\;\left( {\frac{{{N_v}}}{{{N_c}}}} \right)\)   ---(1)

Calculation:

Given V_T = 26 mV and N_V = 2 N_C

Using Equation (1), we can write:

\({E_{{F_i}}} - {E_{mid}} = \frac{1}{2}kT\;ln\left( {\frac{{{N_V}}}{{{N_C}}}} \right)\)

\({E_{{F_i}}} - {E_{mid}} = \frac{{26\;m}}{2}\;ln\left( {\frac{{2{N_C}}}{{{N_C}}}} \right)\)

\({E_{{F_i}}} - {E_{mid}} = 13 \times {10^{ - 3}}\;ln\left( 2 \right)\)

= 9.01 meV

Explanation:

Fermi Level and Its Significance

Definition: The Fermi level is a concept in quantum mechanics that represents the energy level at which the probability of finding an electron is 50% at absolute zero temperature. It serves as a reference energy level, separating occupied energy states from unoccupied energy states in a material. The position of the Fermi level is crucial in determining the electrical and thermal properties of a material.

Working Principle: The Fermi level is not a fixed energy level but rather an indicator of the energy distribution of electrons within a material. In metals, the Fermi level lies within the conduction band, allowing free movement of electrons and making them good conductors. In semiconductors, the Fermi level lies between the valence and conduction bands, and its position can shift based on doping. In insulators, the Fermi level is typically far from the conduction band, making electron movement difficult.

Correct Option Analysis:

The correct answer is:

Option 2: Probability of occupancy of electrons or holes

This option accurately describes the Fermi level. The Fermi level is a measure of the probability that an electron or hole will occupy a given energy state at a specific temperature. This probability is governed by the Fermi-Dirac distribution function, which is given by:

F(E) = 1 / [1 + exp((E - Ef) / (k × T))]

Where:

• F(E): Probability of occupancy of an energy state with energy E
• Ef: Fermi level (energy)
• k: Boltzmann constant
• T: Absolute temperature

At absolute zero (T = 0 K), all energy states below the Fermi level are completely filled, and those above are empty. As the temperature increases, electrons gain thermal energy and can occupy higher energy states, leading to a distribution of electrons and holes around the Fermi level.

The position of the Fermi level in a material determines its conductivity and other electronic properties:

• In metals, the Fermi level resides within the conduction band, allowing for free electron movement and high conductivity.
• In semiconductors, the Fermi level lies in the band gap and shifts with doping. For n-type semiconductors, it moves closer to the conduction band, while for p-type, it moves closer to the valence band.
• In insulators, the Fermi level is far from the conduction band, making electron excitation difficult and reducing conductivity.

Importance in Semiconductors: In semiconductors, the Fermi level provides valuable insights into the type of doping (n-type or p-type) and the carrier concentration. It plays a crucial role in determining the behavior of devices like diodes, transistors, and solar cells.

Additional Information:

To further understand the analysis, let’s evaluate the other options:

Option 1: Doping of electrons

This option is incorrect because doping refers to the process of intentionally introducing impurities into a semiconductor to modify its electrical properties. While doping does affect the position of the Fermi level, the Fermi level itself is not a measure of doping. Instead, it represents the probability of electron or hole occupancy in energy states.

Option 3: Probability of occupancy of photons

This option is incorrect because the Fermi level specifically deals with the energy states of electrons (or holes) in a material. Photons, being quanta of light, do not have a Fermi level associated with them. Their behavior is described by other principles, such as Planck's law and the photon energy equation (E = h × ν).

Option 4: Probability of occupancy of wavelength

This option is incorrect because wavelength is a property of waves, including light and sound, and does not directly relate to the Fermi level. The Fermi level is concerned with electron or hole energy states, not with the occupancy of wavelengths.

Option 5: (No provided content for Option 5)

Since no information is provided for Option 5, we can conclude that it is not relevant to the question.

Conclusion:

The Fermi level is a fundamental concept in understanding the electronic properties of materials. It provides a probabilistic measure of electron and hole occupancy in energy states, which is critical for analyzing the behavior of conductors, semiconductors, and insulators. The correct answer, Option 2, highlights its role as a measure of the probability of occupancy of electrons or holes, distinguishing it from other unrelated concepts such as doping, photons, or wavelengths. By understanding the Fermi level, we can gain deeper insights into the operation of electronic devices and the design of advanced materials.

The correct option is 3

Concept:

\({\phi _{fp}} = {E_i} - {E_f}\)

\( = \frac{{KT}}{q}\ln \left( {\frac{{{N_A}}}{{{n_i}}}} \right)\)

\( = 25 \times {10^{ - 3}} \times \ln \left( {\frac{{4 \times {{10}^{17}}}}{{1.5 \times {{10}^{10}}}}} \right)\)

= 0.427 eV

As the impurity is acceptor type the Fermi level goes down by 0.427 eV than the intrinsic level.

