## How to Calculate and Solve for Electron Mobility of Metal | Electrical Properties

The image of electron mobility of metal is shown below.

To compute for electron mobility of metal, two essential parameters are needed and these parameters are Hall Coefficient (RH) and Electrical Conductivity (σ).

The formula for calculating electron mobiltiy of metal:

μe = |RH

Where:

μe = Electron Mobility of Metals
RH = Hall Coefficient
σ = Electrical Conductivity

Let’s solve an example;
Find the electron mobility of metal when the hall coefficient is 28 and the electrical conductivity is 7.

This implies that;

RH = Hall Coefficient = 28
σ = Electrical Conductivity = 7

μe = |RH
μe = |28|(7)
μe = (28)(7)
μe = 196

Therefore, the electron mobility of metals is 196 m²/(V.s).

Calculating the Hall Coefficient when the Electron Mobility of Metals and the Electrical Conductivity is Given.

RH = μe / σ

Where:

RH = Hall Coefficient
μe = Electron Mobility of Metals
σ = Electrical Conductivity

Let’s solve an example;
Find the hall coefficient when the electron mobility of metals is 24 and the electrical conductivity is 12.

This implies that;

μe = Electron Mobility of Metals = 24
σ = Electrical Conductivity = 12

RH = μe / σ
RH = 24 / 12
RH = 2

Therefore, the hall coefficient is 2.

## How to Calculate and Solve for Hall Voltage | Electrical Properties

The image of hall voltage is shown below.

To compute for hall voltage, four essential parameters are needed and these parameters are Hall Coefficient (RH), Current (Ix), Magnetic Field (Bz) and Thickness (d).

The formula for calculating hall voltage:

VH = RHIxBz/d

Where:

VH = Hall Voltage
RH = Hall Coefficient
Ix = Current
Bz = Magnetic Field
d = Thickness

Let’s solve an example;
Find the hall voltage when the hall coefficient is 21, the current is 7, the magnetic field is 4 and the thickness is 2.

This implies that;

RH = Hall Coefficient = 21
Ix = Current = 7
Bz = Magnetic Field = 4
d = Thickness = 2

VH = RHIxBz/d
VH = (21)(7)(4)/2
VH = (588)/2
VH = 294

Therefore, the hall voltage is 294 V.

Calculating the Hall Coefficient when the Hall Voltage, the Current, the Magnetic Field and the Thickness is Given.

RH = VH x d / Ix x Bz

Where:

RH = Hall Coefficient
VH = Hall Voltage
Ix = Current
Bz = Magnetic Field
d = Thickness

Let’s solve an example;
Find the hall coefficient when the hall voltage is 10, the current is 8, the magnetic field is 4 and the thickness is 2.

This implies that;

VH = Hall Voltage = 10
Ix = Current = 8
Bz = Magnetic Field = 4
d = Thickness = 2

RH = VH x d / Ix x Bz
RH = 10 x 2 / 8 x 4
RH = 20 / 32
RH = 0.625

Therefore, the hall coefficient is 0.625.

## How to Calculate and Solve for Electrical Conductivity for Extrinsic p-type Semiconductor | Electrical Properties

The electrical conductivity for extrinsic p-type semiconductor is illustrated by the image below.

To compute for electrical conductivity for extrinsic p-type semiconductor, three essential parameters are needed and these parameters are Number of Holes (p), Hole Mobility (μh) and Electrical Charge (e).

The formula for calculating electrical conductivity for extrinsic p-type semiconductor:

σ ≅ p|e|μh

Where:

σ = Electrical Conductivity for Extrinsic p-type Semiconductor
p = Number of Holes
μh = Hole Mobility
e = Electrical Charge

Let’s solve an example;
Find the electrical conductivity for extrinsic p-type semiconductor when the number of holes is 24, the hole mobility is 12 and the electrical charge is 4.

