Solid State MCQ Quiz in தமிழ் - Objective Question with Answer for Solid State - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

The density of a crystalline solid can be calculated using the formula:

\( \text{Density} = \frac{Z \cdot M}{N_A \cdot V} \)

where:

• Z is the number of atoms per unit cell.
• M is the molar mass of the element.
• N_A is Avogadro's number (6.022 × 10²³ mol^-1).
• V is the volume of the unit cell.

For a monoclinic lattice, the volume (V) of the unit cell is given by:

\( V = a \cdot b \cdot c \cdot \sin β \)

Given the following values:

• a = 620 pm = 620× 10^-10 cm
• b = 480 pm = 480 × 10^-10 cm
• c = 1100 pm = 1100 × 10^-10 cm
• sin β = 0.98
• Number of atoms per unit cell, Z = 16
• Atomic mass of Plutonium, M = 244 g mol^-1

Calculation of the volume of the unit cell:

\([ V = (620 × 10^{-10} \text{cm}) \cdot (480 × 10^{-10} \text{cm}) \cdot (1100 × 10^{-10} \text{cm}) \cdot 0.98 \)

\( V = 620 × 480 × 1100 × 0.98 × 10^{-30} \text{cm}^3 \)

V = 3.209 × 10^-22 cm³

Calculation of the density:

\( \text{Density} = \frac{Z \cdot M}{N_A \cdot V} \)

\( \text{Density} = \frac{16 × 244}{6.022 × 10^{23} × 3.209 × 10^{-22}} \)

\( \text{Density} = \frac{3904}{193.25} \approx 20.21 \text{g/cm}^3\)

Conclusion:

The density of Plutonium with the given monoclinic lattice parameters is 20.21 g/cm³.

Concept:

• Corner Atom: Each corner atom contributes 1/8 of an atom to the unit cell.
• Face-centered atoms: Each face-centered atom contributes 1/2 of an atom to the unit cell.
• Edge-centered atoms: Each edge-centered atom contributes 1/4 of an atom to the unit cell.

Explanation:
1. Copper Atoms (Cu):

Each unit cell in a cubic structure has 6 faces.
Cu atom occupies the face center and one atom is missing.

So, the number of copper atoms at the face centers:

\(5\times(1/2)\)

\(5/2\)

2. Silver Atoms (Ag):
Each unit cell in a cubic structure has 12 edges.

So, the number of copper atoms at the edge centers:
 \(12 \times \frac{1}{4} = 3 \)

3. Gold Atoms (Au):

Gold atoms are present at the body center.

1 body-centered atom: Each body-centered atom contributes 1 atom to the unit cell.

Number of gold atoms = 1

Now, the formula is:

\(\text{Cu}_{5/2} \text{Ag}_{3} \text{Au}_{1}\)

\(\text{Cu}_{10} \text{Ag}_{6} \text{Au}_{2}\)

So, the correct formula for the alloy with one copper atom missing is: \(\text{Cu}_{10} \text{Ag}_{6} \text{Au}_{2}\) 

CONCEPT:

Properties of ZnS (Zinc Sulfide)

• ZnS shows both cubic and hexagonal structures:
  • ZnS can exist in two distinct crystal structures: cubic (sphalerite) and hexagonal (wurtzite). Therefore, Statement I is correct.
• Sphalerite exhibits the ZnS structure:
  • Sphalerite is the cubic form of ZnS, where Zn and S are tetrahedrally coordinated. This structure is commonly referred to as the ZnS structure. Therefore, Statement II is correct.
• ZnS is a semiconductor:
  • Zinc sulfide is a semiconductor with a wide band gap. It is used in optoelectronic devices such as phosphors and LEDs. Therefore, Statement III is correct.
• Precipitation of ZnS from an acidic solution:
  • This statement is incorrect because ZnS precipitates in basic solutions when H₂S is passed through, but not in acidic solutions. Therefore, Statement IV is incorrect.

CONCLUSION:

• Statements I, II, and III are correct, while Statement IV is incorrect. Thus, the correct option is the one that includes I, II, and III only.

