Physics (3110018) Semester 1 & 2 Winter 2025 GTU Solved Paper
GTU Physics (3110018) Semester 1 & 2 Winter 2025 Solved Paper
Complete, student-friendly worked solutions for GTU Physics (Subject Code 3110018) Winter 2025 examination. Step-by-step answers for all theory, derivations, and numerical questions.
Introduction
This document contains fully worked, step-by-step solutions for the Gujarat Technological University (GTU) Bachelor of Engineering (BE) Semester I & II Physics examination (Subject Code: 3110018) held for Winter 2025. All numerical calculations are worked out with explicit formulas, substitution, and standard units, while theoretical concepts and derivations are presented with clear explanations and diagrams to help students prepare effectively.
Paper Information
- University
- Gujarat Technological University
- Department
- First Year Engineering (Common to All Branches)
- Semester
- Semester 1 & 2
- Subject
- Physics
- Subject Code
- 3110018
- Exam
- Winter 2025
- Year
- 2025
Questions and Solutions
Q.1 (a)
Question
Prove that Probability of occupancy of Fermi energy level is 0.5 at T > 0 K
Solution
Fermi-Dirac Distribution Function
The probability f(E) that an available electron energy state with energy E is occupied by an electron at an absolute temperature T is given by the Fermi-Dirac distribution function:
f(E) = 1 / [1 + exp((E - Ef) / (kB * T))]
Where:
- E = Energy of the given state (Joules or eV)
- Ef = Fermi energy level (Joules or eV)
- kB = Boltzmann constant = 1.38 × 10⁻²³ J/K
- T = Absolute temperature in Kelvin (K)
---
Proof for E = Ef at T > 0 K
- At the Fermi energy level, we set the state energy equal to the Fermi energy:
E = Ef
- Substitute E = Ef into the exponent term:
(E - Ef) / (kB * T) = (Ef - Ef) / (kB * T) = 0 / (kB * T)
- Since T > 0 K, the denominator is non-zero, giving:
(E - Ef) / (kB * T) = 0
- Substituting this value into the exponential term:
exp(0) = 1
- Substitute into the probability function:
f(Ef) = 1 / [1 + exp(0)] = 1 / [1 + 1] = 1 / 2 = 0.5
Conclusion
At any absolute temperature above absolute zero (T > 0 K), the probability of finding an electron in the Fermi energy level is exactly 0.5 (or 50%). Hence proved.
Q.1 (b)
Question
Explain how the materials are classified into conductors, semiconductors and insulators with the help of energy band diagrams.
Solution
Classification of Materials Based on Energy Band Theory
In solid materials, electron energy states form continuous bands separated by forbidden energy gaps (band gap, Eg). Materials are classified into three primary categories depending on the size of the band gap and the filling of the valence band (VB) and conduction band (CB).
Energy Band Diagrams (Schematic, not to scale):
Conductor Semiconductor Insulator
------------- ------------- -------------
|Conduction | |Conduction | |Conduction |
| Band | | Band | | Band |
|===Over====| ------------- -------------
|===lap=====| ↑ ↑
| Valence | Eg ≈ 1.1 eV Eg > 3 to 5 eV
| Band | ↓ ↓
------------- ------------- -------------
| Valence | | Valence |
| Band | | Band |
------------- -------------
---
1. Conductors (Metals)
- Band Gap: There is no forbidden gap (Eg = 0 eV). The valence band and conduction band overlap, or the valence band is partially filled.
- Electrical Conduction: A very large number of free electrons are readily available in the conduction band even at absolute zero (0 K). When a small electric field is applied, electrons easily gain energy and move, yielding very high electrical conductivity (typically 10⁶ to 10⁸ S/m).
- Examples: Copper (Cu), Aluminum (Al), Silver (Ag).
---
2. Semiconductors
- Band Gap: A small forbidden energy gap exists between the valence band and conduction band (typically Eg < 3 eV; for Silicon Eg ≈ 1.12 eV, for Germanium Eg ≈ 0.67 eV at room temperature).
- Electrical Conduction: At 0 K, the valence band is completely full and the conduction band is completely empty, behaving like an insulator. At room temperature, thermal energy excites a small fraction of electrons across the band gap into the conduction band, leaving holes behind. Conductivity is moderate (10⁻⁶ to 10² S/m) and increases with temperature.
- Examples: Silicon (Si), Germanium (Ge), Gallium Arsenide (GaAs).
---
3. Insulators
- Band Gap: A very wide forbidden energy gap exists between the valence band and conduction band (Eg > 3 eV to 9 eV; for diamond Eg ≈ 5.4 eV).
- Electrical Conduction: The valence band is completely filled and the conduction band is empty. Under ordinary conditions, thermal excitation cannot provide sufficient energy for valence electrons to cross the large gap. Therefore, virtually no free charge carriers exist, resulting in extremely low conductivity (< 10⁻¹⁰ S/m).
- Examples: Diamond, Glass, Quartz, Mica.
Q.1 (c)
Question
Explain in detail how Kroning and Penney model explains the origin of band gap in semiconductors.
Solution
Introduction to the Kronig-Penney Model
The Kronig-Penney model is an idealized, one-dimensional quantum-mechanical model that explains the behavior of an electron moving in a periodic crystalline potential. In a crystal lattice, positive ion cores produce periodic potential wells. Kronig and Penney simplified this periodic potential into an array of rectangular potential wells and barriers.
Periodic Rectangular Potential Barrier Setup (schematic, not to scale):
V(x) ^
| V0 V0 V0
| +----+ +----+ +----+
| | | | | | |
0 +---+----+---+----+---+----+-----> x
-b 0 a a+b
Region I | Region II
- Region I (Well): 0 < x < a, where potential V(x) = 0.
- Region II (Barrier): -b < x < 0, where potential V(x) = V0.
- The lattice periodicity is a + b.
---
Wave Equations and Boundary Conditions
According to the Time-Independent Schrödinger Equation:
- For Region I (0 < x < a, V = 0):
d²ψ₁/dx² + α² ψ₁ = 0
where α² = (2 * m * E) / ℏ²
- For Region II (-b < x < 0, V = V0 > E):
d²ψ₂/dx² - β² ψ₂ = 0
where β² = [2 * m * (V0 - E)] / ℏ²
According to Bloch's Theorem, the wave function in a periodic potential must satisfy:
ψ(x) = u(x) * exp(i * k * x)
where u(x) has the periodicity of the lattice: u(x + a + b) = u(x), and k is the electron wave vector.
Applying continuity conditions for ψ(x) and its derivative dψ/dx at boundaries x = 0 and x = a, and applying the Kronig-Penney simplification (letting the barrier height V0 → ∞ and barrier width b → 0 such that the product V0 * b remains finite), the condition for non-trivial solutions reduces to the famous Kronig-Penney equation:
P * [sin(α * a) / (α * a)] + cos(α * a) = cos(k * a)
Where:
- P = (m V0 b * a) / ℏ² is the barrier strength or scattering power.
- α = sqrt(2 m E) / ℏ relates directly to electron energy E.
- k is the crystal wave vector.
---
Explanation of Allowed and Forbidden Bands
Let the left-hand side function be:
F(α * a) = P * [sin(α * a) / (α * a)] + cos(α * a)
Since the right-hand side is cos(k * a), and cosine can only have values between -1 and +1:
-1 ≤ F(α * a) ≤ +1
- Allowed Energy Bands:
- For values of (α a) where
|F(αa)| ≤ 1
k exist.- Forbidden Energy Bands (Band Gaps):
- For values of (α a) where
|F(αa)| > 1
k becomes complex or imaginary, leading to exponentially decaying wavefunctions.- Effect of P (Scattering Strength):
- Case 1: If P → 0 (Completely free electrons):
The equation becomescos(α a) = cos(k a), soα = k. Energy becomesE = ℏ² k² / (2m), which is the continuous energy spectrum of completely free electrons with no band gaps. - Case 2: If P → ∞ (Tightly bound electrons):
sin(α a) = 0impliesα
The spectrum splits into alternating bands of allowed energy states and forbidden band gaps.
