Condensed Matter Physics MCQ Quiz - Objective Question with Answer for Condensed Matter Physics - Download Free PDF

 Explanation:

If the potential is periodic with lattice translations 𝑅, i.e., V(𝑉) = V(𝑉 + 𝑅), then the wavefunction has the form:

ψ_k(𝑉) = e^i_k·𝑉 u_k(𝑉)

where u_k(𝑉) is a function with the same periodicity as the lattice:

u_k(𝑉) = u_k(𝑉 + 𝑅)

1. |ψ_k(𝑉)| is constant
   Incorrect – because u_k(𝑉) varies with position, so the modulus of ψ_k(𝑉) is not constant.

2. ψ_k(𝑉) is also an eigenstate of the momentum operator
   Incorrect – ψ_k is not an eigenstate of the momentum operator due to the periodic modulation u_k(𝑉). Only a plane wave e^i_k·𝑉 would be.

3. ψ_k(𝑉) = ψ_k(𝑉 + 𝑅)
   Incorrect – this is not generally true. However, Bloch’s theorem implies:

   ψ_k(𝑉 + 𝑅) = e^i_k·𝑅 ψ_k(𝑉)

4. |ψ_k(𝑉)| = |ψ_k(𝑉 + 𝑅)|
   Correct – using the property above:

   |ψ_k(𝑉 + 𝑅)| = |e^i_k·𝑅 ψ_k(𝑉)| = |ψ_k(𝑉)| 

Solution:

Type-I superconductors are those which show perfect Meissner effect, i.e., no magnetisation above H_c (only one critical field).

Type-II superconductors are those which do not show perfect Meissner effect, but they exhibit a mixed state/vortex state between the lower critical magnetic field and the upper critical magnetic field.

Satisfied by option-3 exactly.

Given:

The dispersion relation for a one-dimensional monatomic lattice chain is given by the equation:

\( \omega = \frac{2}{a} v \left| \sin\left( \frac{k a}{2} \right) \right| \)

Where: - a is the interatomic spacing, \(- k = \frac{2 \pi}{\lambda}\) is the wave vector, and - v has the dimension of velocity.

The relation between the phase velocity \(v_p\) and the group velocity \(v_g\) in the long wavelength limit is given.

Concept:

• The phase velocity \(v_p\) is defined as \(v_p = \frac{\omega}{k}\) .
• The group velocity \(v_g\) is the derivative of\( \omega\) with respect to k , i.e., \(v_g = \frac{d\omega}{dk}\) .
• In the long wavelength limit, where \(k \to 0 \) , the phase velocity and group velocity become equal for the given dispersion relation.

Calculation:

From the given dispersion relation:

\( \omega = \frac{2}{a} v \left| \sin\left( \frac{k a}{2} \right) \right| \)

For small k (long wavelength limit), \(\sin\left( \frac{k a}{2} \right) \approx \frac{k a}{2}\) , so the dispersion relation simplifies to:

\( \omega = v k \)

Therefore, the phase velocity is:

\( v_p = \frac{\omega}{k} = v \)

The group velocity is the derivative of \omega with respect to k :

\( v_g = \frac{d\omega}{dk} = v \)

∴ In the long wavelength limit, \(v_p = v_g\) .

∴ The correct answer is option 1: \(v_p = v_g\) .

Assumptions:

• The diode voltage remains approximately constant at 0.7 V , as it is close to the 1 mA operating point.

Voltage across the resistor:

Total supply voltage: 1.0 V

Diode voltage: 0.7 V

Voltage across the resistor ( V_R ): \( V_R = 1.0 \, \text{V} - 0.7 \, \text{V} = 0.3 \, \text{V} \)

Current through the resistor (and the diode):

Using Ohm's Law: \( I = \frac{V_R}{R} = \frac{0.3 \, \text{V}}{200 \, \Omega} = 0.0015 \, \text{A} \)

Current: I = 1.5 mA

The estimated diode current is approximately 1.5 mA.

