Condensed Matter Physics MCQ Quiz in తెలుగు - Objective Question with Answer for Condensed Matter Physics - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

For a Debye solid at low temperatures, the heat capacity (C) varies with temperature (T) according to the equation:

where:

• C is the heat capacity.
• β is a material-specific constant.
• T is the temperature in Kelvin.

Solution:

Let's analyze the given information:

• Temperature (T) = 5 K
• Constant (β) = \(2.42 \times 10^{-4} \ \text{J/K}^4\ \)

We need to calculate the heat capacity (C) at 5 K using the given equation:

Plug in the values for β and T:

• \(β = 2.42 \times 10^{-4} \ \text{J/K}^4\ \)
• T = 5 K

First, calculate T³:

Now, plug the value of T³ back into the equation:

Hence, the correct answer is option 2.

Concept:

The relaxation time (τ) is related to the electron mobility (μ) and the effective mass of the electron (m*). The electron mobility is defined as the ratio of the drift velocity of the electron to the applied electric field. It can also be expressed in terms of the relaxation time (τ) as follows:

where:

• μ is the electron mobility.
• e is the elementary charge (e = 1.6 × 10^-19 C).
• τ is the relaxation time.
• m* is the effective mass of the electron.

Rearranging the equation to solve for  the relaxation time (τ):

Solution:

Let's analyze the given information:

• Effective mass (m* = 0.067 m_e)
• Electron mobility (μ= \(0.1 \ \text{m}^2 \text{V}^{-1} \text{s}^{-1}\ \))
• Elementary charge (e = 1.6 × 10^-19 C.
• Electron mass (m_e = 9.10938356 × 10^-31 kg)

First, determine the effective mass (m*):

Therefore, the relaxation time (τ) for the electron in the semiconductor is \(3.81 × 10^{-14}\ \text{s}\ \).

Hence, the correct answer is option 4.

Concept:

The Fermi energy (E_F) of a free electron gas in a metal can be calculated using the electron density. For a metal with atomic mass (A) and density ( ρ), the number density of atoms ( n ) can be calculated. Assuming each atom contributes one free electron, the electron density ( n_e) is equal to the number density of atoms. The formula for the Fermi energy is given by:

where:

• ℏ is the reduced Planck's constant \(ℏ = 1.05 \times 10^{-34} \ \text{J} \cdot \text{s}\ \)
• m_e is the electron mass \(m_e \approx 9.1 \times 10^{-31} \ \text{kg}\ \)
• n_e is the electron density \(\text{electrons/m}^3\ \).

Given the density (ρ) and atomic mass (A), we first determine the number of atoms per unit volume and hence the electron density.

Solution:

Let's analyze the given information:

• Density: \(ρ = 8.92 \ \text{g/cm}^3 = 8.92 \times 10^3 \ \text{kg/m}^3\ \) (converted to SI units)
• Atomic mass: \(A = 63.5 \ \text{u} = 63.5 \times 1.6 \times 10^{-27} \ \text{kg} = 1.05 \times 10^{-25} \ \text{kg}\ \)

First, calculate the number of atoms per unit volume (n):

Assuming each atom contributes one free electron, the electron density (ne) is:

Now, plug n_e into the Fermi energy formula:

Calculate the term (\((3 \pi^2 n_e)^{2/3}\ \)) first:

Now, calculate the Fermi energy:

Therefore, the Fermi energy E_F for the given metal is 1.52 eV.

Hence, the correct answer is option 2.

Concept:

The density of a crystal can be calculated using the formula:

where:

• ρ is the density.
• Z is the number of atoms per unit cell.
• M is the atomic mass of the element.
• N_A is Avogadro's number (6.022 × 10²³ mol^-1).
• V_{unit cell} is the volume of the unit cell.

F or a simple cubic crystal structure, each unit cell contains one atom (Z = 1). In a simple cubic structure, the edge length of the unit cell (a) is twice the atomic radius (r):

The volume of the unit cell V_{unit cell} is then:

Solution:

Let's analyze the given information:

• Atomic radius (r) = 1.241 Å = 1.241 × 10^-10 m (converted to SI units).
• Atomic mass (M) = 55.845 u = 55.845 × 1.66 × 10^-27 kg= 9.270 × 10^-26 kg (converted to SI units).

First, calculate the edge length of the unit cell (a):

Next, calculate the volume of the unit cell :V_{unit cell}

\(ρ = \frac{Z \cdot M}{N_A \cdot V_{\text{unit cell}}}\ \)

For a simple cubic structure, Z = 1, so:

Therefore, the density of the simple cubic crystal structure of iron is \(1 × 10^{3} \ \text{kg/m}^3\ \).