• The bandgap of intrinsic semiconductor material is in order of 0.7 – 1 eV at room temperature
• At absolute zero temperature, the thermal energy is (KT = 25 meV) is not enough to excite electrons from the valence band to the conduction band
• Hence there are no electrons in the conduction band and correspondingly no holes in the valence band
• Thus, an intrinsic semiconductor is an insulator at absolute zero temeprature.

The correct answer is option 1):(close to the conduction band)

Concept:

Fermi Level:

• The Fermi-level in an intrinsic semiconductor is nearly midway between the conductive and valence band.
• Fermi level is the highest energy state occupied by electrons in a material at absolute zero temperature

The Fermi for an n-type semiconductor lies closer to the conduction band as shown:

• Similarly, the Fermi level for a p-type lies near the valence band.
• As the temperature increases above zero degrees, the extrinsic carriers in the conduction band and the valence band increases.
• Since the intrinsic concentration also depends on temperature, ni also increases.
• But for small values of temperature, the extrinsic concentration dominates in comparison to the intrinsic concentration
• As the temperature continues to increase, the semiconductor starts to loose its extrinsic property and becomes intrinsic, as ni becomes comparable to the extrinsic concentration.

Concept:

For a given position of Fermi-level, the concentration of electrons and holes are given by:

\({n_0} = {N_c}{e^{ - \frac{{\left( {{E_c} - {E_F}} \right)}}{{kT}}}}\)

\({p_0} = {N_v}{e^{ - \frac{{\left( {{E_F} - {E_V}} \right)}}{{kT}}}}\)

N­c and Nv are the effective density of states.

We can now calculate the intrinsic Fermi level position.

For the intrinsic semiconductor, the electron and hole concentrations are equal.

∴ we can write:

\({N_c}\;{e^{ - \left( {\frac{{{E_c} - {E_{Fi}}}}{{kT}}} \right)}} = {N_v}\;{e^{ - \left( {\frac{{{E_{Fi}} - {E_v}}}{{kT}}} \right)}}\)

Taking the natural log of both sides of the above equation, we get:

\({E_{{F_i}}} = \frac{1}{2}\left( {{E_c} + {E_v}} \right) + \frac{1}{2}kT\;In\;\left( {\frac{{{N_v}}}{{{N_c}}}} \right)\)

Since, \(\frac{1}{2}\left( {{E_c} + {E_v}} \right) = {E_{midgap}}\)

\({E_{{F_i}}} = {E_{midgap}} + \frac{1}{2}kT\;In\left( {\frac{{{N_v}}}{{{N_c}}}} \right)\)

\({E_{Fi}} - {E_{midgap}} = \frac{1}{2}kT\;In\;\left( {\frac{{{N_v}}}{{{N_c}}}} \right)\)   ---(1)

Calculation:

Given V_T = 26 mV and N_V = 2 N_C

Using Equation (1), we can write:

\({E_{{F_i}}} - {E_{mid}} = \frac{1}{2}kT\;ln\left( {\frac{{{N_V}}}{{{N_C}}}} \right)\)

\({E_{{F_i}}} - {E_{mid}} = \frac{{26\;m}}{2}\;ln\left( {\frac{{2{N_C}}}{{{N_C}}}} \right)\)

\({E_{{F_i}}} - {E_{mid}} = 13 \times {10^{ - 3}}\;ln\left( 2 \right)\)

= 9.01 meV

As the electron moves in the lattice under the influence of applied voltage it suffers a number of collisions and there is transfer of energy from electron to lattice .

Hence its energy decreases from left to right in the conduction band as shown in the option 4

n = n₀ + δ_n

= (10⁵ + 10¹²) / cm³ ≈ δ_n

p = p₀ + δ_p

= (10¹⁶ + 10¹²) / cm³ ≈ p₀

As, p ≈ p₀, \((E_{F_P})\) is same as E_F  [option (1) is correct]

We know,

\((E_{F_n})-E_i=KT~ln\left(\dfrac{n}{n_i}\right)\)

= \(KT~ln\left(\dfrac{n_0+\delta_n}{n_i}\right)\)

= \(KT~ln\left(\dfrac{\delta_n}{n_i}\right)\)

= \(0.026× ln\left(\dfrac{10^{12}}{10^{10}}\right)\)

= 0.026 × ln(100)

= 0.119 eV. [option (4) is correct]

as, \(E_{F_n}-E_i>0\)

\(E_{F_n}\) lies above E_i