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

Explanation:

The Fermi energy in three dimension is defined as 

\(\rm E_F=\frac{\hbar^2}{2 m}\left(\frac{3 \pi^2 N}{V}\right)^{2 / 3}=\) \(\rm \frac{\hbar^2}{2 m}\left(3 \pi^2 n\right)^{2 / 3}\)

Where, n is the electron concentration or density of free Fermi gas.

The total energy of free Fermi gas in 3D is

\(\rm E=\frac{3}{5} N E_F=\) \(\rm \frac{3}{5} N \times \frac{\hbar^2}{2 m}\left(\frac{3 \pi^2 N}{V}\right)^{2 / 3}\)

The pressure of a nonrelativistic free Fermi gas is defined as 

\( p_F =-\left(\frac{\partial E}{\partial V}\right)_N=\) \(-\frac{3}{5} N \times \frac{\hbar^2}{2 m}\left(3 \pi^2 N\right)^{2 / 3} \times\) \(\left(-\frac{2}{3}\right) V^{-5 / 3} \)

\(=\frac{2}{5} n E_F=\frac{2}{5} n \times\)\( \frac{\hbar^2}{2 m}\left(3 \pi^2 n\right)^{2 / 3}=\) \(\frac{2}{5} \frac{\hbar^2}{2 m}\left(3 \pi^2\right)^{2 / 3} n^{5 / 3}\).

The correct option is option (1).

Explanation:

Near the bottom of the band the k → 0

\(\rm \cos k_x a \approx 1-\frac{1}{2}\left(k_x a\right)^2, \) \(\rm \cos k_y a \approx 1-\frac{1}{2}\left(k_y a\right)^2, \) \(\rm \cos k_z a \approx 1-\frac{1}{2}\left(k_z a\right)^2\)

\(\rm E(k)=E_0-B\)\(\rm \left(\cos k_x a+\cos k_y a+\cos k_z a\right)\) = \(\rm E_0-B\)\(\rm \left(1-\frac{1}{2}\left(k_x a\right)^2+1-\frac{1}{2}\left(k_y a\right)^2+1-\frac{1}{2}\left(k_z a\right)^2\right)\)

= \(\rm E_0-B\left(3-\frac{1}{2} a^2\left(k_x+k_x+k_x\right)^2\right)\) = \(\rm E_0-3 B-\frac{1}{2} B a^2 k^2\)

Effective mass of the electron is \(\rm m^*=\frac{\hbar^2}{d^2 E / d k^2}=\frac{\hbar^2}{B a^2}\).

The correct option is option (4).

Explanation:

\(\rm \varepsilon_k=\beta\)\(\left(\cos k_x a+\cos k_y a+\cos k_z a\right)\),

 Effective mass \(m^*=\frac{\hbar^2}{\left(\frac{d^2 \varepsilon_k}{d^2 k}\right)}\).

Brillouin zone boundary is at  \(k_x= \pm \frac{\pi}{a}, k_y=\) \( \pm \frac{\pi}{a}, k_z= \pm \frac{\pi}{a}\).

Hence \(\left.\left(\frac{d^2 \varepsilon_k}{d^2 k}\right)\right|_{\substack{\frac{\pi}{a}, \frac{\pi}{a}, \frac{\pi}{a}}}=\) \(3 \beta a^2 \Rightarrow m^*=\frac{\hbar^2}{3 \beta a^2}\).

The correct option is option (4).

Concept:

Dispersion relation: The dispersion relation gives a direct relation between the frequency(ω) and wave vector of a wave(k), and the group velocity describes the propagation speed of the wave packet. This tells about dynamics of waves in various physical media. 

The particle(v_p) and group velocity(v_g) are as follows:
\(v_p=\frac{ω}{k}\\v_g = \frac{\partial ω}{\partial k}\ \ \)

Explanation:

We have the above dispersion relation.

For large \(\rm \lambda,\left(k_x a, k_y a, k_z a\right)\) are small.

Hence,

 \(\cos k_x a=\left(1-\frac{k_x^2 a^2}{2}\right)\\ \cos k_y a=\left(1-\frac{k_y^2 a^2}{2}\right)\\ \cos k_z a =\left(1-\frac{k_z^2 a^2}{2}\right) \ \)

\(\omega ^2(k)=\omega_0^2 \left[3-\left(1-\frac{k_x^2 a^2}{2}\right)-\left(1-\frac{k_y^2 a^2}{2}\right)-\left(1-\frac{k_z^2 a^2}{2}\right)\right] =\frac{\omega _0^2 a^2}{2}\left(k_x^2+k_y^2+k_z^2\right)\ \ \)

\(\omega ^2(k)=\frac{\omega _0^2 a^2}{2} k^2\\ \Rightarrow \omega=\frac{\omega _0 a}{\sqrt{2}} k\\ \Rightarrow v_g=\frac{d \omega }{d k}=\frac{\omega _0 a}{\sqrt{2}}\ \) .

The correct option is option (4).

Explanation:

Given \(V(r)=-\frac{a}{r^6}+\frac{b}{r^{12}} \). 

