Mobility MCQ Quiz - Objective Question with Answer for Mobility - Download Free PDF

Mobility of electron

Subatomic particles like electrons move in random directions all the time.

When electrons are subjected to an electric field they do move randomly, but they slowly drift in one direction, in the direction of the electric field applied. The net velocity at which these electrons drift is known as drift velocity.

Hence, mobility can be expressed as the ratio of drift velocity to the applied electric field.

\(μ_H={v_d\over E}\)                ............ (i)

where, μ_H = Mobility

v_d = Drift velocity

E = Electric field

The drift velocity is given by:

\(v_d={eE\over m_e}τ\)

Putting the value of drift velocity in equation (i), we get:

\(μ_H={eE\over m_eE}τ\)

\(\mu_H={eτ\over m_e}\) , is the required expression of mobility in terms of relaxation time.

where, e = Charge of electron

τ = Relaxation time

m_e = Mass of the electron

Explanation:

• Ions are immobile charges; So, they don't have any mobility
• Electrons and holes are mobile charge carriers
• The mobility of electrons is 2.5 to 3 times the mobility of holes
• The mobility of electrons and holes depends on their effective masses.
• The effective mass of electrons is less than that of holes hence electrons have higher mobility than holes.

For \({\rm{Ge}} = \frac{{{\mu _e}}}{{{\mu _h}}} = \frac{{3800}}{{1800}} = 2.11\)

For \({\rm{Si}} = \frac{{{\mu _e}}}{{{\mu _h}}} = \frac{{1300}}{{500}} = 2.6\)

For both Ge & Si  μe > 2μh

CONCEPT:

Mobility (μ): It is defined as drift velocity per unit electric field applied i.e.,

\(\mu = \frac{{{v_d}}}{E} = \frac{{q\tau }}{m}\)

Where τ = average relaxation time, m = mass of charge particle and q = charge on a charged particle.

EXPLANATION:

• The effective mass of electrons is 9.11 × 10-31 kg.
• Holes being present in the valence band are closer to the nuclei and experience more attractive force and hence have a higher effective mass.
• So, the mobility of free electrons is higher than that of holes because electrons are lighter

Concept:

The ability of a hole/electron to move through a metal or a semiconductor, in the presence of applied electric field is termed as hole/electron mobility.

Mathematically, this is defined as:

\({v_p} = {\mu _p}E\)

\({\mu _p} = \frac{{{v_p}}}{E}\)

v_p = Drift velocity of holes

μ_p = Mobility of holes

E = applied electric field

Explanation:

• The effective mass of electrons is 9.11 × 10^-31 kg.
• Holes being present in the valence band are closer to the nuclei and experience more attractive force and hence have a higher effective mass.

• Conduction of electrons takes place in the conduction band and conduction of holes (valence electrons) takes place in the valence band, since pull of nucleus is stronger on valence electrons more energy is needed to move holes than electrons.
• Hence mobility of electrons is higher than holes (μ_e > μ_h)
• Mobility of electrons is 2.5 to 3 times the mobility of holes

Mobility of electron

Subatomic particles like electrons move in random directions all the time.

When electrons are subjected to an electric field they do move randomly, but they slowly drift in one direction, in the direction of the electric field applied. The net velocity at which these electrons drift is known as drift velocity.

Hence, mobility can be expressed as the ratio of drift velocity to the applied electric field.

\(μ_H={v_d\over E}\)                ............ (i)

where, μ_H = Mobility

v_d = Drift velocity

E = Electric field

The drift velocity is given by:

\(v_d={eE\over m_e}τ\)

Putting the value of drift velocity in equation (i), we get:

\(μ_H={eE\over m_eE}τ\)

\(\mu_H={eτ\over m_e}\) , is the required expression of mobility in terms of relaxation time.

where, e = Charge of electron

τ = Relaxation time

m_e = Mass of the electron

Explanation:

• Ions are immobile charges; So, they don't have any mobility
• Electrons and holes are mobile charge carriers
• The mobility of electrons is 2.5 to 3 times the mobility of holes
• The mobility of electrons and holes depends on their effective masses.
• The effective mass of electrons is less than that of holes hence electrons have higher mobility than holes.

