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

Last updated on Jun 28, 2025

Latest Condensed Matter Physics MCQ Objective Questions

Condensed Matter Physics Question 1:

Consider N mutually non-interacting electrons moving in a crystal where the ionic potential seen by an electron satisfies the condition  , is one of the lattice translation vectors. The energy eigenstates of the electrons are labelled as where 𝑘⃗ is a vector in the first Brillouin zone. Which of the following is true? 

  1.  is constant
  2.   is also an eigenstate of the momentum operator.

Answer (Detailed Solution Below)

Option 4 :

Condensed Matter Physics Question 1 Detailed Solution

 Explanation:

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

ψk(𝑉) = eik·𝑉 uk(𝑉)

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

uk(𝑉) = uk(𝑉 + 𝑅)

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

  2. ψk(𝑉) is also an eigenstate of the momentum operator
    Incorrect – ψk is not an eigenstate of the momentum operator due to the periodic modulation uk(𝑉). Only a plane wave eik·𝑉 would be.

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

    ψk(𝑉 + 𝑅) = eik·𝑅 ψk(𝑉)

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

    k(𝑉 + 𝑅)| = |eik·𝑅 ψk(𝑉)| = |ψk(𝑉)| 

Condensed Matter Physics Question 2:

Magnetization 𝑀 as a function of applied magnetic field 𝐻 for two different solid samples at temperature 𝑇 are shown below. These samples are known to be superconducting below their respective critical temperatures (𝑇c). 

The correct set of statements is 

  1. Fig. 1: Type I superconductor above 𝑇C;
    Fig. 2: Type II superconductor below 𝑇C and upto upper critical field; 
  2. Fig. 1: Type II superconductor below 𝑇c and upto upper critical field;
    Fig. 2: Type II superconductor below 𝑇c and upto lower critical field. 
  3. Fig. 1: Type I superconductor below 𝑇c and below critical field;
    Fig. 2: Type II superconductor below 𝑇c upto upper critical field; 
  4. Fig. 1: Type I superconductor below 𝑇c and below critical field;
    Fig. 2: Type II superconductor below 𝑇c and below lower critical field.

Answer (Detailed Solution Below)

Option 3 : Fig. 1: Type I superconductor below 𝑇c and below critical field;
Fig. 2: Type II superconductor below 𝑇c upto upper critical field; 

Condensed Matter Physics Question 2 Detailed Solution

Solution:

Type-I superconductors are those which show perfect Meissner effect, i.e., no magnetisation above Hc (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.

Condensed Matter Physics Question 3:

The dispersion relation for a one-dimensional monoatomic lattice chain is given by the equation,  where, ‘ a ’ is the interatomic spacing  and v, has the dimension of velocity. The relation between the phase velocity VP and group velocity Vg in the long wavelength limit is given by

  1. Vp = Vg
  2. Vp = 2Vg
  3. Vp = Vg/2
  4. Vp ≠ Vg

Answer (Detailed Solution Below)

Option 1 : Vp = Vg

Condensed Matter Physics Question 3 Detailed Solution

Given:

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

Where: - a is the interatomic spacing, is the wave vector, and - v has the dimension of velocity.

The relation between the phase velocity and the group velocity in the long wavelength limit is given.

Concept:

  • The phase velocity is defined as .
  • The group velocity is the derivative of with respect to k , i.e., .
  • In the long wavelength limit, where , the phase velocity and group velocity become equal for the given dispersion relation.

Calculation:

From the given dispersion relation:

For small k (long wavelength limit), , so the dispersion relation simplifies to:

Therefore, the phase velocity is:

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

∴ In the long wavelength limit, .

∴ The correct answer is option 1: .

Condensed Matter Physics Question 4:

  1. 0.5 mA
  2. 1.0 mA
  3. 8.5 mA
  4. 1.5 mA

Answer (Detailed Solution Below)

Option 4 : 1.5 mA

Condensed Matter Physics Question 4 Detailed Solution

Explanation:

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 ( VR ):

Current through the resistor (and the diode):

Using Ohm's Law:

Current: I = 1.5 mA

The estimated diode current is approximately 1.5 mA.

The correct option is 4).

Condensed Matter Physics Question 5:

Answer (Detailed Solution Below)

Option 2 :

Condensed Matter Physics Question 5 Detailed Solution

Top Condensed Matter Physics MCQ Objective Questions

For a crystal, let ϕ denote the energy required to create a pair of vacancy and interstitial defects. If n pairs of such defects are formed, and n N, N', where N and N' are, respectively, the total number of lattice and interstitial sites, then n is approximately

Answer (Detailed Solution Below)

Option 1 :

Condensed Matter Physics Question 6 Detailed Solution

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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 

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 

EXPLANATION:

The probability of such frenkel defects 

So, the change in entropy 

⇒ ΔS = k ln Ω = 

⇒ 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 

Since, n NN'

Hence the correct answer is option 1.