At equilibrium radius, \(\left.\frac{d V(r)}{d r}\right|_{r=r_0}=0\) 

\(∴ \frac{d V(r)}{d r}=+\frac{6 a}{r_0^7}-\frac{12 b}{r_0^{13}}=\)

 \(0 \Rightarrow \frac{r_0^{13}}{r_0^7}=\frac{12 b}{6 a}=\) \(\frac{2 b}{a} \Rightarrow r_0^6=\frac{2 b}{a}\).

∴ The value of potential at equilibrium is \(V\left(r_0\right)=-\frac{a}{r_0^6}+\frac{b}{r_0^{12}}=\) \(-\frac{a^2}{2 b}+\frac{a^2}{4 b}=\frac{-a^2}{4 b}\).

The correct option is option (4).

Explanation:

The reciprocal lattice is a lattice of points in momentum space that is dual to the direct lattice of points in real space.

It is defined as the set of all vectors \(\mathbf{G} \) that satisfy the equation.

\(\mathbf{G} \cdot \mathbf{R} = 2 \pi n , \)

where, \(\mathbf{R} \) is any vector in the direct lattice,

and n is an integer.

In two dimensions, the reciprocal lattice is also a two-dimensional lattice.

For a square lattice with lattice constant a, the direct lattice is defined by the vectors.

\(a_1= a\begin{pmatrix} 1 \\ 0 \end{pmatrix} , a_2= a\begin{pmatrix} 0 \\ 1 \end{pmatrix} \).

The reciprocal lattice vectors are given by;

\(b_1= \frac{2 \pi}{a}\begin{pmatrix} 1 \\ 0 \end{pmatrix} , b_2= \frac{2 \pi}{a}\begin{pmatrix} 0 \\ 1 \end{pmatrix} \).

To find the reciprocal lattice vector corresponding to the (1,1) direction in the direct lattice,

we take a linear combination of the reciprocal lattice vectors.

Thus, the reciprocal lattice vector corresponding to the (1,1) direction in the direct lattice is.

\( \mathbf{G} = \frac{2 \pi}{a}\begin{pmatrix} 1 \\ 1 \end{pmatrix} \).

The correct option is option (4).

Explanation:

The structure factor for a reflection can be calculated using the following formula:

\(F(hkl) = Σf(j) e^{(-2πi(hjx + kjy + ljz))} \).

where F(hkl) is the structure factor for the (hkl) reflection, f_j is the atomic scattering factor for the j^th atom in the unit cell,

and (h, k, l) are the Miller indices for the reflection.

The sum is taken over all atoms in the unit cell.

For a simple cubic lattice, the positions of the atoms are given by:

(x, y, z) = (na, mb, lc),

where a is the lattice constant, and n, m, and l are integers that specify the position of the atom in the lattice.

For the (111) reflection, the Miller indices are (1,1,1).

The atomic scattering factor is given as 1.0.

Substituting these values into the formula, we get:

\(F(hkl) = Σf(j) e^{(-2πi(jx + jy + jz))} \).

Expanding we get:

\(1.0[e^{(-2πi(a + b + c))} + e^{(-4πi( a + b + c)) }+ ...] = 1.0[e^{(-2πi(a + b + c))}]/ (1 - e^{(-2πi(a + b + c))}) = 1.0[\cos(2π(a + b + c)) - i \sin(2π(a + b + c))] / [1 - \cos(2π(a + b + c)) - i \sin(2π(a + b + c))] \).

Here we have taken x, y, z as a,b,c.

Simplifying this expression, we get:

F(111) = 1.0 / 3=0.33

The correct option is option (2).

Explanation:

The primitive unit cell in a Bravais lattice is defined as the smallest repeating unit that can fill up space by translation without leaving voids. Every lattice point in a Bravais lattice has an identical environment.

• For a rectangle or square, which is indeed a special type of rectangle, it is clear that they can serve as primitive unit cells as they can fill up space through simple translations.
• For a regular hexagon, it can be translated along two non-collinear vectors (the sides of the hexagon) and can completely fill a plane without gaps or overlaps. Thus, it can serve as a primitive cell of a 2D lattice.
• An equilateral triangle, however, cannot serve as a primitive unit cell for a Bravais lattice. If we use equilateral triangles to tile a plane, each lattice point (triangle vertex) will not have an identical environment. This is because each lattice point is either connected to three triangles or six triangles, meaning the environments for vertices aren't the same and hence it can't form a Bravais lattice.

​So, only the square (or rectangle) and the regular hexagon can serve as primitive unit cells of a Bravais lattice in two dimensions, not an equilateral triangle.

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:

Lattice Parameter is

a3 = \(n_{ca} × m \over N_A × δ\)

where N_A is Avogadro number, δ is density, n is the number of atom present and m is the atomic weight.

Calculation:

Lattice Parameter is

a³ = \(n_{ca} × m \over N_A × δ\)

= \(4 × (39.1 + 35.5) \over 6.023 × 10^{23} × 1.98\)

= 2.5 × 10^-22

a = 6.3 × 10^-8 

= 6.3 A

Bragg's law is 

2d sin θ = λ 

⇒ sin θ = \({\lambda \over 2a} \sqrt{h^2+k^2+l^2}\)

= \({4 \over 2 \times 6.3} \sqrt{2^2 + 0 + 0} \)

= 0.63

∴ θ = sin^-1 0.63

= 39.4 

The correct answer is option (1).