For \({\rm{Ge}} = \frac{{{\mu _e}}}{{{\mu _h}}} = \frac{{3800}}{{1800}} = 2.11\)

For \({\rm{Si}} = \frac{{{\mu _e}}}{{{\mu _h}}} = \frac{{1300}}{{500}} = 2.6\)

For both Ge & Si  μe > 2μh

CONCEPT:

Mobility (μ): It is defined as drift velocity per unit electric field applied i.e.,

\(\mu = \frac{{{v_d}}}{E} = \frac{{q\tau }}{m}\)

Where τ = average relaxation time, m = mass of charge particle and q = charge on a charged particle.

EXPLANATION:

• The effective mass of electrons is 9.11 × 10-31 kg.
• Holes being present in the valence band are closer to the nuclei and experience more attractive force and hence have a higher effective mass.
• So, the mobility of free electrons is higher than that of holes because electrons are lighter

Mobility of electron

Subatomic particles like electrons move in random directions all the time.

When electrons are subjected to an electric field they do move randomly, but they slowly drift in one direction, in the direction of the electric field applied. The net velocity at which these electrons drift is known as drift velocity.

Hence, mobility can be expressed as the ratio of drift velocity to the applied electric field.

\(μ_H={v_d\over E}\)                ............ (i)

where, μ_H = Mobility

v_d = Drift velocity

E = Electric field

The drift velocity is given by:

\(v_d={eE\over m_e}τ\)

Putting the value of drift velocity in equation (i), we get:

\(μ_H={eE\over m_eE}τ\)

\(\mu_H={eτ\over m_e}\) , is the required expression of mobility in terms of relaxation time.

where, e = Charge of electron

τ = Relaxation time

m_e = Mass of the electron

Explanation:

• Ions are immobile charges; So, they don't have any mobility
• Electrons and holes are mobile charge carriers
• The mobility of electrons is 2.5 to 3 times the mobility of holes
• The mobility of electrons and holes depends on their effective masses.
• The effective mass of electrons is less than that of holes hence electrons have higher mobility than holes.

For \({\rm{Ge}} = \frac{{{\mu _e}}}{{{\mu _h}}} = \frac{{3800}}{{1800}} = 2.11\)

For \({\rm{Si}} = \frac{{{\mu _e}}}{{{\mu _h}}} = \frac{{1300}}{{500}} = 2.6\)

For both Ge & Si  μe > 2μh

CONCEPT:

Mobility (μ): It is defined as drift velocity per unit electric field applied i.e.,

\(\mu = \frac{{{v_d}}}{E} = \frac{{q\tau }}{m}\)

Where τ = average relaxation time, m = mass of charge particle and q = charge on a charged particle.

EXPLANATION:

• The effective mass of electrons is 9.11 × 10-31 kg.
• Holes being present in the valence band are closer to the nuclei and experience more attractive force and hence have a higher effective mass.
• So, the mobility of free electrons is higher than that of holes because electrons are lighter

Concept:

The ability of a hole/electron to move through a metal or a semiconductor, in the presence of applied electric field is termed as hole/electron mobility.

Mathematically, this is defined as:

\({v_p} = {\mu _p}E\)

\({\mu _p} = \frac{{{v_p}}}{E}\)

v_p = Drift velocity of holes

μ_p = Mobility of holes

E = applied electric field

Explanation:

• The effective mass of electrons is 9.11 × 10^-31 kg.
• Holes being present in the valence band are closer to the nuclei and experience more attractive force and hence have a higher effective mass.

• Conduction of electrons takes place in the conduction band and conduction of holes (valence electrons) takes place in the valence band, since pull of nucleus is stronger on valence electrons more energy is needed to move holes than electrons.
• Hence mobility of electrons is higher than holes (μ_e > μ_h)
• Mobility of electrons is 2.5 to 3 times the mobility of holes

Explanation:

Hall Mobility (μH)

Definition: Hall mobility is a parameter that quantifies the ability of charge carriers, such as electrons or holes, to move through a semiconductor material under the influence of an electric field. It is directly related to the charge carrier’s drift velocity in response to the applied electric field and is a critical parameter in characterizing the electrical properties of materials.