This implies that;

p = Number of Holes = 24
μh = Hole Mobility = 12
e = Electrical Charge = 4

σ ≅ p|e|μh
σ ≅ (24)|4|(12)
σ ≅ (24)(4)(12)
σ ≅ 1152

Therefore, the electrical conductivity for extrinsic p-type semiconductor is 1152 S/m.

Calculating the Number of Holes when the Electrical Conductivity for extrinsic p-type semiconductor, the Hole Mobility and the Electrical Charge is Given.

p = σ / μh(e)

Where:

p = Number of Holes
σ = Electrical Conductivity for Extrinsic p-type Semiconductor
μh = Hole Mobility
e = Electrical Charge

Let’s solve an example;
Find the number of holes when the electrical conductivity for extrinsic p-type semiconductor is 42, the hole mobility is 3 and the electrical charge is 1.

This implies that;

σ = Electrical Conductivity for Extrinsic p-type Semiconductor = 42
μh = Hole Mobility = 3
e = Electrical Charge = 1

p = σ / μh(e)
p = 42 / 3(1)
p = 42 / 3
p = 14

Therefore, the number of holes is 14.

## How to Calculate and Solve for Electrical Conductivity for Extrinsic n-type Semiconductor | Electrical Properties

The image of electrical conductivity for extrinsic n-type semiconductor is shown below.

To compute for electrical conductivity for extrinsic n-type semiconductor, three essential parameters are needed and these parameters are Number of Free Conducting Electrons (n), Electron Mobility (μe) and Electrical Charge (e).

The formula for calculating the electrical conductivity for extrinsic n-type semiconductor:

σ ≅ n|e|μe

Where:

σ = Electrical Conductivity for Extrinsic n-type Semiconductor
n = Number of Free Conducting Electrons
μe = Electron Mobility
e = Electrical Charge

Let’s solve an example;
Find the electrical conductivity for extrinsic n-type semiconductor when the number of free conducting electrons is 12, the electron mobility is 6 and the electrical charge is 4.

This implies that;

n = Number of Free Conducting Electrons = 12
μe = Electron Mobility = 6
e = Electrical Charge = 4

σ ≅ n|e|μe
σ ≅ (12)|4|(6)
σ ≅ (12)(4)(6)
σ ≅ 288

Therefore, the electrical conductivity for extrinsic n-type semiconductor is 288 S/m.

Calculating the Number of Free Conducting Electrons when the Electrical Conductivity for Extrinsic n-type Semiconductor, the Electron Mobility and the Electrical Charge is Given.

n = σ / μe(e)

Where:

n = Number of Free Conducting Electrons
σ = Electrical Conductivity for Extrinsic n-type Semiconductor
μe = Electron Mobility
e = Electrical Charge

Let’s solve an example;
Given that, the electrical conductivity for extrinsic n-type semiconductor is 40, the electron mobility is 4 and the electrical charge is 2. Find the number of free conducting electrons?

This implies that;

σ = Electrical Conductivity for Extrinsic n-type Semiconductor = 40
μe = Electron Mobility = 4
e = Electrical Charge = 2

n = σ / μe(e)
n = 40 / 4(2)
n = 40 / 8
n = 5

Therefore, the number of free conducting electrons is 5.

## How to Calculate and Solve for Electrical Conductivity for Intrinsic Semiconductor | Electrical Properties

The image of electrical conductivity for intrinsic semiconductor is shown below.

To compute for electrical conductivity for intrinsic semiconductor, five essential parameters are needed and these parameters are Number of Holes (p), Number of Free Conducting Electrons (n), Hole Mobility (μh), Electron Mobility (μe) and Electrical Charge (e).

The formula for calculating electrical conductivity for intrinsic semiconductor:

σ = n|e|μe + p|e|μh

Where:

σ = Electrical Conductivity for Intrinsic Semiconductor
p = Number of Holes
n = Number of Free Conducting Electrons
μh = Hole Mobility
μe = Electron Mobility
e = Electrical Charge

Let’s solve an example;
Find the electrical conductivity for intrinsic semiconductor when the number of holes is 14, the number of free conducting electrons is 12, the hole mobility is 8, the electron mobility is 4 and the electrical charge is 2.