Conclusion
The periodic potential of the atomic lattice reflects the electron waves destructively at the Brillouin zone boundaries (k = ±n * π / a), leading to energy discontinuities that naturally create the band gaps observed in semiconductors and insulators.
Q.2 (a)
Question
Define Following Terms (i) Population Inversion , (ii) Meta stable state (iii) Active Medium
Solution
(i) Population Inversion
Population Inversion is a non-equilibrium state of an atomic or molecular system in which the number of atoms or electrons in an excited higher energy state (N₂) exceeds the number of atoms in a lower energy state (N₁), i.e., N₂ > N₁.
- Under normal thermal equilibrium conditions described by Boltzmann statistics, N₁ is always greater than N₂.
- Population inversion is an essential prerequisite for stimulated emission to dominate over absorption, enabling laser action.
---
(ii) Metastable State
A metastable state is an excited atomic state that possesses an unusually long lifetime compared to ordinary excited states.
- While standard excited states typically have lifetimes on the order of 10⁻⁸ seconds, a metastable state has an extended lifetime of 10⁻⁶ to 10⁻³ seconds (or longer).
- This extended lifetime allows excited atoms to accumulate in the higher energy level, making population inversion achievable.
---
(iii) Active Medium
An active medium is the material or substance inside a laser system that contains the specific atoms, ions, or molecules capable of undergoing excitation and stimulated emission to amplify light.
- When supplied with energy from an external source (pumping), the active medium supports population inversion and emits coherent laser radiation.
- Examples: Helium-Neon gas mixture (He-Ne laser), Chromium-doped Sapphire crystal (Ruby laser), Nd:YAG crystal, or a semiconductor diode junction.
Q.2 (b)
Question
Define carrier transport and explore how charge carriers move within semiconductor materials, emphasizing the mechanisms of drift and diffusion.
Solution
Carrier Transport in Semiconductors
Carrier transport refers to the directional movement of charge carriers (free electrons in the conduction band and holes in the valence band) through a semiconductor lattice, resulting in an electric current.
Charge carriers move via two independent physical mechanisms:
- Drift Mechanism
- Diffusion Mechanism
---
1. Drift Mechanism
- Definition: Drift is the directional motion of charge carriers under the influence of an applied external electric field (E).
- Mechanism: In the absence of an electric field, carriers move randomly due to thermal energy with zero net current. When an electric field is applied, electrons experience an electrostatic force opposite to the field, and holes experience a force in the direction of the field, acquiring a net drift velocity (vd):
vd = μ * E
where μ is the carrier mobility.
- Drift Current Densities:
- Electron drift current density:
Jn,drift = n * e * μn * E - Hole drift current density:
Jp,drift = p * e * μp * E - Total drift current density:
Jdrift = (n * μn + p μ_p) e E = σ E
whereσis electrical conductivity.
---
2. Diffusion Mechanism
- Definition: Diffusion is the movement of charge carriers from a region of higher carrier concentration to a region of lower carrier concentration due to thermal motion and concentration gradients, independent of any electric field.
- Mechanism: When a non-uniform distribution of carriers exists (such as localized photoexcitation or carrier injection at a junction), carriers spread out to achieve spatial equilibrium, obeying Fick's first law.
- Diffusion Current Densities:
- Electron diffusion current density:
Jn,diff = e * Dn * (dn/dx) - Hole diffusion current density:
Jp,diff = -e * Dp * (dp/dx)
whereDnandD
dn/dx and dp/dx are the concentration gradients.
---
Total Current Density in a Semiconductor
The total current is the algebraic sum of drift and diffusion components:
J_total = J_n + J_p
J_total = [n * e * μ_n * E + e * D_n * (dn/dx)] + [p * e * μ_p * E - e * D_p * (dp/dx)]Q.2 (c)
Question
Describe how to formulate the density of holes in the valence band for intrinsic semiconductors.
Solution
Density of Holes in the Valence Band
In a semiconductor, the density of holes in the valence band, p, represents the number of unoccupied electron states per unit volume in the valence band at a given temperature T.
---
Mathematical Formulation
- Basic Integral Formulation:
The total hole concentration is given by the integral of the product of the density of states in the valence band, g_v(E), and the probability of a state being vacant (occupied by a hole), [1 - f(E)], from the lower energy boundary (-∞) to the top of the valence band (Ev):
p = ∫_{-∞}^{Ev} g_v(E) * [1 - f(E)] dE
- Density of States in the Valence Band:
Near the valence band edge (Ev), the density of quantum states per unit volume is:
g_v(E) = [4 * π * (2 * m_h*)^(3/2) / h³] * sqrt(Ev - E)
where m_h* is the effective mass of a hole and h is Planck's constant.
- Hole Occupancy Probability:
The Fermi-Dirac distribution function for electrons is:
f(E) = 1 / [1 + exp((E - Ef) / (kB * T))]
The probability that a state is empty of an electron (hence contains a hole) is:
1 - f(E) = 1 - 1 / [1 + exp((E - Ef) / (kB * T))]
= exp((E - Ef) / (kB * T)) / [1 + exp((E - Ef) / (kB * T))]
For an intrinsic semiconductor, the Fermi level Ef lies within the band gap well above Ev, such that (Ef - E) >> kB T (or Ef - Ev ≥ 3 kB
1 - f(E) ≈ exp(-(Ef - E) / (kB * T)) = exp((E - Ef) / (kB * T))
- Setting up the Integral:
Substitute g_v(E) and 1 - f(E) into the integral:
p = [4 * π * (2 * m_h*)^(3/2) / h³] * ∫_{-∞}^{Ev} sqrt(Ev - E) * exp((E - Ef) / (kB * T)) dE
Rewrite exp((E - Ef) / (kB T)) as exp(-(Ef - Ev) / (kB
p = [4 * π * (2 * m_h*)^(3/2) / h³] * exp(-(Ef - Ev) / (kB * T)) * ∫_{-∞}^{Ev} sqrt(Ev - E) * exp(-(Ev - E) / (kB * T)) dE
- Evaluating the Integral:
Let x = (Ev - E) / (kB * T). Then:
- When
E = Ev,x = 0 - When
E → -∞,x → ∞ Ev - E = kB T xdE = -(kB * T) dx
Substituting these variables:
∫_{-∞}^{Ev} sqrt(Ev - E) * exp(-(Ev - E) / (kB * T)) dE = (kB * T)^(3/2) * ∫_{0}^{∞} x^(1/2) * exp(-x) dx
The standard gamma function integral evaluates to:
∫_{0}^{∞} x^(1/2) * exp(-x) dx = Γ(3/2) = (1/2) * sqrt(π)
- Final Expression for Hole Concentration:
Multiplying back:
p = [4 * π * (2 * m_h*)^(3/2) / h³] * (kB * T)^(3/2) * [(sqrt(π) / 2)] * exp(-(Ef - Ev) / (kB * T))
Simplifying the numerical factors:
p = 2 * [(2 * π * m_h* * kB * T) / h²]^(3/2) * exp(-(Ef - Ev) / (kB * T))
We define the effective density of states in the valence band (Nv) as:
Nv = 2 * [(2 * π * m_h* * kB * T) / h²]^(3/2)
Thus, the expression for the hole density in the valence band is:
p = Nv * exp(-(Ef - Ev) / (kB * T))Q.2 (c) OR
Question
Describe the definition of carrier concentration in intrinsic semiconductors and show how to derive the formula for electron density in the conduction band.