The correct option is 4).

Derivation of Forward Current Equation:

The total current in a p⁺-n diode is:

\( I = I_p + I_n \)

In a p⁺-n diode, the hole current I_p dominates. The expression for I_p is:

\( I_p = A \cdot q \cdot n_i^2 \cdot \left(\frac{D_p}{L_p N_D}\right) \cdot \left(e^{V / V_T} - 1\right) \)

Here:

• q : Electronic charge
• D_p : Diffusion coefficient of holes
• \(L_p = \sqrt{D_p \cdot \tau_p}\) : Hole diffusion length
• \(V_T = k_B T / q\) : Thermal voltage

Simplifying, we get:

\( I = I_s \cdot \left(e^{V / V_T} - 1\right) \)

Where \(I_s\) , the reverse saturation current, is:

\( I_s = A \cdot q \cdot n_i^2 \cdot \left(\frac{D_p}{L_p N_D}\right) \)

Using the given parameters:

\( I_s = 0.72 \times 10^{-15} \, \text{A} \)

To calculate V when I = 0.2 mA :

\( V = V_T \cdot \ln\left(\frac{I}{I_s} + 1\right) = 0.6798 \, \text{V} \)

Excess Minority-Carrier Charge ( Q_p ) and Diffusion Capacitance ( C_d ):

\( Q_p = \tau_p \cdot I_p = 2 \times 10^{-11} \, \text{C} = 20 \, \text{pC} \)

\( C_d = \frac{dQ_p}{dV} \approx \frac{\tau_p \cdot I}{V_T} = 0.7752 \, \text{nF} \)

The correct Option is 2).

CONCEPT:

The relation between entropy S and themodynamic probability Ω is 

​→ S = k ln Ω 

Frenkel defect: A Frenkel defect is a type of point defect in crystalline solids. The defect forms when an atom or smaller ion leaves its place in the lattice, creating a vacancy and becomes an interstitial by lodging in a nearby location.

Thermodynamic probability of the Frenkel defects is 

\( Ω = \frac{N!}{(N-n)! n!}\frac{N'!}{(N'-n)! n!}\)

Where, N, N' and n are total number of lattice and interstitial sites and number of defects respectively.

change in free energy in creating 'n' Frenkel defects 

\( Δ G = nE - TΔ S\)

EXPLANATION:

The probability of such frenkel defects 

\( Ω = \frac{N!}{(N-n)! n!}\frac{N'!}{(N'-n)! n!}\)

So, the change in entropy 

⇒ ΔS = k ln Ω = \( k \ln \left( \frac{N!}{(N-n)! n!}\frac{N'!}{(N'-n)! n!} \right)\)

⇒ ΔS = k ln [ N ln N + N' ln N' - (N - n) - (N' - n)ln(N' - n) - 2n ln n ]

[ Here we have used stirling's approximation i.e ln N! = N ln N - N ]

Now, change in free energy in creating n Frenkel defects 

\(\begin{aligned} & ⇒ \Delta G=n \phi-T \Delta s \\ & \Rightarrow \Delta G=n \phi-T\left\{k \ln \left[N \ln N+N^{\prime} \ln N^{\prime}-(N-n)-\left(N^{\prime}-n\right) \ln \left(N^{\prime}-n\right)-2 n \ln n\right]\right\} \\ & \text { Since, } \frac{\partial(\Delta G)}{\partial n}=0 \\ & \Rightarrow \frac{\partial(\Delta G)}{\partial n}=\phi-k T[0+0+\ln (N-n)+1-2 \ln n-2] \\ &\Rightarrow \phi-k T \ln \left[\frac{N-n\left(N^{\prime}-n\right)}{n^2}\right]=0 \end{aligned}\)

Since, n <<< N, N'

\(⇒ N-n \cong N \\ ⇒ N^{\prime}-n \cong N^{\prime}\)

\(\begin{aligned} & \therefore \phi-k T \ln \left(\frac{N N^{\prime}}{n^2}\right)=0\\ & \Rightarrow \phi-k T \ln \left[\frac{\sqrt{N N^{\prime}}}{n}\right]^2=0 \\ & \Rightarrow \ln \left[\frac{\sqrt{N N^{\prime}}}{n}\right]=\frac{\phi}{2 k T} \\ & \Rightarrow n=\sqrt{N N^{\prime}} e^{\frac{-\phi}{2 k T}} \end{aligned}\)

Hence the correct answer is option 1.

