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

Calculation:

The number of states in the momentum interval p to p + dp is given by:

8πV / (h3) p2 dp.

From energy ε = cp, the number of states in the energy interval ε to ε + dε is:

N(ε) dε = 8πV / (c3 h3) ε2 dε.

The total energy is found by integrating over the energy spectrum:

E = 8πV / (c3 h3) ∫0∞ ε3 / (eβ(E-μ) + 1) dε.

The equation of state is then derived from the thermodynamic potential Ξ and is given by pV = E/3.

Final Answer: The equation of state pV = E/3 is derived from the thermodynamic potential Ξ.

The possible arrangements of spins for atoms are

Therefore, partition function is \(\rm Q=2+e^{-\beta g \mu_{B} B}+e^{\beta g \mu_{B} B}\) ...(i)

Now, there are two arrangements (microstates) where two atoms have same spin. Corresponding probabilities are

\(\rm P\left(+\frac{1}{2},+\frac{1}{2}\right)=\frac{e^{\beta g \mu_{B} B}}{Q}\) and \(\rm P\left(-\frac{1}{2},-\frac{1}{2}\right)=\frac{e^{-\beta g \mu_{B} B}}{Q}\)

∴ Net probability of two atoms having same spin is \(\rm P=\frac{e^{\beta g \mu_{B} B}+e^{-\beta g \mu_{B} B}}{2+e^{-\beta g \mu_{B} B}+e^{\beta g \mu_{B} B}}\)

Calculation:

For 80% of ions to be in the lowest energy state:

Let the ratio of ions in lowest energy state to higher energy state = 80:20.

⇒ Probability ratio = 0.8 / 0.2 = 4.

From Boltzmann distribution:

⇒ e^{(μ_B × B₁) / kT} = 4

⇒ (μ_B × B₁) / kT = ln(4)

For 60% of ions to be in the lowest energy state:

Let the ratio of ions in lowest energy state to higher energy state = 60:40.

⇒ Probability ratio = 0.6 / 0.4 = 1.5.

⇒ e^{(μ_B × B₂) / kT} = 1.5

⇒ (μ_B × B₂) / kT = ln(1.5)

Ratio of magnetic fields:

⇒ B₁ / B₂ = ln(4) / ln(1.5)

Using ln(4) = 2 × ln(2):

⇒ B₁ / B₂ = (2 × ln(2)) / ln(1.5)

Calculation:

From the relationship:

(E / E_F)^3/2 = 1 / 2

Taking the cube root on both sides:

⇒ (E / E_F) = (1 / 2)^2/3

Therefore,

E = (1 / 2)^2/3 × E_F

From the given energy formula:

E = 2ⁿ × E_F

Equating both expressions for E:

2ⁿ × E_F = (1 / 2)^2/3 × E_F

2ⁿ = (1 / 2)^2/3

⇒ 2ⁿ = 2^-2/3

n = -2/3

Concept Used:

The Bravais lattice type can be determined by analyzing the primitive vectors.

The volume of the primitive cell is given by the scalar triple product of the primitive vectors:
V_cell = |a₁ • (a₂ × a₃)|.

The given basis and primitive vectors align with the structure of Body-Centered Cubic (BCC) or Face-Centered Cubic (FCC) lattices depending on the arrangement.

Calculation:

Let the primitive vectors be:

a₁ = a î

a₂ = a ĵ

a₃ = (a / 2) (î + ĵ + k̂)

The volume of the primitive cell:

⇒ V_cell = |a₁ • (a₂ × a₃)|

⇒ a₁ • (a₂ × a₃) = a î • [(a / 2) (k̂ - î)]

⇒ a₁ • (a₂ × a₃) = a × (a / 2) × (-1)

⇒ a₁ • (a₂ × a₃) = -(a³ / 2)

⇒ V_cell = |-(a³ / 2)| = a³ / 2

∴ The Bravais lattice is Body-Centered Cubic (BCC), and the volume of the primitive cell is a³ / 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:

• 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.

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).

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.