Hence, the correct answer is option 2.

Concept:

The Hall Effect can be used to determine the carrier concentration in a material. When a current-carrying conductor is placed in a perpendicular magnetic field, a voltage (Hall voltage) is developed across the conductor. The Hall coefficient (R_H) is related to the carrier concentration (n) by the equation:

where:

• R_H is the Hall coefficient.
• n is the carrier concentration.
• q is the charge of the carrier (for electrons, q = e = 1.6 × 10^-19 C).

The Hall coefficient R_H can be determined using the formula:

where:

• V_H is the Hall voltage.
• t is the thickness (width) of the specimen.
• I is the current through the specimen.
• B is the magnetic field.

Solution:

Let's analyze the given information:

• Width of the specimen (t) = 2 cm = 2 × 10^-2) m (converted to meters)
• Current (I) = 1 mA = 1 × 10^-3 A (converted to amperes)
• Hall voltage: V_H = 2 mV = 2 × 10^-3 V (converted to volts)
• Magnetic field (B) = 0.1 T

First, calculate the Hall coefficient (R_H):

Plug in the values:

• V_H = 2 × 10^-3 V
• t = 2 × 10^-2 m
• I = 1 × 10^-3 A
• B = 0.1 T

\(R_H = \frac{2 × 10^{-3} \ \text{V} × 2 × 10^{-2} \ \text{m}}{1 × 10^{-3} \ \text{A} × 0.1 \ \text{T}} = 0.4 \ \frac{\text{V} \cdot \text{m}}{\text{A} \cdot \text{T}} \ \)

Now, use the Hall coefficient to find the carrier concentration (n):
\(R_H = \frac{1}{nq} \\ n = \frac{1}{R_H q}\ \)

where q = 1.6 × 10^-19 C.

\(n = \frac{1}{0.4 × 1.6 × 10^{-19}} \\ n \approx 1.56 × 10^{19} \ \text{m}^{-3} \ \)

Therefore, the carrier concentration (n) is \(1.56 × 10^{19} \ \text{m}^{-3}\ \).

Hence, the correct answer is option 1.

Concept:

The magnetic moment per unit volume, also known as the magnetization (M), of a paramagnetic material can be calculated using the magnetic susceptibility (χ_m) and the external magnetic field (B). The relationship is given by:

where:

• M is the magnetization, or magnetic moment per unit volume.
• χ_m is the magnetic susceptibility.
• H is the magnetic field strength.

The magnetic field strength (H) is related to the magnetic flux density (B) and the permeability of free space (μ₀) by:

This can be rearranged to solve for (H):

Here, μ₀ is the permeability of free space and has a value of \(4\pi \times 10^{-7} \ \text{T} \text{m/A}\ \).

Solution:

Let's analyze the given information:

• Magnetic susceptibility χ_m = 0.001
• External magnetic field (B) = 0.2 T
• Temperature (T) = 300 K (Though it affects (χ_m), it is not needed in the direct calculation given (χ_m) is provided.)

First, calculate the magnetic field strength (H):

Plug in the values:

• B = 0.2 T
• \(μ_0 = 4\pi \times 10^{-7} \ \text{T} \ \text{m/A}\ \)

 \(H = \frac{0.2 \ \text{T}}{4\pi \times 10^{-7} \ \text{T} \ \text{m/A}} \approx 1.592 \times 10^{5} \ \text{A/m} \)

Next, calculate the magnetization (M) using the susceptibility (χ _m) and the magnetic field strength (H):

Plug in the values:

• χ_m = 0.001
• \(H \approx 1.592 \times 10^{5} \ \text{A/m} \ \)

Hence, the correct answer is option 4.

Concept:

The specific heat capacity (C_v) of a solid at low temperatures can be approximated using the Debye model. According to this model, the specific heat capacity follows a T³ dependence and is given by:

where, T is the temperature and θD is the Debye temperature. The Debye model provides a good approximation for the specific heat at temperatures much lower than the Debye temperature.

Given specific values, the calculation involves substituting the temperature (T) and the Debye temperature θD into the formula.

Solution:

Let's analyze the given information:

• Debye temperature:θD = 428 K.
• Temperature:T = 20 K.

First, let's write down the formula again for clarity:

Now, we substitute the given values into the formula:

Calculate the numerator and the denominator separately:

Numerator:

Denominator:

Combine both terms to calculate the specific heat:

Therefore, the specific heat capacity at T = 20 K is \(1.03 \times 10^{-3}\ \text{J/K}\ \).

Hence, the correct answer is option 1.