Solution
Carrier Concentration in Intrinsic Semiconductors
Definition: In an intrinsic (pure) semiconductor without any dopant impurities, charge carriers are generated solely through thermal excitation across the energy band gap. This process produces equal numbers of free electrons in the conduction band and holes in the valence band:
n = p = ni
where:
n= Electron density in the conduction band (m⁻³ or cm⁻³)p= Hole density in the valence band (m⁻³ or cm⁻³)ni= Intrinsic carrier concentration
---
Derivation of Electron Density in the Conduction Band (n)
- Fundamental Integral:
The concentration of electrons in the conduction band is obtained by integrating the product of the density of available quantum states g_c(E) and the probability of occupation f(E) from the conduction band edge (Ec) to the top of the band (∞):
n = ∫_{Ec}^{∞} g_c(E) * f(E) dE
- Density of States in Conduction Band:
For electrons near the bottom of the conduction band:
g_c(E) = [4 * π * (2 * m_e*)^(3/2) / h³] * sqrt(E - Ec)
where m_e* is the electron effective mass and h is Planck's constant.
- Fermi-Dirac Distribution:
f(E) = 1 / [1 + exp((E - Ef) / (kB * T))]
For intrinsic and non-degenerate semiconductors, the Fermi level Ef is situated within the band gap such that (Ec - Ef) >> kB T (at least 3 kB
f(E) ≈ exp(-(E - Ef) / (kB * T))
- Integration:
Substitute g_c(E) and f(E) into the integral:
n = [4 * π * (2 * m_e*)^(3/2) / h³] * ∫_{Ec}^{∞} sqrt(E - Ec) * exp(-(E - Ef) / (kB * T)) dE
We can split the exponential:
exp(-(E - Ef) / (kB * T)) = exp(-(Ec - Ef) / (kB * T)) * exp(-(E - Ec) / (kB * T))
Taking constants outside the integral:
n = [4 * π * (2 * m_e*)^(3/2) / h³] * exp(-(Ec - Ef) / (kB * T)) * ∫_{Ec}^{∞} sqrt(E - Ec) * exp(-(E - Ec) / (kB * T)) dE
- Variable Substitution:
Let y = (E - Ec) / (kB * T). Then:
E - Ec = kB T ydE = kB T dy- Limits: When
E = Ec,y = 0; whenE → ∞,y → ∞
The integral becomes:
(kB * T)^(3/2) * ∫_{0}^{∞} y^(1/2) * exp(-y) dy
Using the standard Gamma function identity:
∫_{0}^{∞} y^(1/2) * exp(-y) dy = Γ(3/2) = (1/2) * sqrt(π)
- Final Expression:
n = [4 * π * (2 * m_e*)^(3/2) / h³] * (kB * T)^(3/2) * [(sqrt(π) / 2)] * exp(-(Ec - Ef) / (kB * T))
Simplifying terms:
n = 2 * [(2 * π * m_e* * kB * T) / h²]^(3/2) * exp(-(Ec - Ef) / (kB * T))
Defining the effective density of states in the conduction band (Nc) as:
Nc = 2 * [(2 * π * m_e* * kB * T) / h²]^(3/2)
We obtain the formula for electron density in the conduction band:
n = Nc * exp(-(Ec - Ef) / (kB * T))Q.3 (a)
Question
Determine the band gap of the material when the light source composed of it emits radiation with a wavelength of 8628 Å.
Solution
1. Given Information
- Emitted wavelength (λ) = 8628 Å = 8628 × 10⁻¹⁰ m = 8.628 × 10⁻⁷ m
- Planck's constant (h) = 6.626 × 10⁻³⁴ J·s
- Speed of light (c) = 3 × 10⁸ m/s
- Electronic charge (e) = 1.602 × 10⁻¹⁹ C
2. What is Required
- Energy band gap of the material (Eg) in Joules (J) and electron-volts (eV).
3. Formula
Radiation emitted during interband radiative recombination corresponds to the band gap energy:
Eg = h * c / λ
To express the band gap directly in electron-volts (eV):
Eg (in eV) = (h * c) / (λ * e) ≈ 12400 / λ (in Å)
4. Calculation Steps
- Calculation in Joules (J):
Eg = (6.626 × 10⁻³⁴ J·s × 3 × 10⁸ m/s) / (8.628 × 10⁻⁷ m)
Eg = (1.9878 × 10⁻²⁵) / (8.628 × 10⁻⁷)
Eg = 2.3039 × 10⁻¹⁹ J
- Conversion to electron-volts (eV):
Eg = (2.3039 × 10⁻¹⁹ J) / (1.602 × 10⁻¹⁹ J/eV)
Eg = 1.438 eV
5. Final Answer
The band gap of the material is 1.44 eV (or 2.30 × 10⁻¹⁹ J).
Q.3 (b)
Question
Explain construction and working of PV solar cell.
Solution
Photovoltaic (PV) Solar Cell
A photovoltaic (PV) solar cell is a specialized semiconductor p-n junction diode that directly converts solar/light energy into electrical energy using the photovoltaic effect.
---
Construction
Schematic Diagram of a Silicon PV Solar Cell (not to scale):
Sunlight (Photons)
↓↓↓↓ ↓↓↓↓ ↓↓↓↓
============================== <-- Top Metal Grid / Finger Contacts
------------------------------ <-- Anti-reflection Coating (ARC)
++++++++++++++++++++++++++++++ <-- Thin n-type Emitter Layer (~0.3 μm)
------------------------------ <-- Depletion Region (W)
.............................. <-- Thick p-type Base Layer (~300 μm)
============================== <-- Continuous Bottom Metal Contact
- Substrate / Base: A relatively thick (around 200–300 µm) p-type silicon wafer forms the base.
- Emitter Layer: A very thin, heavily doped n-type silicon layer (~0.3 µm) is formed on the top surface by phosphorus diffusion. The top layer is kept thin to ensure incident photons penetrate directly into and near the depletion layer.
- Depletion Region: A built-in electric field exists naturally at the p-n junction.
- Anti-Reflection Coating (ARC): A layer of silicon nitride (Si₃N₄) or titanium dioxide (TiO₂) is deposited on top to reduce surface reflection of incoming sunlight.
- Contacts: A metallic grid pattern is applied to the front surface to collect electrons while allowing light to pass, and a full-area metallic layer (typically aluminum or silver) forms the back contact.
---
Working Principle
The operation of a solar cell involves three sequential steps:
- Generation of Electron-Hole Pairs:
Incident photons with energy greater than the band gap of silicon (hν ≥ Eg ≈ 1.12 eV) are absorbed near the junction, exciting electrons from the valence band to the conduction band and creating electron-hole (e-h) pairs.
- Separation of Charge Carriers:
The strong built-in electric field in the depletion region sweeps generated electrons toward the n-side and holes toward the p-side before they can recombine.
- Collection and Current Flow:
- Electrons accumulate in the n-layer (making it negatively charged), and holes accumulate in the p-layer (making it positively charged).
- This builds an open-circuit photovoltage (Voc) across the terminals.
- When an external load resistor (RL) is connected, photo-generated electrons flow through the external circuit, producing a continuous direct current (photocurrent, Iph).
Q.3 (c)
Question
Interpret the process of deriving the Fermi level expression for n-Type semiconductors, and subsequently, utilize this expression to compute the electron density in the conduction band.
Solution
1. Introduction and Charge Neutrality Condition
In an n-type semiconductor, donor impurities of concentration Nd are added to an intrinsic host crystal. At ordinary operating temperatures, complete ionization occurs, so almost all donor atoms donate an electron:
Nd⁺ ≈ Nd
From the condition of electrical neutrality:
n = p + Nd⁺
Since donor doping makes the electron concentration substantially larger than the hole concentration (n >> p), we can safely approximate:
n ≈ Nd
---
2. Derivation of the Fermi Level Expression for n-Type Semiconductors
The electron density in the conduction band is given by:
n = Nc * exp(-(Ec - Ef) / (kB * T))
where:
Nc= Effective density of states in the conduction band = 2 [(2 π m_e kB T) / h²]^(3/2)Ec= Energy of the conduction band edgeEf= Fermi energy level in the n-type semiconductor (Efn)kB= Boltzmann constantT= Temperature in Kelvin
Equating n to the active donor concentration Nd:
Nd = Nc * exp(-(Ec - Ef) / (kB * T))
Divide both sides by Nc:
exp(-(Ec - Ef) / (kB * T)) = Nd / Nc
Invert the exponential by taking the reciprocal:
exp((Ec - Ef) / (kB * T)) = Nc / Nd
Take the natural logarithm (ln) on both sides:
(Ec - Ef) / (kB * T) = ln(Nc / Nd)
Multiply by kB * T:
Ec - Ef = kB * T * ln(Nc / Nd)
Rearrange to solve for the Fermi level Ef:
Ef = Ec - kB * T * ln(Nc / Nd)
Alternatively, written using properties of logarithms:
Ef = Ec + kB * T * ln(Nd / Nc)
- Interpretation: Since
Nc > Ndunder non-degenerate conditions,ln(Nc / Nd) > 0. Therefore,Eflies slightly below the bottom of the conduction bandEc. As the donor concentrationNdincreases,Efmoves closer toEc.
