FSc Pre-Engineering Physics Physics of Solids — Set 3

Physics of Solids MCQs set 3 for FSc Pre-Engineering Physics — 20 solved questions.

FSc Pre-Engineering Physics Physics of Solids — Set 3

  1. Question 1

    Q1. In a simple cubic lattice, the distance between (1 1 1) planes is 2 Å. What is the lattice constant?

    • A) 2√3 Å
    • B) 3.46 Å
    • C) 4 Å
    • D) 2 Å

    Answer: 3.46 Å

    Explanation: Using the formula d = a / √(h² + k² + l²), we get a = d * √3 = 2 * √3 = 3.46 Å.

  2. Question 2

    Q2. The Hall coefficient of a semiconductor is 10^(-4) m³/C. What is the carrier concentration?

    • A) 10²2 m^(-3)
    • B) 6.25 * 10²2 m^(-3)
    • C) 10²0 m^(-3)
    • D) 10^18 m^(-3)

    Answer: 6.25 * 10²2 m^(-3)

    Explanation: Using R_H = 1 / (nq), we can find n = 1 / (R_H * q) = 6.25 * 10²2 m^(-3) for q = 1.6 * 10^(-19) C.

  3. Question 3

    Q3. A crystal has a dielectric constant of 10. What is the polarization if the electric field is 10⁴ V/m?

    • A) 8.85 * 10^(-7) C/m²
    • B) 7.96 * 10^(-7) C/m²
    • C) 8 * 10^(-7) C/m²
    • D) 9 * 10^(-7) C/m²

    Answer: 8.85 * 10^(-7) C/m²

    Explanation: Using P = ε₀(ε_r - 1)E, we get P = 8.85 * 10^(-12) * (10 - 1) * 10⁴ = 8.85 * 10^(-7) C/m².

  4. Question 4

    Q4. The effective mass of an electron in a crystal is 0.1m₀. What is the mobility if the relaxation time is 10^(-13) s?

    • A) 1.76 m²/Vs
    • B) 1.76 * 10^(-3) m²/Vs
    • C) 0.176 m²/Vs
    • D) 17.6 m²/Vs

    Answer: 1.76 m²/Vs

    Explanation: Using μ = eτ / m*, we get μ = (1.6 * 10^(-19) * 10^(-13)) / (0.1 * 9.1 * 10^(-31)) = 1.76 m²/Vs.

  5. Question 5

    Q5. A ferromagnetic material has a saturation magnetization of 10^6 A/m. What is the magnetic moment per atom if the lattice constant is 2 Å?

    • A) 1.6 * 10^(-23) Am²
    • B) 3.2 * 10^(-23) Am²
    • C) 2 * 10^(-23) Am²
    • D) 4 * 10^(-23) Am²

    Answer: 1.6 * 10^(-23) Am²

    Explanation: For a simple cubic lattice, the number of atoms per unit volume is 1/a³. Using M_s = n * μ, we can find μ.

  6. Question 6

    Q6. The London penetration depth of a superconductor is 100 nm. What is the superfluid density?

    • A) 10²5 m^(-3)
    • B) 10²6 m^(-3)
    • C) 2.5 * 10²5 m^(-3)
    • D) 5 * 10²5 m^(-3)

    Answer: 2.5 * 10²5 m^(-3)

    Explanation: Using λ_L = √(m / (μ₀ * n_s * e²)), we can find n_s = m / (μ₀ * λ_L² * e²).

  7. Question 7

    Q7. In a metal, the resistivity is 10^(-8) Ωm at 300 K. What is the mean free time if the Fermi velocity is 10^6 m/s?

    • A) 2.5 * 10^(-14) s
    • B) 3.2 * 10^(-14) s
    • C) 4 * 10^(-14) s
    • D) 5 * 10^(-14) s

    Answer: 2.5 * 10^(-14) s

    Explanation: Using ρ = m / (n * e² * τ) and v_f = √(2E_f/m), we can find τ = m * v_f / (n * e² * ρ).

  8. Question 8

    Q8. A semiconductor has an energy gap of 1 eV. At what temperature will the intrinsic carrier concentration be 10^16 m^(-3)?

    • A) 300 K
    • B) 400 K
    • C) 500 K
    • D) 600 K

    Answer: 500 K

    Explanation: Using n_i = √(N_c * N_v) * exp(-E_g / (2kT)), we can find T for given n_i.

  9. Question 9

    Q9. The Debye temperature of a crystal is 300 K. What is the specific heat at 100 K?