The dispersion relation of a gas of non-interacting bosons in two dimensions is E(k) = , where c is a positive constant. At low temperatures, the leading dependence of the specific heat on temperature T, is

  1. T4
  2. T3
  3. T2
  4. T3/2

Answer (Detailed Solution Below)

Option 1 : T4

Condensed Matter Physics Question 7 Detailed Solution

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Concept

We are using the dispersion relation here which is given by

  •  Ek =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
  • Equate this with given equation s=1/2
  • At low-temperature Cv ∝ Td/s
  • Here d is dimension = 2 (Given in the question)
  •    
  • Cv∝ T4


So, the correct answer is option (1).

In the AC Josephson effect, a supercurrent flows across two superconductors separated by a thin insulating layer and kept at an electric potential difference ΔV. The angular frequency of the resultant supercurrent is given by

Answer (Detailed Solution Below)

Option 1 :

Condensed Matter Physics Question 8 Detailed Solution

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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 = . This is called the A.C. Josephson Effect.

EXPLANATION:

Current density through thin insulating layer is

∴ The angular frequency of the supercurrent is 

⇒ 

Hence the correct answer is option 1.

The dispersion relation of electrons in three dimensions is ϵ(k) = ℏvFk, where vF is the Fermi velocity. If at low temperatures (T F) the Fermi energy ϵF depends on the number density n as ϵF (n) ~ nα, the value of α is

  1. 1/3
  2. 2/3
  3. 1
  4. 3/5

Answer (Detailed Solution Below)

Option 1 : 1/3

Condensed Matter Physics Question 9 Detailed Solution

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Explanation:

  • This is the dispersion relation for electrons where  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  in three-dimension can be represented as 
  • Since  from the original dispersion relation: Thus, we arrive at  which indicates that  

A particle hops randomly from a site to its nearest neighbour in each step on a square lattice of unit lattice constant. The probability of hopping to the positive x-direction is 0.3, to the negative x-direction is 0.2, to the positive y-direction is 0.2 and to the negative y-direction is 0.3. If a particle starts from the origin, its mean position after N steps is

Answer (Detailed Solution Below)

Option 2 :

Condensed Matter Physics Question 10 Detailed Solution

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Concept:

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

Calculation:

i> ∑ piri

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

= 0.1i - 0.1j

For N steps, = [i - j]

The correct answer is option (2).

A lattice A consists of all points in three-dimensional space with coordinates (nx, ny, nz) where nx, ny and nz are integers with n+ ny + nz being odd integers. In another lattice B, n+ ny + nz are even integers. The lattices A and B are

  1. both BCC
  2. both FCC
  3. BCC and FCC, respectively
  4. FCC and BCC, respectively

Answer (Detailed Solution Below)

Option 2 : both FCC

Condensed Matter Physics Question 11 Detailed Solution

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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.

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

where p and n are the carrier densities of the holes and electrons, µp and µn are their respective mobilities. For a p-type semiconductor in which the mobility of holes is less than that of electrons, which of the following graphs best describes the variation of the Hall coefficient with temperature?

Answer (Detailed Solution Below)

Option 4 :

Condensed Matter Physics Question 12 Detailed Solution

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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 = 

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, μp n

⇒ pμp2 > nμn2 

⇒ pμp2 - nμn2 > 0 ⇒ RH = positive.

Case II: At moderate temperature: p > n

⇒ pμp2 ≈ nμn2 (since, μp n)

⇒ RH > 0

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

⇒ pμp2 - nμn2 

⇒ RH 

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

Hence the correct answer is option 4.

Condensed Matter Physics Question 13:

Answer (Detailed Solution Below)

Option 3 :

Condensed Matter Physics Question 13 Detailed Solution

Condensed Matter Physics Question 14:

A paramagnetic salt with magnetic moment per ion μ± = ±μB (where μB is the Bohr magneton) is in thermal equilibrium at temperature T in a constant magnetic field B. The average magnetic moment ⟨M⟩, as a function of  is best represented by

Answer (Detailed Solution Below)

Option 3 :

Condensed Matter Physics Question 14 Detailed Solution

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 (kB) 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 .
  • The probability of being in state with energy is given by the Boltzmann factor: 
  •  We then normalize by dividing by the sum of the probabilities of all possible states (the partition function Z): 
  • This gives us: 
  • The average magnetic moment ⟨M⟩ is the sum of each possible magnetic moment multiplied by its Boltzmann-weighted probability:

Simplify to get: 

Now, 

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.

Condensed Matter Physics Question 15:

For a crystal, let ϕ denote the energy required to create a pair of vacancy and interstitial defects. If n pairs of such defects are formed, and n N, N', where N and N' are, respectively, the total number of lattice and interstitial sites, then n is approximately

Answer (Detailed Solution Below)

Option 1 :

Condensed Matter Physics Question 15 Detailed 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 

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 

EXPLANATION:

The probability of such frenkel defects 

So, the change in entropy 

⇒ ΔS = k ln Ω = 

⇒ 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 

Since, n NN'

Hence the correct answer is option 1.

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