---
3. Computation of Electron Density in the Conduction Band
Substitute the derived expression for Ef back into the equation for electron density:
n = Nc * exp(-(Ec - Ef) / (kB * T))
Substitute (Ec - Ef) = kB T ln(Nc / Nd):
n = Nc * exp(-[kB * T * ln(Nc / Nd)] / (kB * T))
n = Nc * exp(-ln(Nc / Nd))
n = Nc * exp(ln(Nd / Nc))
n = Nc * (Nd / Nc)
n = Nd
Conclusion
This confirms that at room temperature, all donor atoms are ionized, and the total free electron density in the conduction band is directly determined by the donor doping concentration: n = Nd.
Q.3 (a) OR
Question
The intrinsic carrier concentration at room temp in Germanium is 4.17×1018 m-3. The mobility of Electron and Hole are 0.29 m2/VSec and 0.16 m2/vSec respectively. Find out its Resistivity
Solution
1. Given Information
- Intrinsic carrier concentration of Germanium (ni) = 4.17 × 10¹⁸ m⁻³
- Electron mobility (μe) = 0.29 m²/(V·s)
- Hole mobility (μh) = 0.16 m²/(V·s)
- Electronic charge (e) = 1.602 × 10⁻¹⁹ C
2. What is Required
- Electrical resistivity (ρ) of intrinsic Germanium in Ω·m.
3. Formulas
- Electrical conductivity of an intrinsic semiconductor (σ):
σ = ni * e * (μe + μh)
- Electrical resistivity (ρ) is the reciprocal of conductivity:
ρ = 1 / σ = 1 / [ni * e * (μe + μh)]
4. Calculation Steps
- Calculate the sum of mobilities:
μe + μh = 0.29 + 0.16 = 0.45 m²/(V·s)
- Calculate electrical conductivity (σ):
σ = (4.17 × 10¹⁸ m⁻³) × (1.602 × 10⁻¹⁹ C) × (0.45 m²/(V·s))
σ = 4.17 × 1.602 × 0.45 × 10⁻¹
σ = 3.0062 × 10⁻¹ = 0.3006 S/m (or Ʊ/m)
- Calculate resistivity (ρ):
ρ = 1 / σ = 1 / 0.3006
ρ = 3.327 Ω·m
5. Final Answer
The resistivity of the Germanium sample is 3.33 Ω·m.
Q.3 (b) OR
Question
The conductivity of a semiconductor at 30 ℃ is 742 Ʊ/m and at 90 ℃ is 2100 Ʊ/m. Calculate the Energy Band Gap.
Solution
1. Given Information
- Temperature T1 = 30 °C = 30 + 273.15 = 303.15 K
- Conductivity at T1 (σ1) = 742 Ʊ/m (S/m)
- Temperature T2 = 90 °C = 90 + 273.15 = 363.15 K
- Conductivity at T2 (σ2) = 2100 Ʊ/m (S/m)
- Boltzmann constant (kB) = 1.38 × 10⁻²³ J/K = 8.617 × 10⁻⁵ eV/K
2. What is Required
- Energy Band Gap (Eg) of the semiconductor.
3. Formula
The temperature dependence of electrical conductivity in an intrinsic semiconductor is governed by:
σ(T) = σ0 * exp(-Eg / (2 * kB * T))
Taking the ratio of conductivities at temperatures T2 and T1:
σ2 / σ1 = exp[ (Eg / (2 * kB)) * (1/T1 - 1/T2) ]
Taking natural logarithms on both sides:
ln(σ2 / σ1) = (Eg / (2 * kB)) * [ (T2 - T1) / (T1 * T2) ]
Solving for Eg:
Eg = [ 2 * kB * ln(σ2 / σ1) * T1 * T2 ] / (T2 - T1)
4. Calculation Steps
- Calculate the ratio of conductivities and its logarithm:
σ2 / σ1 = 2100 / 742 ≈ 2.83019
ln(2.83019) = 1.04034
- Calculate temperature parameters:
T2 - T1 = 363.15 - 303.15 = 60.0 K
T1 * T2 = 303.15 × 363.15 = 110088.92 K²
(T1 * T2) / (T2 - T1) = 110088.92 / 60 = 1834.815 K
- Compute Eg in eV using kB = 8.617 × 10⁻⁵ eV/K:
Eg = 2 × (8.617 × 10⁻⁵ eV/K) × 1.04034 × 1834.815 K
Eg = 0.17929 × 1.834815
Eg = 0.329 eV
- Equivalent calculation in Joules:
Eg = 0.329 eV × 1.602 × 10⁻¹⁹ J/eV = 5.27 × 10⁻²⁰ J
5. Final Answer
The energy band gap of the semiconductor is 0.329 eV (or 5.27 × 10⁻²⁰ J).
Q.3 (c) OR
Question
Discuss the Einstein’s theory about interaction between matter and radiation in detail. Also discuss the absorption and emission with necessary equation.
Solution
Einstein's Theory of Matter-Radiation Interaction
In 1917, Albert Einstein proposed that the interaction between electromagnetic radiation and matter involves three fundamental transition processes between two atomic energy levels (E₁ and E₂ with E₂ > E₁):
- Stimulated Absorption
- Spontaneous Emission
- Stimulated Emission
---
1. Three Interaction Processes
Transition Diagrams (schematic, not to scale):
(a) Induced Absorption (b) Spontaneous Emission (c) Stimulated Emission
E2 ----------- E2 ---•------- E2 ---•-------
↑ ↓ ↓
hν ~> • hν ~> hν ~> ~~> hν
~~> hν
E1 ---•------- E1 ----------- E1 -----------
A. Stimulated (Induced) Absorption
An atom in the lower energy state E₁ absorbs an incident photon of energy hν = E₂ - E₁ and transitions to the higher excited state E₂.
- The rate of stimulated absorption is proportional to the number of atoms in level 1 (N₁) and the radiation energy density
u(ν):
R_abs = B₁₂ * N₁ * u(ν)
where B₁₂ is the Einstein coefficient of stimulated absorption.
B. Spontaneous Emission
An excited atom in state E₂ transitions spontaneously to lower state E₁ without any external influence, emitting a photon of energy hν = E₂ - E₁. The emitted photons have random phase, direction, and polarization (incoherent light).
- The rate of spontaneous emission depends solely on the population N₂:
R_spont = A₂₁ * N₂
where A₂₁ is the Einstein coefficient of spontaneous emission.
C. Stimulated Emission
An incident photon of energy hν = E₂ - E₁ triggers an excited atom in state E₂ to drop to state E₁, emitting a second photon identical to the incident photon in frequency, phase, direction, and polarization (coherent light).
- The rate of stimulated emission is proportional to the population N₂ and radiation density
u(ν):
R_stim = B₂₁ * N₂ * u(ν)
where B₂₁ is the Einstein coefficient of stimulated emission.