Concept- 

We are using the dispersion relation here which is given by

•  E_k =C√k here k is wave number
• At low temperatures for bosons, E  ∝ ks  
• At low-temperature entropy at constant volume is given by Cv ∝ Td/s  
• s =power of wavenumber in energy dispersion relation and d =dimension

Explanation:

• Ek =C√k
• At low temperatures for bosons, E  ∝ k^s 
• Equate this with given equation s=1/2
• At low-temperature C_v ∝ T^d/s
• ^{\(C_v = \frac{d \langle E \rangle}{dT} \propto \frac{d}{dT} (kT)^5 \propto T^4 \)}
• Here d is dimension = 2 (Given in the question)
• \(Cv ∝ T^{\frac{2} {1/2} } \)   
• C_v∝ T⁴

So, the correct answer is option (1).

CONCEPT:

Josephson junction: Two superconductors separated by a very thin strip of an insulator forms a Josephson junction.

A.C. Josephson Effect: when a potential difference V is applied between the two sides of the junction, there will be an oscillation of the tunneling current with angular frequency ω = \(\frac{2eV}{h} \). This is called the A.C. Josephson Effect.

EXPLANATION:

Current density through thin insulating layer is

\(J = J_0 \sin \left[ \delta(0) -\frac{2e\Delta v}{h}t \right] = J_0\sin \left[ \delta(0) - ω t \right] \)

∴ The angular frequency of the supercurrent is 

⇒ \(ω = \frac{2e\Delta v}{h} \)

Hence the correct answer is option 1.

Explanation:

\(ϵ(k) = ℏv_Fk\)

• This is the dispersion relation for electrons where \(v_F\) is the speed of particles near the Fermi level in an electron gas. This is actually the case with graphene or a very similar system where we are using a linear dispersion relation.
• The total number of particles (or electrons in this case) with given k up to Fermi wavenumber \(k_F\) in three-dimension can be represented as \(n = ∫d³k = V{(4π/3)}{(k_F^3)}.\)
• Since \( k_F = \frac{ϵ_F}{(ℏv_F)}\) from the original dispersion relation: \(n ∝ (\frac{ϵ_F} {ℏv_F})^3.\)Thus, we arrive at \(ϵ_F ∝ n^{(1/3)}\) which indicates that \(α = \frac13.\) 

Concept:

The hopping probability is exponential in the height only of the (free) energy barrier between sites.

Calculation:

i> ∑ p_ir_i

= 0.3i - 0.2i + 0.2j - 0.3j

= 0.1i - 0.1j

For N steps, = \({N \over 10}\)[i - j]

The correct answer is option (2).

Explanation:

• The two lattices A and B are essentially shifted versions of each other. A can be thought of as the checkerboard pattern where we select all the black squares (those that represent odd sums of indices), whereas B is the lattice that consists of all the white squares (those representing the even sums of indices).
• In a three-dimensional setting, if you translate (shift) lattice A by one unit in any direction (along the x, y or z axis), you will get the lattice points of lattice B.
• Similarly, if you translate lattice B by one unit you will get lattice A. So we can say that lattices A and B are in a relationship called "dual" or "face-centered" where one can be obtained by shifting the other by one unit. This concept is often used in crystal structures where the atoms reside on the vertices of such a lattice point.