The correct answer is 111

Concept:-

1. Miller Indices and Bragg's Law
Miller indices (hkl) represent the intercepts of the crystal planes with the axes of the crystal lattice, inverted and reduced to the smallest integer values. Bragg's Law, which relates the angle of incidence to the lattice spacing and the wavelength of X-rays, is given by:
nλ=2d _hkl sinθ

Explanation:-

In FCC crystals, planes where indices are either all odd or all even are typically the ones that satisfy Bragg's condition first. This is because:

If all indices are odd or all are even, the planes do not pass through lattice points that are missing in the FCC structure, such as in the center of the cube edges.
Planes indexed by mixed odd and even numbers (like (101)) typically do not have dense packing and therefore do not contribute to the first Bragg’s peak.

first bragg’s peak for FCC lattice is either all hkl even or all odd, since in option (b) we have (h.k.l) = (1.1.1)

Thus it belngs to FCC lattice

Conclusion:-

So, the miller indices for the first Bragg’s peak is 111

The correct answer is sin2θ = \(\frac{\lambda^2}{4 a^2 c^2}\left[c^2\left(h^2+k^2\right)+a^2 l^2\right]\)

Concept: 

Bragg's law: when the X-ray is incident onto a crystal surface, its angle of incidence, θ, will reflect with the same angle of scattering, θ.

Explanation: 

Braggs law 
nd= 2d sinθ 
squaring above equation 
sin ²θ =n²d² / 4d².......................A

we know 
\(\frac{1}{d^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2}\)

For tetragonal (a = b ≠ c ; α = β = γ = 90°)

\(\frac{1}{d^2} = \frac{h^2+k^2}{a^2} + \frac{l^2}{c^2}\)

\(\frac{1}{d^2} = \frac{c^2(h^2+k^2) + l^2 a^2}{a^2 + c^2} \)

putting in equation A

sin2θ = \(\frac{\lambda^2}{4 a^2 c^2}\left[c^2\left(h^2+k^2\right)+a^2 l^2\right]\)
Conclusion:

With the help of Bragg's law, we can calculate the value of sinθ is sin2θ = \(\frac{\lambda^2}{4 a^2 c^2}\left[c^2\left(h^2+k^2\right)+a^2 l^2\right]\)

Explanation:-

• Li+ ion is so small that all the ions in the structure touch as shown:

• Let, a = distance between the centers of Li⁺ ion and Cl^- ions, then

a = 5.14/2 

= 2.57 Å

• Now, b = distance between the centers of Cl- ions, then

​b = \(\sqrt{a^2+a^2}\)

= \(\sqrt{2}a\)

= \(\sqrt{2}\times 2.57 Å\)

= 3.63 Å

• Thus, the ionic radius for the chloride ions (r) = b/2

= 3.63 Å / 2

= 1.81 Å

Conclusion:-

• Hence, the ionic radius for the chloride ions is 1.81 Å

The correct answer is 30⁰

Concept:

The angle at which the first-order Bragg reflection occurs is determined by Bragg's law. Bragg's Law, named after W. H. Bragg and his son W. L. Bragg who proposed it, is used to calculate the angle for coherent and incoherent scattering from a crystal lattice.

Bragg's Law is typically written as:

nλ = 2d sin θ

Explanation:

We can use Bragg's law to determine the angle at which first-order Bragg reflection is observed from the (110) plane in a simple cubic unit cell:

nλ = 2dsinθ

where n is the order of reflection (n=1 for first order), λ is the wavelength of X-rays used, d is the interplanar spacing for the (110) plane in a simple cubic unit cell, and θ is the angle of incidence.

The interplanar spacing for the (110) plane in a simple cubic unit cell can be calculated using the formula:

d = a/√(h²+k²+l²)

where a is the side length of the unit cell, and h, k, l are the Miller indices for the plane. For (110) plane, h=1, k=1, l=0

So, d = a/√(1²+1²+0²)

d = a/√2

Given a = 3.238 Å,

We can calculate d:

d = 3.238/√2

The X-ray wavelength used is Å = 2.290 Å(typical for Cu Kα radiation), we can now calculate the angle of incidence θ:

nλ = 2dsinθ

2.29Å = 2(2.290 Å)sinθ

sinθ = 1/2

θ = sin⁻¹(1/2) = 30°

Concept:

• A lattice point is a point at the intersection of two or more grid lines in a regularly spaced array of points, which is a point lattice.
• The lattice is a 2D representation of a 3D crystal on the plane of a paper. It is constructed by using certain points known as lattice points and lattice lines.
• The unit cell is a smaller part of a crystal that is unique for a particular crystal substance.
• In crystallography, miller indices form a notation system for lattice planes in crystal lattices. In particular, a family of lattice planes of a given Bravais lattice is determined by three integers h, k, and ℓ, the Miller indices.

​Explanation:

• In crystallography, x-rays are allowed to be passed through a crystal in a series of parallel beams so that they diffract from different sets of parallel miller planes.
• The allowed cubic planes for a BCC (body-centered cubic) lattice in crystallography depend upon miller indices (h, k, ℓ) as, that the sum of h, k, and ℓ, must be even (\({\rm{h + k + l\; = \;even}}\)).
• Now, for the (100) plane, the value of h, k, and ℓ is 1, 0, and 0. The sum of h, k, and ℓ is an odd number and this cubic plane is not allowed for powder diffraction.