---
2. Derivation of Einstein's Relations
Under thermal equilibrium at temperature T:
Rate of upward transitions (Absorption) = Rate of downward transitions (Total Emission)
B₁₂ * N₁ * u(ν) = A₂₁ * N₂ + B₂₁ * N₂ * u(ν)
Rearranging for u(ν):
u(ν) * [B₁₂ * N₁ - B₂₁ * N₂] = A₂₁ * N₂
u(ν) = (A₂₁ * N₂) / [B₁₂ * N₁ - B₂₁ * N₂]
Divide numerator and denominator by B₂₁ * N₂:
u(ν) = (A₂₁ / B₂₁) / [ (B₁₂ / B₂₁) * (N₁ / N₂) - 1 ]
According to Boltzmann's Distribution Law:
N₁ / N₂ = exp((E₂ - E₁) / (kB * T)) = exp(hν / (kB * T))
Substitute this into the expression for u(ν):
u(ν) = (A₂₁ / B₂₁) / [ (B₁₂ / B₂₁) * exp(hν / (kB * T)) - 1 ] ---- (Equation 1)
According to Planck's Radiation Law, the energy density of blackbody radiation is:
u(ν) = (8 * π * h * ν³ / c³) / [ exp(hν / (kB * T)) - 1 ] ---- (Equation 2)
Comparing Equation (1) with Equation (2):
- The coefficients of stimulated processes must be equal:
B₁₂ / B₂₁ = 1 ==> B₁₂ = B₂₁
(Probability of stimulated absorption is equal to probability of stimulated emission.)
- The ratio of spontaneous to stimulated emission coefficients is:
A₂₁ / B₂₁ = (8 * π * h * ν³) / c³
Significance
- The ratio
A₂₁ / B₂₁is directly proportional toν³. Hence, as photon frequency increases (e.g., in X-rays or gamma rays), spontaneous emission dominates strongly, making it harder to establish laser action at very high frequencies. - The equality
B₁₂ = B₂₁provides the physical foundation for optical amplification and lasers.
Q.4 (a)
Question
A Silicon plate of thickness 1mm, breadth 10 mm and length 100 mm is placed in a magnetic field of 0.5 Wb/m2 acting along its thickness. If 0.01 A current flows along its length, calculate the Hall coefficient if developed Hall voltage is 1.83 mV.
Solution
1. Given Information
- Thickness of plate along the magnetic field direction (t) = 1 mm = 1 × 10⁻³ m
- Breadth of the plate (b) = 10 mm = 10 × 10⁻³ m = 0.01 m
- Length of the plate (L) = 100 mm = 100 × 10⁻³ m = 0.1 m
- Applied magnetic field (B) = 0.5 Wb/m² (Tesla)
- Current flowing through length (I) = 0.01 A
- Developed Hall voltage (VH) = 1.83 mV = 1.83 × 10⁻³ V
2. What is Required
- Hall coefficient (RH) of the Silicon plate.
3. Formula
The Hall voltage developed across a rectangular conductor or semiconductor is given by:
VH = (RH * I * B) / t
where t is the thickness of the sample parallel to the magnetic field vector B.
Rearranging to solve for the Hall coefficient RH:
RH = (VH * t) / (I * B)
4. Calculation Steps
- Substitute given values:
RH = (1.83 × 10⁻³ V × 1 × 10⁻³ m) / (0.01 A × 0.5 Wb/m²)
- Evaluate the numerator:
Numerator = 1.83 × 10⁻⁶ V·m
- Evaluate the denominator:
Denominator = 0.01 × 0.5 = 0.005 A·Wb/m² = 5 × 10⁻³
- Divide to find RH:
RH = (1.83 × 10⁻⁶) / (5 × 10⁻³)
RH = 0.366 × 10⁻³ m³/C
RH = 3.66 × 10⁻⁴ m³/C
5. Final Answer
The Hall coefficient of the Silicon plate is 3.66 × 10⁻⁴ m³/C.
Q.4 (b)
Question
Discuss in detail Four probe method used for resistivity measurements in thin sheet.
Solution
Four-Probe Method for Resistivity Measurement
The Four-Probe (or collinear Kelvin probe) method is an accurate experimental technique used to measure the electrical resistivity of semiconductors and thin conducting sheets without errors introduced by probe contact resistance.
---
Experimental Setup
Four-Probe Setup Diagram (schematic, not to scale):
Constant Current Source (I)
[ + ]------------[ - ]
| |
| Voltmeter(V) |
| [+] [-] |
| | | |
↓ ↓ ↓ ↓
Probe1 Probe2 Probe3 Probe4
| | | |
──●──────●────●────●──
| s s s |
| | <-- Semiconductor Sheet (Thickness = t)
----------------------
- The setup consists of four equally spaced, collinear, spring-loaded tungsten metal probes placed in contact with the flat surface of the thin sheet.
- The probe separation is designated as
s(typically 1 mm to 2 mm apart). - Outer Probes (1 and 4): Connected to a high-precision constant DC current source that injects current
I. - Inner Probes (2 and 3): Connected to a high-impedance digital voltmeter (DVM) to measure the potential drop
V.
---
Why Four Probes Instead of Two?
In a conventional two-probe measurement, the measured resistance includes contact resistance between the metal tips and the semiconductor surface, spreading resistance, and lead wire resistance. In the four-probe method:
- The current passes through the outer probes.
- The voltmeter draws virtually zero current due to its extremely high input impedance (~10¹² Ω).
- Consequently, no voltage drop occurs across the contact resistance at the inner probes, providing an accurate reading of the bulk material resistance.
---
Mathematical Formulation for Thin Sheets
For a thin conducting sheet where the thickness t is much smaller than the probe spacing s (i.e., t << s), the current expands radially in two dimensions:
- Sheet Resistance (Rs):
The sheet resistance is related to the measured voltageVand currentIby:
Rs = [ π / ln(2) ] * (V / I) ≈ 4.532 * (V / I)
- Bulk Resistivity (ρ):
Since bulk resistivity equals sheet resistance multiplied by thicknesst:
ρ = Rs * t = [ π / ln(2) ] * (V / I) * t
ρ = 4.532 * (V / I) * t (Ω·m or Ω·cm)
If the sample has finite lateral dimensions (boundaries close to the probes), a geometric correction factor F is included: ρ = 4.532 (V / I) t * F.
Q.4 (c)
Question
Define density of states and obtain expression for the density of states in terms of mass of electron(m), Planck’s constant (h) and Energy (E).
Solution
1. Definition of Density of States
The Density of States, denoted by g(E) or Z(E), is defined as the number of available quantum electron states per unit energy range per unit volume of the material at energy level E:
g(E) = (1 / V) * (dN / dE)
where dN is the number of available quantum states in the energy interval between E and E + dE, and V is the crystal volume.
---
2. Derivation of Expression for Density of States
- Electron in a Three-Dimensional Potential Well:
Consider a free electron enclosed in a cubical crystal of side length L (Volume V = L³). The energy of an electron in terms of quantum numbers (nx, ny, nz) is:
E = [h² / (8 * m * L²)] * (nx² + ny² + nz²)
where m is the mass of the electron and h is Planck's constant.
- Radius in Quantum Number Space:
Let n² = nx² + ny² + nz². Then:
E = [h² / (8 * m * L²)] * n²
Solving for n:
n² = (8 * m * L² * E) / h²
n = (2 * L / h) * sqrt(2 * m * E) ---- (Equation 1)
- Volume in n-space:
Each set of positive integers (nx, ny, nz ≥ 1) represents one distinct spatial orbital state.
- All states with energy less than or equal to
Elie within the positive octant (nx > 0, ny > 0, nz > 0) of a sphere of radiusnin quantum-number space. - The total number of spatial states within energy
Eis:
N_spatial = (1/8) * [ (4/3) * π * n³ ] = (π / 6) * n³
- Inclusion of Electron Spin:
According to the Pauli Exclusion Principle, each spatial state can hold two electrons of opposite spins (spin up and spin down):
N(E) = 2 * N_spatial = 2 * [ (π / 6) * n³ ] = (π / 3) * n³
- Substitute
nfrom Equation (1):
N(E) = (π / 3) * [ (2 * L / h) * sqrt(2 * m * E) ]³
N(E) = (π / 3) * [ 8 * L³ / h³ ] * (2 * m * E)^(3/2)
N(E) = [ 8 * π * V / (3 * h³) ] * (2 * m)^(3/2) * E^(3/2)
where V = L³ is the volume of the material.