CONCEPT:

Hall coefficient: the quotient of the potential difference per unit width of a metal strip in the Hall effect divided by the product of the magnetic intensity and the longitudinal current density

The hall coefficient for a semiconductor having both types of carriers is given as

⇒ RH = \(\frac{{p\mu _p^2 - n\mu _n^2}}{{\left| e \right|{{\left( {p{\mu _p} + n{\mu _n}} \right)}^2}}}\)

When the temperature is low p (the no. of holes) is greater than n (no. of electrons)

with the increase in temperature, the no. of holes decreases, and the no. of electrons increases. 

Case I: At low temperature: p >> n, μ_p < μ_n

⇒ pμp² > nμn2 

⇒ pμp2 - nμn2 > 0 ⇒ R_H = positive.

Case II: At moderate temperature: p > n

⇒ pμp2 ≈ nμn2 (since, μp < μn)

⇒ R_H > 0

Case III: At high temperature: p/n ≈ 1

⇒ pμp2 - nμn2 < 0

⇒ RH < 0

Only, option (4) matches with the above conclusions. 

Hence the correct answer is option 4.

Explanation:

1. Fermi Level Position:

\( f(E) \approx \exp\left(-\frac{E - E_F}{k_B T}\right) \)

\( E_F = \frac{\epsilon_C + \epsilon_V}{2} \)

• For an intrinsic semiconductor, the number of electrons in the conduction band ( n ) equals the number of holes in the valence band ( p ).
• Under the condition \( E - E_F \gg k_B T\) , the Fermi Dirac distribution simplifies to:
• Equating n and p leads to the Fermi level ( \(E_F\) ) being located at:
• Here, \(\epsilon_C\) is the bottom of the conduction band, and \(\epsilon_V\) is the top of the valence band.

The Fermi level is therefore at the midpoint of the bandgap.

2. Fermi Distribution Function:

\( f(E) = \frac{1}{\exp\left(\frac{E - E_F}{k_B T}\right) + 1} \)

• The Fermi Dirac distribution function ( f(E) ) describes the probability of an electron occupying an energy state E :
• Assuming N(E) already includes the spin degeneracy factor of 2, this function is directly applicable.

3. Density of Conduction Band Electrons:

\( n \approx n_0 k_B T \exp\left(-\frac{\epsilon_C - E_F}{k_B T}\right) \)

\( n \approx 2 \times 10^{21} \cdot \frac{1}{40} \cdot \exp\left(-\frac{0.75}{0.025}\right) \)

\( n \approx 4.68 \times 10^6 \, \text{cm}^{-3} \)

The correct option is 3).

Explanation:

• For a paramagnetic system at thermal equilibrium, the average magnetic moment 〈M〉 can be derived using the Boltzmann distribution. The Boltzmann distribution tells us that the probability of a system being in a particular state is proportional to the exponential of the negative energy of that state divided by the product of Boltzmann's constant (k_B) and the temperature (T).
• The magnetic dipole moment μ of an atom or ion in the presence of a magnetic field B has energy -μB. If the magnetic moment per ion (μ±) can take on values ±μB, then the energies of the two states are \( E± = ∓μ_BB\).
• The probability of being in state with energy E± is given by the Boltzmann factor: \( P_{± }= e^{(-E_{±/}kBT)} = e^{\frac{∓μ_BB}{k_BT}}\)
•  We then normalize by dividing by the sum of the probabilities of all possible states (the partition function Z): \(P± = \frac{e^{∓\frac{μ_BB}{k_BT}}} { [e^{\frac{μ_BB}{k_BT}} + e^{\frac{-μ_BB}{k_BT}}]}\)
• This gives us: \(P_- = \frac{e^{\frac{μ_BB}{k_BT}}} { [e^{\frac{μ_BB}{k_BT}} + e^{\frac{-μ_BB}{k_BT}}]}\)
• \(P_+ = \frac{e^{-\frac{μ_BB}{k_BT}}} { [e^{\frac{μ_BB}{k_BT}} + e^{\frac{-μ_BB}{k_BT}}]}\)
• The average magnetic moment 〈M〉 is the sum of each possible magnetic moment multiplied by its Boltzmann-weighted probability:

