Condensed Matter Physics MCQ Quiz in తెలుగు - Objective Question with Answer for Condensed Matter Physics - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Mar 28, 2025
Latest Condensed Matter Physics MCQ Objective Questions
Top Condensed Matter Physics MCQ Objective Questions
Condensed Matter Physics Question 1:
For a Debye solid at low temperatures, the heat capacity (C) varies as
Answer (Detailed Solution Below)
Condensed Matter Physics Question 1 Detailed Solution
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) =
We need to calculate the heat capacity (C) at 5 K using the given equation:
C = β T3
Plug in the values for β and T:
- T = 5 K
First, calculate T3:
Now, plug the value of T3 back into the equation:
Hence, the correct answer is option 2.
Condensed Matter Physics Question 2:
An electron in a semiconductor has an effective mass of 0.067 me. If the electron mobility is
Answer (Detailed Solution Below)
Condensed Matter Physics Question 2 Detailed Solution
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 me)
- Electron mobility (μ=
) - Elementary charge (e = 1.6 × 10-19 C.
- Electron mass (me = 9.10938356 × 10-31 kg)
First, determine the effective mass (m*):
Therefore, the relaxation time (τ) for the electron in the semiconductor is
Hence, the correct answer is option 4.
Condensed Matter Physics Question 3:
Calculate the Fermi energy for a free electron gas in a metal with a density of
Answer (Detailed Solution Below)
Condensed Matter Physics Question 3 Detailed Solution
Concept:
The Fermi energy (EF) 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 ( ne) 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
- me is the electron mass
- ne is the electron density
.
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:
(converted to SI units) - Atomic mass:
First, calculate the number of atoms per unit volume (n):
Assuming each atom contributes one free electron, the electron density (ne) is:
Now, plug ne into the Fermi energy formula:
Calculate the term (
Now, calculate the Fermi energy:
Therefore, the Fermi energy EF for the given metal is 1.52 eV.
Hence, the correct answer is option 2.
Condensed Matter Physics Question 4:
Calculate the density of a simple cubic crystal structure of iron (Fe) with an atomic radius of 1.241 Å and atomic mass of 55.845 u.
Answer (Detailed Solution Below)
Condensed Matter Physics Question 4 Detailed Solution
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.
- NA is Avogadro's number (6.022 × 1023 mol-1).
- Vunit 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):
a = 2r
The volume of the unit cell Vunit 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 :Vunit cell
For a simple cubic structure, Z = 1, so:
Therefore, the density of the simple cubic crystal structure of iron is
Hence, the correct answer is option 2.
Condensed Matter Physics Question 5:
In a Hall Effect experiment, a specimen with a width of 2 cm, carrying a current of 1 mA, produces a Hall voltage of 2 mV under a magnetic field of 0.1 T. Calculate the carrier concentration assuming a single type of charge carrier.
Answer (Detailed Solution Below)
Condensed Matter Physics Question 5 Detailed Solution
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 (RH) is related to the carrier concentration (n) by the equation:
where:
- RH 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 RH can be determined using the formula:
where:
- VH 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: VH = 2 mV = 2 × 10-3 V (converted to volts)
- Magnetic field (B) = 0.1 T
First, calculate the Hall coefficient (RH):
Plug in the values:
- VH = 2 × 10-3 V
- t = 2 × 10-2 m
- I = 1 × 10-3 A
- B = 0.1 T
Now, use the Hall coefficient to find the carrier concentration (n):
where q = 1.6 × 10-19 C.
Therefore, the carrier concentration (n) is
Hence, the correct answer is option 1.
Condensed Matter Physics Question 6:
A paramagnetic material with a magnetic susceptibility of 0.001 at 300 K is subjected to an external magnetic field of 0.2 T. Calculate the magnetic moment per unit volume.
Answer (Detailed Solution Below)
Condensed Matter Physics Question 6 Detailed Solution
Concept:
The magnetic moment per unit volume, also known as the magnetization (M), of a paramagnetic material can be calculated using the magnetic susceptibility (χm) and the external magnetic field (B). The relationship is given by:
where:
- M is the magnetization, or magnetic moment per unit volume.
- χm 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
Solution:
Let's analyze the given information:
- Magnetic susceptibility χm = 0.001
- External magnetic field (B) = 0.2 T
- Temperature (T) = 300 K (Though it affects (χm), it is not needed in the direct calculation given (χm) is provided.)
First, calculate the magnetic field strength (H):
Plug in the values:
- B = 0.2 T
Next, calculate the magnetization (M) using the susceptibility (χ m) and the magnetic field strength (H):
Plug in the values:
- χm = 0.001
Hence, the correct answer is option 4.
Condensed Matter Physics Question 7:
Consider the specific heat capacity of a solid at low temperatures for a Debye temperature θD = 428 K, calculate the specific heat at T = 20 K.
Answer (Detailed Solution Below)
Condensed Matter Physics Question 7 Detailed Solution
Concept:
The specific heat capacity (Cv) 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
Hence, the correct answer is option 1.
Condensed Matter Physics Question 8:
Consider a triatomic molecule of the shape shown in the figure below in three dimension. The ratio of heat capacity at room temperature and at high temperature(much higher but lower than the bond dissociation energy ) is ....... upto one decimal point.
Answer (Detailed Solution Below) 0.5
Condensed Matter Physics Question 8 Detailed Solution
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
Given :
N=3 (no of atom)
K=3 (no of bond)
The effective degrees of freedom ( f ) at room temperature are:
The heat capacity at constant volume (
High Temperature:
At high temperatures, all degrees of freedom become active, including rotational and vibrational modes. And every bond will give contribution of
The heat capacity at constant volume ( C_V ) at high temperature is:
Ratio of Heat Capacities:
Now, we can find the ratio of the heat capacities:
Condensed Matter Physics Question 9:
The energy of a ferromagnet as a function of magnetization M is given by F(M) = F0 + 2(T - Tc)M2 + M4, F0 > 0. The number of minima in the function F(M) for T > Tc is
Answer (Detailed Solution Below)
Condensed Matter Physics Question 9 Detailed Solution
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,
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,
That means it is a minima.
Then it is a maxima.
Explanation:
Given function: F(M) = F0 + 2(T - Tc)M2 + M4, F0 > 0
Hence, taking its first derivative:
Hence, the only real solution exists is for M=0.
Hence, for M=0
As, T>Tc this value is positive which means it is a minima.
The correct answer is option 2.
Condensed Matter Physics Question 10:
A diffraction grating used at normal incidence gives a green line (540 nm) at a certain order super-imposed on a violet line (405 nm) of the next higher order. If the angle of diffraction be 30°, the number of lines per cm on the grating is
Answer (Detailed Solution Below)
Condensed Matter Physics Question 10 Detailed Solution
Concept:
According to Bragg's equation,
- Condition for overlapping is
Explanation:
A diffraction grating follows the Bragg's law as,
, (for green line) , (for violet line)
Angle of Diffraction given is 30º
Now,
Equations become,
Substitute value of n from equation 1 to equation 2, we get,
Number of lines per cm is
Put all the given values in the above equation, we get,
Hence, the correct answer is Option-4-