- Differentiation with respect to Energy E:
The density of states per unit volume is g(E) = (1 / V) * (dN / dE):
dN/dE = [ 8 * π * V / (3 * h³) ] * (2 * m)^(3/2) * [ (3/2) * E^(1/2) ]
dN/dE = [ 4 * π * V / h³ ] * (2 * m)^(3/2) * E^(1/2)
Dividing by volume V:
g(E) = (1 / V) * (dN / dE) = [ 4 * π / h³ ] * (2 * m)^(3/2) * sqrt(E)
Alternatively, expanding (2 m)^(3/2) = 2 sqrt(2) * m^(3/2):
g(E) = [ 8 * sqrt(2) * π * m^(3/2) / h³ ] * E^(1/2)
Conclusion
The density of available energy states is proportional to the square root of energy: g(E) ∝ E^(1/2).
Q.4 (a) OR
Question
What is hot point probe method? Explain
Solution
Hot Point Probe Method
The hot point probe method (also known as the thermal probe method) is a simple, non-destructive experimental technique used to determine whether a semiconductor sample is n-type or p-type.
---
Experimental Setup and Working Principle
Hot Probe Setup (schematic, not to scale):
Hot Probe (Heated) Cold Probe (Room Temp)
(+) (-)
| |
↓ ↓
───●─────────────────────────────●───
| Semiconductor Sample |
─────────────────────────────────────
| |
+---[ μV ]-+ (Sensitive Microvoltmeter / Galvanometer)
- Setup:
- Two metal contact probes are placed on the surface of the semiconductor specimen.
- One probe is heated (using an internal resistance coil or soldering iron tip) to maintain a higher temperature (Hot Probe).
- The other probe remains at ambient room temperature (Cold Probe).
- A sensitive center-zero microvoltmeter or galvanometer is connected between the two probes.
- Operating Principle (Seebeck Effect):
- The localized temperature difference creates a thermal gradient across the sample.
- Charge carriers near the hot probe acquire higher thermal kinetic energy than those near the cold probe.
- Consequently, the majority charge carriers diffuse away from the hot probe toward the cold probe.
---
Identification of Carrier Type
- For n-type Semiconductors:
- Majority carriers are electrons (negative charge).
- Electrons diffuse from the hot probe to the cold probe.
- The cold probe accumulates negative charge (becomes negative potential), leaving the hot probe with a net positive potential.
- Therefore, the hot probe is positive (+) with respect to the cold probe.
- For p-type Semiconductors:
- Majority carriers are holes (positive charge).
- Holes diffuse from the hot probe toward the cold probe.
- The cold probe accumulates positive charge (becomes positive potential), leaving the hot probe with a net negative potential.
- Therefore, the hot probe is negative (-) with respect to the cold probe.
Q.4 (b) OR
Question
List the condition to be satisfied to use Vander Pauw technique and derive expression of bulk resistivity measurement using it.
Solution
Van der Pauw Technique
The Van der Pauw method is an experimental technique used to measure the resistivity and Hall coefficient of an arbitrary, flat, two-dimensional sample of uniform thickness.
---
1. Conditions to be Satisfied
To apply the Van der Pauw technique, the specimen must meet the following four conditions:
- The sample must have a flat shape of uniform thickness (t).
- The sample surface must be singly connected (it must contain no through-holes or isolated non-conducting islands).
- The four electrical contacts must be located at the exact periphery (edges) of the sample.
- The contact areas must be infinitesimally small compared to the total surface area of the sample.
---
2. Derivation of Resistivity Expression
Arbitrary Van der Pauw Sample (schematic, not to scale):
Contact 1 Contact 2
(●)───────────(●)
/ \
| Flat Sample |
| Thickness = t |
\ /
(●)───────────(●)
Contact 4 Contact 3
Let four small contacts placed consecutively along the perimeter be numbered 1, 2, 3, and 4.
- Characteristic Resistances:
- Inject a direct current
I₁₂through contacts 1 and 2, and measure the resulting voltageV₄₃across contacts 4 and 3. Define the resistance:
R_A = R₁₂,₄₃ = |V₄₃| / I₁₂
- Inject a direct current
I₂₃through contacts 2 and 3, and measure the resulting voltageV₁₄across contacts 1 and 4. Define the resistance:
R_B = R₂₃,₁₄ = |V₁₄| / I₂₃
- Van der Pauw Theorem:
Van der Pauw proved using conformal mapping that these two characteristic resistances satisfy the relation:
exp(-π * R_A * t / ρ) + exp(-π * R_B * t / ρ) = 1
where:
ρ= Bulk resistivity of the material (Ω·m)t= Thickness of the sample (m)
- Derivation for Symmetrical Geometry (RA ≈ RB = R):
When the contacts are symmetrically spaced (e.g., square or circular sample):
R_A = R_B = R
Substitute R into the theorem:
exp(-π * R * t / ρ) + exp(-π * R * t / ρ) = 1
2 * exp(-π * R * t / ρ) = 1
exp(-π * R * t / ρ) = 1 / 2
Take the natural logarithm of both sides:
-π * R * t / ρ = ln(1/2) = -ln(2)
Cancel negative signs:
π * R * t / ρ = ln(2)
Rearrange to solve for bulk resistivity ρ:
ρ = [ π * t / ln(2) ] * R
Since π / ln(2) ≈ 4.5323:
ρ = [ π * t / ln(2) ] * [ (R_A + R_B) / 2 ] * f
where f is a geometric correction factor that equals 1 when RA = RB.
Q.4 (c) OR
Question
What is Band gap? How can we use UV-VIS spectroscopy for band gap measurement for semiconductors explain in detail.
Solution
1. Definition of Band Gap
The band gap (Eg), or forbidden energy gap, is the difference in energy between the lowest point of the conduction band (conduction band edge, Ec) and the highest point of the valence band (valence band edge, Ev) in a solid material:
Eg = Ec - Ev
It represents the minimum energy required to excite a valence electron to become a free charge carrier in the conduction band.
---
2. UV-Vis Spectroscopy Principle
Ultraviolet-Visible (UV-Vis) absorption spectroscopy measures the fraction of light absorbed by a semiconductor sample as a function of incident photon wavelength λ (typically from 200 nm to 1100 nm).
- When the incident photon energy
hν < Eg, the photons lack sufficient energy to excite electrons across the gap; the material is largely transparent and absorbance is near zero. - When
hν ≥ Eg, fundamental absorption occurs as electrons transition from the valence band to the conduction band, causing a sharp rise in optical absorption (the absorption edge).
Absorption Edge and Tauc Plot (schematic, not to scale):
Absorbance (A) (α * hν)^(1/n)
^
| / | /
| / | / Extrapolated
| / | / linear fit
| / | /
|___/_______> λ |______/_____
λ_edge Eg --> Photon Energy hν (eV)
---
3. Step-by-Step Procedure to Determine Band Gap Using Tauc's Relation
- Data Acquisition:
Measure optical absorbanceA(or transmittanceT) as a function of wavelengthλusing a UV-Vis spectrophotometer.
- Calculate Absorption Coefficient (α):
Using the Beer-Lambert law, calculate the absorption coefficientα:
α = (2.303 * A) / d
where d is the sample thickness.
- Calculate Photon Energy (hν):
Convert the wavelengthλ(in nm) to photon energyhν(in eV):
hν (in eV) = 1240 / λ (in nm)
- Apply Tauc's Equation:
The relationship between the absorption coefficientαand photon energyhνnear the band edge is given by Tauc's relation:
(α * hν) = B * (hν - Eg)^n
where:
Bis an energy-independent proportionality constant.Egis the optical band gap energy.- The exponent
ndepends on the nature of the electronic transition: - n = 1/2 for a direct allowed transition (e.g., GaAs, CdTe, ZnO).