This implies that;

p = Number of Holes = 14
n = Number of Free Conducting Electrons = 12
μh = Hole Mobility = 8
μe = Electron Mobility = 4
e = Electrical Charge = 2

σ = n|e|μe + p|e|μh
σ = (12)|2|(4) + (14)|2|(8)
σ = (12)(2)(4) + (14)(2)(8)
σ = 96 + 224
σ = 320

Therefore, the electrical conductivity for intrinsic semiconductor is 320 S/m.

## How to Calculate and Solve for Electron Mobility | Electrical Properties

The electron mobility is illustrated by the image below.

To compute for electron mobility, three essential parameters are needed and these parameters are Electrical Conductivity (σ), Number of Free Conductivity Electrons (n) and Electrical Charge (e).

The formula for calculating electron mobility:

μe = σ/n|e|

Where:

σ = Electron Mobility
σ = Electrical Conductivity
n = Number of Free Conductivity Electrons
e = Electrical Charge

Let’s solve an example;
Find the electron mobility when the electrical conductivity is 28, the number of free conductivity electrons is 7 and the electrical charge is 2.

This implies that;

σ = Electrical Conductivity = 28
n = Number of Free Conductivity Electrons = 7
e = Electrical Charge = 2

μe = σ/n|e|
μe = 28/7|2|
μe = 28/7(2)
μe = 28/14
μe = 2

Therefore, the electron mobility is 2 m²/(V.s).

## How to Calculate and Solve for Impurity Resistivity for Two phase Impurities | Electrical Properties

The image of impurity resistivity for two phase impurities is shown below.

To compute for impurity resistivity for two phase impurities, four essential parameters are needed and these parameters are Resistivity of Phase α (ρα), Resistivity of Phase β (ρβ), Volume Fraction of Phase α (Vα) and Volume Fraction of Phase β (Vβ).

The formula for calculating impurity resistivity for two pahse impurities:

ρi = ραVα + ρβVβ

Where:

ρi = Impurity Resistivity for Two Phase Impurities
ρα = Resistivity of Phase α
ρβ = Resistivity of Phase β
Vα = Volume Fraction of Phase α
Vβ = Volume Fraction of Phase β

Let’s solve an example;
Find the impurity resistivity for two phase impurities when the resistivity of phase α is 8, the resistivity of phase β is 6, the volume fraction of phase α is 4 and the volume fraction of phase β is 2.

This implies that;

ρα = Resistivity of Phase α = 8
ρβ = Resistivity of Phase β = 6
Vα = Volume Fraction of Phase α = 4
Vβ = Volume Fraction of Phase β = 2

ρi = ραVα + ρβVβ
ρi = (8)(4) + (6)(2)
ρi = (32) + (12)
ρi = 44

Therefore, the impurity resistivity for two phase impurities is 44 Ωm.

## How to Calculate and Solve for Impurity Resistivity | Electrical Properties

The impurity resistivity is illustrated by the image below.

To compute for impurity resistivity, two essential parameters are needed and these parameters are Impurity Concentration (Ci) and Composition Independent Constant (A).

The formula for calculating impurity resistivity:

ρi = ACi(1 – Ci)

Where:

ρi = Impurity Resistivity
Ci = Impurity Concentration
A = Composition Independent Constant

Given an example;
Find the impurity resistivity when the impurity concentration is 35 and the composition independent constant is 7.

This implies that;

Ci = Impurity Concentration = 35
A = Composition Independent Constant = 7

ρi = ACi(1 – Ci)
ρi = 7(35)(1 – 35)
ρi = (245)(-34)
ρi = -8330

Therefore, the impurity resistivity is -8330 Ωm.

Calculating the Impurity Concentration when the Impurity Resistivity and the Composition Independent Constant is Given.