Formula Derivation:

Hall mobility can be expressed mathematically as:

μH=σne" id="MathJax-Element-29-Frame" role="presentation" style="position: relative;" tabindex="0">μH=σneμH=σne $\mu_{\mathrm{H}} = \frac{\sigma}{n e}$

Where:

• σ" id="MathJax-Element-30-Frame" role="presentation" style="position: relative;" tabindex="0">σσ $\sigma$ is the electrical conductivity of the material.
• n" id="MathJax-Element-31-Frame" role="presentation" style="position: relative;" tabindex="0">nn $n$ is the charge carrier concentration (number of charge carriers per unit volume).
• e" id="MathJax-Element-32-Frame" role="presentation" style="position: relative;" tabindex="0">ee $e$ is the elementary charge of an electron (approximately 1.6×10−19" id="MathJax-Element-33-Frame" role="presentation" style="position: relative;" tabindex="0">1.6×10−191.6×10−19 $1.6 \times 10^{-19}$ C).

Alternatively, Hall mobility can also be expressed in terms of the relaxation time (τ" id="MathJax-Element-34-Frame" role="presentation" style="position: relative;" tabindex="0">ττ $\tau$ ) and the effective mass of the charge carrier (m" id="MathJax-Element-35-Frame" role="presentation" style="position: relative;" tabindex="0">mm $m$ ):

μH=eτm" id="MathJax-Element-36-Frame" role="presentation" style="position: relative;" tabindex="0">μH=eτmμH=eτm $\mu_{\mathrm{H}} = \frac{e \tau}{m}$

Here:

• e" id="MathJax-Element-37-Frame" role="presentation" style="position: relative;" tabindex="0">ee $e$ is the charge of the carrier.
• τ" id="MathJax-Element-38-Frame" role="presentation" style="position: relative;" tabindex="0">ττ $\tau$ is the average time between two successive collisions of the charge carriers (relaxation time).
• m" id="MathJax-Element-39-Frame" role="presentation" style="position: relative;" tabindex="0">mm $m$ is the effective mass of the charge carrier.

This equation is derived from the relationship between drift velocity (vd" id="MathJax-Element-40-Frame" role="presentation" style="position: relative;" tabindex="0">vdvd $v_d$ ), mobility (μ" id="MathJax-Element-41-Frame" role="presentation" style="position: relative;" tabindex="0">μμ $\mu$ ), and the electric field (E" id="MathJax-Element-42-Frame" role="presentation" style="position: relative;" tabindex="0">EE $E$ ), as well as the fundamental definitions of charge carrier dynamics.

Correct Option Analysis:

The correct answer is:

Option 1: μH=eτm" id="MathJax-Element-43-Frame" role="presentation" style="position: relative;" tabindex="0">μH=eτmμH=eτm $\mu_{\mathrm{H}} = \frac{e \tau}{m}$

This is the standard and accurate formula for Hall mobility. It directly reflects the dependence of Hall mobility on the charge of the carrier, the relaxation time, and the effective mass of the charge carrier. The higher the relaxation time or the smaller the effective mass, the greater the mobility of the charge carriers.

Important Information:

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

Option 2: μH=e×τ/m" id="MathJax-Element-44-Frame" role="presentation" style="position: relative;" tabindex="0">μH=e×τ/mμH=e×τ/m $\mu_{\mathrm{H}} = e \times \tau / m$

This option appears to represent the formula correctly, but the use of ambiguous formatting (e.g., missing parentheses or improper division syntax) can lead to misinterpretation. For clarity and standardization in scientific communication, it is essential to use proper mathematical formatting, such as the one provided in Option 1.