- n = 2 for an indirect allowed transition (e.g., Si, Ge).
- n = 3/2 for a direct forbidden transition.
- n = 3 for an indirect forbidden transition.
- Construct the Tauc Plot:
- For a direct band gap semiconductor, plot
(α * hν)²on the y-axis versushν(in eV) on the x-axis. - For an indirect band gap semiconductor, plot
(α * hν)^(1/2)on the y-axis versushνon the x-axis.
- Extrapolation to Find Eg:
Identify the linear portion of the curve at the onset of strong absorption. Draw a straight tangent line through this linear region and extrapolate it to intersect the x-axis (where y = 0):
(α * hν)^(1/n) = 0 ==> hν = Eg
The value of photon energy hν at the x-axis intercept yields the experimental optical band gap Eg.
Q.5 (a)
Question
Distinguish between Type-I and Type-II Superconductivity
Solution
Comparison between Type-I and Type-II Superconductors
| Parameter / Feature | Type-I Superconductors (Soft) | Type-II Superconductors (Hard) | | :--- | :--- | :--- | | Critical Fields | Exhibits only one critical magnetic field (Hc). | Exhibits two critical magnetic fields: Lower critical field (Hc1) and Upper critical field (Hc2). | | Meissner Effect | Exhibits a complete Meissner effect; diamagnetism is completely lost above Hc abruptly. | Exhibits complete Meissner effect only up to Hc1. Between Hc1 and Hc2, it enters a vortex / mixed state. | | Critical Field Strength | Critical field Hc is very low (typically < 0.1 Tesla). | Upper critical field Hc2 is very high (often > 10 to 100 Tesla). | | Transition Nature | Abrupt transition from superconducting state to normal state at Hc. | Gradual transition from pure superconducting state to mixed state at Hc1, then to normal state at Hc2. | | Flux Penetration | Magnetic flux does not penetrate the bulk material below Hc. | Magnetic flux lines penetrate the material as quantized flux vortices between Hc1 and Hc2. | | Material Examples | Pure elemental metals such as Lead (Pb), Mercury (Hg), Tin (Sn), Aluminum (Al). | Alloys and complex ceramics such as Nb-Ti, Nb₃Sn, YBa₂Cu₃O₇ (YBCO). | | Applications | Not suitable for high-field electromagnets due to low Hc. | Widely used for strong electromagnets (MRI machines, particle accelerators, Maglev). |
Q.5 (b)
Question
The Fermi level in given material is 2.8 eV. What are the energies for which the probabilities of occupancy at 270K are 0.89,0.20, 0.02 and 0.5
Solution
1. Given Information
- Fermi energy level (Ef) = 2.8 eV
- Absolute temperature (T) = 270 K
- Occupancy probabilities f(E) = 0.89, 0.20, 0.02, and 0.50
- Boltzmann constant (kB) = 8.6173 × 10⁻⁵ eV/K
2. What is Required
- Energy values E corresponding to each occupancy probability f(E).
3. Formula
The Fermi-Dirac distribution function is:
f(E) = 1 / [ 1 + exp((E - Ef) / (kB * T)) ]
Rearranging to express E explicitly:
1 + exp((E - Ef) / (kB * T)) = 1 / f(E)
exp((E - Ef) / (kB * T)) = [1 - f(E)] / f(E)
(E - Ef) / (kB * T) = ln( [1 - f(E)] / f(E) )
E = Ef + kB * T * ln( [1 - f(E)] / f(E) )
4. Calculation Steps
- Compute thermal energy term (kB * T):
kB * T = (8.6173 × 10⁻⁵ eV/K) × 270 K = 0.023267 eV
- Case 1: For f(E) = 0.50
[1 - 0.50] / 0.50 = 1.0
ln(1.0) = 0
E = 2.8 eV + 0.023267 × 0 = 2.800 eV
- Case 2: For f(E) = 0.89
[1 - 0.89] / 0.89 = 0.11 / 0.89 ≈ 0.123596
ln(0.123596) = -2.0907
E = 2.8 eV + (0.023267 eV × (-2.0907))
E = 2.8 - 0.04864 = 2.751 eV
- Case 3: For f(E) = 0.20
[1 - 0.20] / 0.20 = 0.80 / 0.20 = 4.0
ln(4.0) = 1.3863
E = 2.8 eV + (0.023267 eV × 1.3863)
E = 2.8 + 0.03225 = 2.832 eV
- Case 4: For f(E) = 0.02
[1 - 0.02] / 0.02 = 0.98 / 0.02 = 49.0
ln(49.0) = 3.8918
E = 2.8 eV + (0.023267 eV × 3.8918)
E = 2.8 + 0.09055 = 2.891 eV
5. Final Answer
The corresponding energies are:
- For f(E) = 0.50: E = 2.800 eV
- For f(E) = 0.89: E = 2.751 eV
- For f(E) = 0.20: E = 2.832 eV
- For f(E) = 0.02: E = 2.891 eV
Q.5 (c)
Question
(i) Discuss BCS theory of superconductivity in brief (ii)What is cryotron? Discus
Solution
(i) BCS Theory of Superconductivity
The microscopic theory of superconductivity was proposed by John Bardeen, Leon Cooper, and John Robert Schrieffer in 1957 (known as the BCS Theory).
Electron-Phonon Interaction (Cooper Pair Formation):
Electron 1 (k↑) Electron 2 (-k↓)
● ●
\ /
\ Lattice /
\ Distortion /
\+---+---+---+---+--/
| + | + | + | + |
(Virtual Phonon)
- Electron-Lattice-Electron Interaction: An electron moving through the positive ion lattice attracts nearby positive ions, creating a localized positive charge polarization (lattice deformation / virtual phonon emission).
- Cooper Pair Formation: A second electron with opposite momentum and opposite spin is attracted to this passing positive distortion before the lattice relaxes. This indirect, phonon-mediated attractive force overcomes the Coulomb repulsion between the two electrons at temperatures below Tc, binding them into a correlated pair known as a Cooper Pair (wave vectors
k↑and-k↓). - Bose-Einstein Condensation: While individual electrons are fermions (spin-1/2), a Cooper pair has zero total spin (spin = 0 or 1), behaving as a boson. All Cooper pairs condense into the same macroscopic ground quantum state.
- Zero Electrical Resistance: The condensed pairs require a finite minimum energy (the superconducting energy gap,
2Δ) to be broken apart. At temperatures below Tc, thermal lattice vibrations cannot break the pairs, allowing them to flow without scattering and resulting in zero electrical resistance.
---
(ii) What is a Cryotron? Discuss
A cryotron is a switching and memory device that operates at cryogenic temperatures, based on the principle that superconductivity can be destroyed by an applied magnetic field exceeding the critical magnetic field (Hc).
Schematic Construction of a Wire-wound Cryotron (not to scale):
Gate Current (Ig)
----------------->
===============================
| GATE WIRE (Tantalum, Ta) |
===============================
| | | | | | | | |
| | | | | | | | | <-- CONTROL COIL (Niobium, Nb)
|_|_|_|_|_|_|_|_|
----------------->
Control Current (Ic)
- Construction:
- Gate Wire: A straight wire made of a superconductor with a lower critical field (e.g., Tantalum, Ta, with Tc ≈ 4.4 K, lower Hc).
- Control Coil: A single-layer coil wrapped around the gate wire, made of a superconductor with a significantly higher critical field (e.g., Niobium, Nb, with Tc ≈ 9.2 K, higher Hc).
- The entire device is submerged in liquid helium (T ≈ 4.2 K).
- Working Mechanism:
- ON State (Superconducting / Zero Resistance): When no current flows through the control coil, the gate wire remains superconducting below its critical temperature. Current flows freely through the gate with zero voltage drop.