\(⟨M⟩ = μ_B × [P_+ - P_-]\)

\(⟨M⟩ = μ_B × [ \frac{e^{\frac{μ_BB}{k_BT}}} { [e^{\frac{μ_BB}{k_BT}} + e^{\frac{-μ_BB}{k_BT}}]}- \frac{e^{-\frac{μ_BB}{k_BT}}} { [e^{\frac{μ_BB}{k_BT}} + e^{\frac{-μ_BB}{k_BT}}]}]\)

Simplify to get: \(⟨M⟩ = μ_B \times \tanh(\frac{μ_BB}{k_BT})\)

Now, \(⟨M⟩ = μ_B \times \tanh({\frac{μ_BB}{k_BT}}) \)

\(⟨M⟩ = μ_B × \left( \frac{1-e^{-2\frac{\mu_BB}{ k_B T}}} { 1+e^{-2\frac{\mu_BB}{ k_B T}}}\right)\)

This has following behavior 

as T → 0 ,  → max ,

as T → ∞ ,   → 0,

The curve will be an inverse of hyperbolic tangent.

So, the option (C) curve is the solution.

CONCEPT:

The relation between entropy S and themodynamic probability Ω is 

​→ S = k ln Ω 

Frenkel defect: A Frenkel defect is a type of point defect in crystalline solids. The defect forms when an atom or smaller ion leaves its place in the lattice, creating a vacancy and becomes an interstitial by lodging in a nearby location.

Thermodynamic probability of the Frenkel defects is 

\( Ω = \frac{N!}{(N-n)! n!}\frac{N'!}{(N'-n)! n!}\)

Where, N, N' and n are total number of lattice and interstitial sites and number of defects respectively.

change in free energy in creating 'n' Frenkel defects 

\( Δ G = nE - TΔ S\)

EXPLANATION:

The probability of such frenkel defects 

\( Ω = \frac{N!}{(N-n)! n!}\frac{N'!}{(N'-n)! n!}\)

So, the change in entropy 

⇒ ΔS = k ln Ω = \( k \ln \left( \frac{N!}{(N-n)! n!}\frac{N'!}{(N'-n)! n!} \right)\)

⇒ ΔS = k ln [ N ln N + N' ln N' - (N - n) - (N' - n)ln(N' - n) - 2n ln n ]

[ Here we have used stirling's approximation i.e ln N! = N ln N - N ]

Now, change in free energy in creating n Frenkel defects 

\(\begin{aligned} & ⇒ \Delta G=n \phi-T \Delta s \\ & \Rightarrow \Delta G=n \phi-T\left\{k \ln \left[N \ln N+N^{\prime} \ln N^{\prime}-(N-n)-\left(N^{\prime}-n\right) \ln \left(N^{\prime}-n\right)-2 n \ln n\right]\right\} \\ & \text { Since, } \frac{\partial(\Delta G)}{\partial n}=0 \\ & \Rightarrow \frac{\partial(\Delta G)}{\partial n}=\phi-k T[0+0+\ln (N-n)+1-2 \ln n-2] \\ &\Rightarrow \phi-k T \ln \left[\frac{N-n\left(N^{\prime}-n\right)}{n^2}\right]=0 \end{aligned}\)

Since, n <<< N, N'

\(⇒ N-n \cong N \\ ⇒ N^{\prime}-n \cong N^{\prime}\)

\(\begin{aligned} & \therefore \phi-k T \ln \left(\frac{N N^{\prime}}{n^2}\right)=0\\ & \Rightarrow \phi-k T \ln \left[\frac{\sqrt{N N^{\prime}}}{n}\right]^2=0 \\ & \Rightarrow \ln \left[\frac{\sqrt{N N^{\prime}}}{n}\right]=\frac{\phi}{2 k T} \\ & \Rightarrow n=\sqrt{N N^{\prime}} e^{\frac{-\phi}{2 k T}} \end{aligned}\)

Hence the correct answer is option 1.