Ci = 1 + ρi / A

Where:

Ci = Impurity Concentration
ρi = Impurity Resistivity
A = Composition Independent Constant

Let’s solve an example;
Find the impurity concentration when the impurity resistivity is 20 and the composition independent constant is 10.

This implies that;

ρi = Impurity Resistivity = 20
A = Composition Independent Constant = 10

Ci = 1 + ρi / A
Ci = 1 + 20 / 10
Ci = 1 + 10
Ci = 11

Therefore, the impurity concentration is 11.

## How to Calculate and Solve for Relationship between Thermal Resistivity and Temperature | Electrical Properties

The relationship betweeen thermal resistivity and temperature is illustrated by the image below.

To compute for the relationship between thermal resistivity and temperature, three essential parameters are needed and these parameters are Constant of the Metal (a), Constant of the Metal (ρo) and Temperature (T).

The formula for calculating relationship between thermal resistivity and temperature:

ρt = ρo + aT

Where:

ρt = Thermal Resistivity
ρo = Constant of the Metal
a = Constant of the Metal
T = Temperature

Let’s solve an example;
Find the relationship between thermal resistivity and temperature when the constant of the metal is 32, the constant of the metal is 8 and the temperature is 2.

This implies that;

ρo = Constant of the Metal = 32
a = Constant of the Metal = 8
T = Temperature = 2

ρt = ρo + aT
ρt = 32 + (8)(2)
ρt = 32 + (16)
ρt = 48

Therefore, the relationship between thermal resistivity and temperature is 48 Ωm.

Calculating the Constant of the Metal when the Relationship between Thermal Resistivity and Temperature, the Constant of the Metal and the Temperature is Given.

ρo = ρt – aT

Where:

ρo = Constant of the Metal
ρt = Thermal Resistivity
a = Constant of the Metal
T = Temperature

Let’s solve an example;
Find the constant of the metal when the thermal resistivity is 32, the constant of the metal is 2 and the temperature is 4.

This implies that;

ρt = Thermal Resistivity = 32
a = Constant of the Metal = 2
T = Temperature = 4

ρo = ρt – aT
ρo = 32 – 2(4)
ρo = 32 – 8
ρo = 24

Therefore, the constant of the metal is 24.

## How to Calculate and Solve for Total Resistivity of Metal | Electrical Properties

The total resistivity of metal is illustrated by the image below.

To compute for total resistivity of metal, three essential parameters are needed and these parameters are Thermal Resistivity (ρt), Impurity Resistivity (ρi) and Deformation Resistivity (ρd).

The formula for calculating the total resistivity of metal:

ρtotal = ρt + ρi + ρd

Where:

ρtotal = Total Resistivity of Metal
ρt = Thermal Resistivity
ρi = Impurity Resistivity
ρd = Deformation Resistivity

Let’s solve an example;
Find the total resistivity of metal when the thermal resistivity is 20, the impurity resistivity is 10 and the deformation resistivity is 2.

This implies that;

ρt = Thermal Resistivity = 20
ρi = Impurity Resistivity = 10
ρd = Deformation Resistivity = 2

ρtotal = ρt + ρi + ρd
ρtotal = 20 + 10 + 2
ρtotal = 32

Therefore, the total resistivity of metal is 32 Ω-m.

Calculating the Thermal Resistivity when the Total Resistivity of Metal, the Impurity Resistivity and the Deformation Resistivity is Given.

ρt = ρtotal – ρi – ρd

Where:

ρt = Thermal Resistivity
ρtotal = Total Resistivity of Metal
ρi = Impurity Resistivity
ρd = Deformation Resistivity

Let’s solve an example;
Find the thermal resistivity when the total resistivity of metal is 40, the impurity resistivity is 10 and the deformation resistivity is 5.

This implies that;

ρtotal = Total Resistivity of Metal = 40
ρi = Impurity Resistivity = 10
ρd = Deformation Resistivity = 5

ρt = ρtotal – ρi – ρd
ρt = 40 – 10 – 5
ρt = 25

Therefore, the thermal resistivity is 25.