- OFF State (Resistive / Normal State): When a control current
Icis passed through the niobium coil, it establishes a magnetic field. Because the critical field of the gate wire (Ta) is small, this field easily exceeds Hc(gate), switching the gate wire into its normal resistive state. The control coil itself remains superconducting throughout because its critical field Hc(Nb) is much higher.
- Applications: Used as high-speed cryogenic electronic switches, logical gates, and flip-flops in low-temperature memory circuits.
Q.5 (a) OR
Question
Explain the Fermi’s Golden Rule in brief.
Solution
Fermi's Golden Rule
Fermi's Golden Rule is an essential formula in quantum mechanics derived using time-dependent perturbation theory. It calculates the transition probability per unit time (the transition rate) from an initial stationary state |i⟩ to a continuum of final eigenstates |f⟩ under the influence of a weak, time-harmonic or constant perturbation Hamiltonian H'.
---
Mathematical Expression
The transition probability per unit time, W_if, is given by:
W_if = (2 * π / ℏ) * |M_fi|² * ρ(Ef)
Where:
- W_if = Transition rate (transitions per second, s⁻¹).
- ℏ = h / (2 * π) = Reduced Planck's constant.
- M_fi = ⟨f| H' |i⟩ = Transition matrix element connecting initial state
|i⟩and final state|f⟩via the perturbation HamiltonianH'. It reflects the coupling strength between the two states. - ρ(Ef) = Density of final quantum states at the transition energy
Ef.
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Physical Significance and Applications
- Conservation of Energy: Transitions occur predominantly between states of essentially equal energy (or energy matched to absorbed/emitted photon energy:
Ef = Ei ± ℏω). - Dependence on Matrix Element: If
|M_fi| = 0, the transition is forbidden by quantum-mechanical selection rules. - Applications: Widely used in solid-state physics to calculate carrier scattering rates (electron-phonon, electron-impurity scattering), optical absorption and emission coefficients in semiconductors, and nuclear decay rates.
Q.5 (b) OR
Question
The critical magnetic field of a given superconductor is 9.5 x 10^3 A/m at 0 K. At what temperature does the critical magnetic field of the sample drop to 5 x 10^3 A/m, given that the critical temperature of the sample is 7.2 K? Also, calculate the critical current density at that temperature if the radius of the wire is 0.4 mm.
Solution
1. Given Information
- Critical magnetic field at 0 K:
Hc(0) = 9.5 × 10³ A/m - Critical temperature:
Tc = 7.2 K - Critical magnetic field at temperature T:
Hc(T) = 5.0 × 10³ A/m - Radius of the superconducting wire:
r = 0.4 mm = 0.4 × 10⁻³ m = 4.0 × 10⁻⁴ m
2. What is Required
- Temperature
Tat which the critical magnetic field is5.0 × 10³ A/m. - Critical current density
Jcat that temperature.
3. Formulas
- Temperature Dependence of Critical Magnetic Field:
Hc(T) = Hc(0) * [ 1 - (T / Tc)² ]
- Silsbee's Rule (Critical Current for a Cylindrical Wire):
Ic = 2 * π * r * Hc(T)
- Critical Current Density (Jc):
Jc = Ic / A = (2 * π * r * Hc(T)) / (π * r²) = (2 * Hc(T)) / r
4. Calculation Steps
Step A: Determine the Temperature T
- Substitute values into the critical field formula:
5.0 × 10³ = 9.5 × 10³ * [ 1 - (T / 7.2)² ]
- Divide both sides by
9.5 × 10³:
1 - (T / 7.2)² = (5.0 × 10³) / (9.5 × 10³) = 5 / 9.5 ≈ 0.52632
- Solve for
(T / 7.2)²:
(T / 7.2)² = 1 - 0.52632 = 0.47368
- Take the square root:
T / 7.2 = sqrt(0.47368) ≈ 0.68824
- Calculate
T:
T = 7.2 × 0.68824 = 4.955 K ≈ 4.96 K
Step B: Determine the Critical Current Density Jc
- Cross-sectional area of wire
A = π * r². - Using the direct formula for
Jc:
Jc = (2 * Hc(T)) / r
- Substitute given values:
Jc = [ 2 × (5.0 × 10³ A/m) ] / (4.0 × 10⁻⁴ m)
Jc = (1.0 × 10⁴) / (4.0 × 10⁻⁴)
Jc = 0.25 × 10⁸ A/m² = 2.5 × 10⁷ A/m²
(For completeness, the total critical current is: Ic = 2 π (4.0 × 10⁻⁴ m) (5.0 × 10³ A/m) = 4 π ≈ 12.57 A
5. Final Answer
- The temperature at which the field drops to 5 × 10³ A/m is 4.96 K.
- The critical current density at this temperature is 2.5 × 10⁷ A/m² (or 25 A/mm²).
Q.5 (c) OR
Question
List various properties of Superconductors and explain them in detail.
Solution
Properties of Superconductors
Superconductivity is a distinct quantum state of matter discovered by Heike Kamerlingh Onnes in 1911. Superconductors exhibit several unique electrical, magnetic, thermal, and structural properties below a critical temperature (Tc).
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1. Zero Electrical Resistance (Infinite Conductivity)
- Below the critical transition temperature
Tc, the direct current (DC) electrical resistance of a superconductor abruptly drops to absolute zero (R = 0,σ → ∞). - An electric current induced in a closed superconducting loop will persist indefinitely without decay as long as the temperature is kept below Tc; such currents are known as persistent currents.
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2. Meissner Effect (Perfect Diamagnetism)
- When a superconducting material is cooled below its critical temperature in the presence of an external magnetic field, it completely expels all magnetic flux from its interior.
- Inside the bulk superconductor, the magnetic induction is zero: B = 0.
- Since
B = μ₀ * (H + M) = 0, it follows that the magnetization isM = -H. - The magnetic susceptibility is exactly: χ = M / H = -1.
- This demonstrates that superconductivity is not merely infinite conductivity, but a distinct thermodynamic state.
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3. Critical Magnetic Field (Hc)
- Superconductivity cannot persist in arbitrarily high magnetic fields. If the applied magnetic field exceeds a threshold value known as the critical magnetic field (Hc), the superconducting state is destroyed and the material reverts to its normal state.
- The critical field follows a parabolic temperature dependence:
Hc(T) = Hc(0) * [ 1 - (T / Tc)² ]
where Hc(0) is the critical field at absolute zero (0 K).
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4. Critical Current and Critical Current Density (Silsbee's Effect)
- When an electric current flows through a superconductor, it produces its own magnetic field around the surface.
- According to Silsbee's rule, when this self-induced magnetic field reaches
Hc(T), superconductivity is destroyed. The maximum current that can be carried without resistance is the critical current (Ic):
Ic = 2 * π * r * Hc
for a round wire of radius r.
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5. Energy Gap (Eg = 2Δ)
- The electronic excitation spectrum of a superconductor contains an energy gap
Eg = 2Δcentered at the Fermi level. - This gap corresponds to the binding energy required to break a Cooper pair into two single quasiparticle electrons. At 0 K, BCS theory predicts:
Eg(0) = 2 * Δ(0) ≈ 3.52 * kB * Tc
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6. Isotope Effect
- For different isotopes of the same superconducting element, the critical temperature
Tcvaries inversely with the square root of the isotopic massM:
Tc * M^α = constant
where the isotope coefficient α ≈ 0.5 for many simple metals.
- This experimental discovery confirmed that superconductivity involves interactions between electrons and lattice vibrations (phonons).
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7. Discontinuity in Specific Heat
- At the critical transition temperature
Tcin zero magnetic field, the electronic specific heat exhibits a sharp, finite jump (discontinuity) without any latent heat, characteristic of a second-order thermodynamic phase transition. - Below
Tc, the electronic specific heat decreases exponentially with temperature (Ce ∝ exp(-Δ / (kB * T))), confirming the presence of the energy gap.
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