Concept:

The heat capacity of a molecule varies with temperature due to the activation of different degrees of freedom. At lower temperatures, only translational contribute significantly, resulting in a heat capacity that reflects these 3N k degrees of freedom. As temperature increases, vibrational modes and rotational modes also become excited, adding additional contributions to heat capacity. This shift in active degrees of freedom highlights the relationship between temperature and molecular energy states, illustrating how thermal energy influences molecular behavior and heat absorption.

Expalantion:

Degrees of Freedom and Heat Capacity Calculation

Room Temperature:
    For triatomic molecule, the total number of degrees of freedom at room temperature includes only translational , The degree of freedom is \(f=3N-k\), where N is the no of atom and k is the no of constraints.

Given :

N=3 (no of atom)

K=3 (no of bond)

The effective degrees of freedom ( f ) at room temperature are:
\(\ f_{\text{room}} = 3 N-k = 6\)

The heat capacity at constant volume ( \(C_V \)) at room temperature is given by:
\(\ C_V^{\text{room}} = \frac{f_{\text{room}}}{2} k_B = \frac{6}{2} k_B \)

 High Temperature:
    At high temperatures, all degrees of freedom become active, including rotational and  vibrational modes. And every bond will give contribution of  \( K_BT\) from vibrational and rotational modes, The total contribution is 

The heat capacity at constant volume ( C_V ) at high temperature is:
\(\ C_V^{\text{high}} = \frac{6K_B}{2}+K_B+K_B +K_B= 6K_B\\ \quad \quad = 6 k_B\)

 Ratio of Heat Capacities:
Now, we can find the ratio of the heat capacities:
\(\ \frac{C_V^{\text{room}}}{C_V^{\text{high}}} = \frac{ 3 k_B}{6 k_B} = 0.5 \)

Concept:

Calculation of Maxima/Minima of a function: The number of minima or maxima of any function(F) can be calculated by taking the first derivative and keeping that equal to zero.

Let F be a function of x then,

\(\frac{\partial F}{\partial x}=0 \)

will give us number of minima/maxima.

Now to find out whether it is a minima or maxima we take its double derivate and put the values from first derivate in it. Such that,

\(\frac{\partial^2 F}{\partial x^2}>0 \)

That means it is a minima.

\(\frac{\partial^2 F}{\partial x^2}<0 \)

Then it is a maxima.

Explanation:

Given function: F(M) = F0 + 2(T - Tc)M2 + M4, F0 > 0

Hence, taking its first derivative:

\(\frac{\partial F(M)}{\partial M}=0 \\ \frac{\partial [F_0+2(T-T_c)M^2+M^4]}{\partial M} =0\\ 2(T-T_c)M+4M^3=0\\ 2M[(T-T_c)+2M^2]=0\\ \text{either} \ \ M=0 \ \ \text{or} \ \ M^2=-(T-T_c)\)

Hence, the only real solution exists is for M=0.

\(\frac{\partial^2 [F_0+2(T-T_c)M^2+M^4]}{\partial M^2} =2(T-T_c)+12M^2\)

Hence, for M=0

\(2(T-T_c)+12M^2=2(T-T_c)\)

As, T>T_c this value is positive which means it is a minima.

The correct answer is option 2.

Concept:

According to Bragg's equation,

• \(n\lambda=2dsin\theta\)
• Condition for overlapping is \(n\lambda=dsin\theta\)

Explanation:

A diffraction grating follows the Bragg's law as,

• \(n\lambda_g=2dsin\theta\), (for green line)
• \((n+1)\lambda_v=2dsin\theta\) , (for violet line)

Angle of Diffraction given is 30º

• \(n\lambda_g=dsin30^0\)
• \((n+1)\lambda_v=dsin30^0\)

Now, \(sin30^0=\frac{1}{2}\)

Equations become, 

\(n\lambda_g=\frac{d}{2}\)----------------1

\((n+1)\lambda_v=\frac{d}{2}\)-------------2

Substitute value of n from equation 1 to equation 2, we get,

\((\frac{d}{2\lambda_g}+1)\lambda_v=\frac{d}{2}\)

\(\frac{d}{2}(1-\frac{\lambda_v}{\lambda_g})=\lambda_v\)

\(d=\frac{2\lambda_g\lambda_v}{\lambda_g-\lambda_v}\)

Number of lines per cm is \(N=\frac{1}{d}=\frac{\lambda_g-\lambda_v}{2\lambda_v\lambda_g}\)

Put all the given values in the above equation, we get,

\(N=\frac{1}{d}=\frac{540-405}{2\times540\times405\times10^{-9}\times10^{2}}\)

\(N=3086/cm\)

Hence, the correct answer is Option-4-\(N=3086/cm\).