Option 3: μH=mτe" id="MathJax-Element-45-Frame" role="presentation" style="position: relative;" tabindex="0">μH=mτeμH=mτe $\mu_{\mathrm{H}} = \frac{m \tau}{e}$

This option is incorrect because it suggests that Hall mobility is directly proportional to the product of the effective mass (m" id="MathJax-Element-46-Frame" role="presentation" style="position: relative;" tabindex="0">mm $m$ ) and the relaxation time (τ" id="MathJax-Element-47-Frame" role="presentation" style="position: relative;" tabindex="0">ττ $\tau$ ) and inversely proportional to the charge (e" id="MathJax-Element-48-Frame" role="presentation" style="position: relative;" tabindex="0">ee $e$ ). This contradicts the actual relationship, where mobility is inversely proportional to the effective mass and directly proportional to both the charge and the relaxation time.

Option 4: μH=emτ" id="MathJax-Element-49-Frame" role="presentation" style="position: relative;" tabindex="0">μH=emτμH=emτ $\mu_{\mathrm{H}} = \frac{e m}{\tau}$

This option is incorrect because it implies that Hall mobility is proportional to the product of charge (e" id="MathJax-Element-50-Frame" role="presentation" style="position: relative;" tabindex="0">ee $e$ ) and effective mass (m" id="MathJax-Element-51-Frame" role="presentation" style="position: relative;" tabindex="0">mm $m$ ) and inversely proportional to the relaxation time (τ" id="MathJax-Element-52-Frame" role="presentation" style="position: relative;" tabindex="0">ττ $\tau$ ). This is inconsistent with the correct formula, which states that mobility is proportional to charge and relaxation time but inversely proportional to the effective mass.

Conclusion:

Hall mobility is a fundamental parameter in semiconductor physics, directly influencing the performance of electronic devices. The correct formula, μH=eτm" id="MathJax-Element-53-Frame" role="presentation" style="position: relative;" tabindex="0">μH=eτmμH=eτm $\mu_{\mathrm{H}} = \frac{e \tau}{m}$ , highlights the interplay between charge, relaxation time, and effective mass in determining the mobility of charge carriers. A clear understanding of this relationship is critical for the analysis and design of materials and devices in electronics and optoelectronics.

Concept:-

Drift Velocity (Vd)- It is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

Mobility of Electron (μ )- The drift velocity of an electron for a unit electric field is known as mobility of the electron.

Current Density

J = σE

σ is the conductivity.

\({\rm{J}} = \frac{{\rm{I}}}{{\rm{A}}}\)

Current

 I= neAVd

∴ \({{\rm{V}}_{\rm{d}}} = \frac{{\rm{I}}}{{{\bf{neA}}}}\)

\({{\rm{V}}_{\rm{d}}} = \frac{{\rm{J}}}{{{\rm{ne}}}}\)

\({{\rm{V}}_{\rm{d}}} = \frac{{{\rm{\sigma E}}}}{{{\rm{ne}}}}\)

If E is the Electric Field

Also, Vd = μE 

So, mobility

\({\rm{\mu }} = \frac{{{{\rm{V}}_{\rm{d}}}}}{{\rm{E}}}\)

Putting the value of drift velocity from above 

\({\rm{\mu }} = \frac{{{\rm{\sigma E}}}}{{{\rm{ne}}}} \times \frac{1}{{\rm{E}}}\)

\({\rm{\mu }} = \frac{{\rm{\sigma }}}{{{\rm{ne}}}}\)

\({\rm{\mu }} = \frac{1}{{{\rm{\rho ne}}}}\)

Calculation:-

Given

Resistivity (ρ ) = 2.5×10-4 Ω /m

Diameter = 1.03 mm

Concentration of free electron (n) = 8.4×1028 /m3

Current density (J) = 2.1×106 A/m2

\({\rm{\mu }} = \frac{1}{{{\rm{\rho ne}}}}\)

\({\rm{\mu }} = \frac{1}{{\left( {2.5 \times {{10}^{ - 4}} \times 8.4 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}}} \right){\rm{\;}}}}\) m2/V-sec

μ = 2.97×10-7 m2/V-sec

Data of diameter and current density is given but not used in the calculation, We may waste a lot of time thinking where to use them and even we can adopt the wrong approach.