\({\rm{h + k + l = 1 + 0 + 0}}\)

\({\rm{ = 1}}\)

\({\rm{ = odd}}\)

• For the (111) plane, the value of h, k, and ℓ is 1, 1, and 1. The sum of h, k, and ℓ is an odd number and this cubic plane is not allowed for powder diffraction.

\({\rm{h + k + l = 1 + 1 + 1}}\)

\({\rm{ = 3}}\)

​\({\rm{ = odd}}\)

• For the (200) plane, the value of h, k, and ℓ is 2, 0, and 0. The sum of h, k, and ℓ is an even number and this cubic plane is allowed for powder diffraction.

\({\rm{h + k + l = 2 + 0 + 0}}\)

\({\rm{ = 2}}\)

\({\rm{ = even}}\)

• For the (210) plane, the value of h, k, and ℓ is 2, 1, and 0. The sum of h, k, and ℓ is an odd number and this cubic plane is not allowed for powder diffraction.

\({\rm{h + k + l = 2 + 1 + 0}}\)

\({\rm{ = 3}}\)

​\({\rm{ = odd}}\)

Conclusion:

• Hence, the Miller indices of the second allowed reflection in the powder diffraction pattern of iron would be (200)

Concept:

Cubic Close Paking: 

• The cubic close packing (ccp) is the name given to a crystal structure. When we put atoms in the octahedral void, the packing is the of the form ABCABC, the packing is known as cubic close packing (ccp).  

Explanation:

• In case of (100) plane, 5 lattice points are present in the area is a². Thus, the area occupied by 1 lattice point is (a²/4)=0.20a².
• In case of (110) plane, 6 lattice points are present in the area is \(\sqrt 2 {a^2}\) Thus, the area occupied by 1 lattice point is \({{\sqrt 2 {a^2}} \over 6}\)=0.23a². 
• In case of (111) plane, 6 lattice points are present in the area \({{\sqrt 3 } \over 4} \times \sqrt 2 a \times \sqrt 2 a = {{\sqrt 3 } \over 2}{a^2}\). Thus, the area occupied by 1 lattice point is \({{\sqrt 3 } \over {12}}{a^2}\) = 0.14a².
• Thus, In the ccp packing, the number of lattice points per unit area in the planes is in the order (111) > (100) > (110).

​

Conclusion:-

So, In the ccp packing, the number of lattice points per unit area in the planes is in the order (111) > (100) > (110)

Concept:

Defects in ionic solids:

• The irregularities in the arrangement of the constituent particles of the crystal system are known as defects in ionic solids.
• Such type of defects can be line defects or point defects.
• Such defects arise due to difference in the speed of formation of crystals during crystallization.

Types of defects in ionic solids -

The types of defects that arises at increasing temperature (>0K) in ionic solids are of 3 types -

1. Stoichiometric defects 
2. Non-stoichiometric defects
3. Impurity defects

→ Stoichiometric defects - The defects in which the stoichiometry of the compounds remains the same are known as stoichiometric defects.

The stoichiometric defects in ionic solids are majorly of two types -

1. Frenkel defect
2. Schottky defects 

Explanation:

Frenkel defects - In Frenkel defect,

• Some cations are missing from their original lattice site and occupy the interstitial site of the lattice.
• This is shown by the low radius ratio and low coordination number ionic solids.
• there is a large difference in the size of ions i.e. cations and anions in the compounds.
• The density of the crystals having Frenkel defects remain the same as no ion is removed from the crystal system, only the position of the cation (smaller ion) is changed from the lattice site to the interstitial site.
• Ex - AgBr.

→ZnS having a large difference in the constituent ions i.e Zn²⁺ and S^2- exhibit the Frenkel defect. It does not show Schottky defect as there should be a high coordination number of the ions but  Zns has a lower coordination number.

Therefore , it only shows Frenkel defect.

Conclusion:

∴ The stoichiometric defect shown by ZnS is Frenkel defect.

Hence, the correct answer is option 2.

Additional Information

Schottky defects -

• The same number of cations and anions are missing from the lattice site. 
• This is shown by the Crystals having high coordination number and a high radius ratio.
• The neutrality of the system kept maintained in Schottky's defect.
• The density of the ionic solid having Schottky defect is decreases due to vacancy created by missed cations and anions.
• AgBr and KBr show Schottky defect.