    • A) 1.2 J/molK
    • B) 2.4 J/molK
    • C) 0.6 J/molK
    • D) 3 J/molK

    Answer: 2.4 J/molK

    Explanation: Using the Debye model, C_v = (12/5) * π⁴ * N * k * (T/θ_D)³ for T << θ_D.

  10. Question 10

    Q10. The thermal conductivity of a crystal is 10 W/mK. What is the phonon mean free path if the specific heat is 100 J/kgK and the sound velocity is 5000 m/s?

    • A) 10 nm
    • B) 20 nm
    • C) 50 nm
    • D) 100 nm

    Answer: 20 nm

    Explanation: Using K = (1/3) * C_v * v * λ, we can find λ = 3K / (C_v * v).

  11. Question 11

    Q11. A crystal has a coefficient of thermal expansion of 10^(-5) K^(-1). What is the change in lattice constant if the temperature changes by 100 K?

    • A) 0.1%
    • B) 0.2%
    • C) 0.05%
    • D) 0.01%

    Answer: 0.1%

    Explanation: Using Δa / a = α * ΔT, we get Δa / a = 10^(-5) * 100 = 0.1%.

  12. Question 12

    Q12. The Fermi temperature of a metal is 5000 K. What is the Fermi energy?

    • A) 0.43 eV
    • B) 0.62 eV
    • C) 0.86 eV
    • D) 1 eV

    Answer: 0.43 eV

    Explanation: Using E_f = k * T_f, we get E_f = 8.62 * 10^(-5) * 5000 = 0.43 eV.

  13. Question 13

    Q13. A superconductor has a critical current density of 10^6 A/m². What is the critical magnetic field if the radius of the wire is 1 mm?

    • A) 0.2 T
    • B) 0.5 T
    • C) 1 T
    • D) 2 T

    Answer: 0.5 T

    Explanation: Using J_c = 2H_c / r, we can find H_c = J_c * r / 2 = 0.5 T.

  14. Question 14

    Q14. What is the primary bonding type in solids like NaCl?

    • A) Covalent
    • B) Ionic
    • C) Van der Waals
    • D) Hydrogen

    Answer: Ionic

    Explanation: NaCl is an ionic solid, held together by electrostatic forces between oppositely charged ions, following Coulomb's law.

  15. Question 15

    Q15. The lattice constant 'a' for a cubic crystal is related to its density by?

    • A) a³ = nM / N_A ρ
    • B) a³ = nρ / N_A M
    • C) a = nM / N_A ρ
    • D) a = nρ / N_A M

    Answer: a³ = nM / N_A ρ

    Explanation: The density ρ = nM / N_A a³, where n is the number of atoms per unit cell, M is the molar mass, and N_A is Avogadro's number.

  16. Question 16

    Q16. In a simple cubic lattice, the coordination number is?

    • A) 6
    • B) 8
    • C) 12
    • D) 4

    Answer: 6

    Explanation: Each atom in a simple cubic lattice is surrounded by 6 nearest neighbors, hence the coordination number is 6.

  17. Question 17

    Q17. The energy gap between the valence and conduction bands in insulators is typically?

    • A) > 3 eV
    • B) < 1 eV
    • C) 1-3 eV
    • D) 0 eV

    Answer: > 3 eV

    Explanation: Insulators have a large energy gap (typically > 3 eV) between the valence and conduction bands, making it difficult for electrons to be excited.

  18. Question 18

    Q18. The Hall effect is used to determine the?

    • A) Type of charge carriers
    • B) Density of charge carriers
    • C) Both A and B
    • D) None of these

    Answer: Both A and B

    Explanation: The Hall effect measures the Hall voltage, which depends on the type and density of charge carriers, given by RH = 1 / nq.

  19. Question 19

    Q19. The Wiedemann-Franz law relates the?

    • A) Thermal conductivity to electrical conductivity
    • B) Specific heat to electrical conductivity
    • C) Thermal conductivity to specific heat
    • D) None of these

    Answer: Thermal conductivity to electrical conductivity

    Explanation: The Wiedemann-Franz law states that K / σ = LT, where K is thermal conductivity, σ is electrical conductivity, and L is the Lorenz number.

  20. Question 20

    Q20. In a p-type semiconductor, the majority charge carriers are?

    • A) Electrons
    • B) Holes
    • C) Both
    • D) None of these

    Answer: Holes

    Explanation: In p-type semiconductors, acceptor impurities create holes, making them the majority charge carriers.