Solid State Physics, Devices and Electronics MCQ Quiz in मल्याळम - Objective Question with Answer for Solid State Physics, Devices and Electronics - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

• The depletion region in a semiconductor diode is the region where there are no free charge carriers (neither electrons nor holes).
• In an unbiased (or zero bias) condition, the free electrons from the n-type region and holes from the p-type region combine (recombine) at the junction. This results in the depletion of charge carriers near the junction, forming the depletion region. The depletion region contains only immobile ions, with no free charge carriers.
• Hence, the depletion region consists of neither free electrons nor holes.

Explanation of Options:

• Option 1: Incorrect. The depletion region does not contain free electrons and holes. The free charge carriers recombine to form the depletion region.
• Option 2: Correct. The depletion region contains neither free electrons nor holes, as the charge carriers recombine in this region.
• Option 3: Incorrect. The depletion region does not contain only free electrons.
• Option 4: Incorrect. The depletion region does not contain only holes.

∴ The correct answer is: Option 2 (neither free electrons nor holes).

Solution:

The problem involves determining the electronic pressure in a white dwarf star at $T=0K$, sing the average energy per electron, ${ϵ}_0$.

Average Energy per Electron:

${ϵ}_0=\frac{3{\hslash }^2}{10m}{\left(3{\pi }^{2}n\right)}^{2/3},$

 where;

$n=\frac{N}{V}$ is the electron density, $\hslash$ is the reduced Planck's constant, and m  is the electron mass. 

Fermi Pressure Relation:

 The pressure due to electrons at $T=0K$ is derived from the energy density $U=n{ϵ}_0$ using the thermodynamic relation for a degenerate Fermi gas: 

$P=\frac{2}{3}U.$

Energy Density:

$U=n{ϵ}_0.$

Electronic Pressure:

Using the relation $P=\frac{2}{3}U$

$P=\frac{2}{3}n{ϵ}_0.$

The electronic pressure in the star is:  $\boxed{{\displaystyle \frac{2}{3}n{ϵ}_0}}$

Thus, option '4' is correct.

Explanation:

To accurately determine the curve representing the variation of conductivity (σ) with temperature (T) for a metal, the relationship typically follows the behavior that conductivity decreases as the temperature increases. This is because, in metals, as temperature increases, the thermal vibrations of atoms increase, leading to more scattering of electrons, which reduces conductivity.

Therefore, the schematic curve for conductivity (σ) as a function of temperature (T) would show a decreasing trend. It should be a curve that slopes downward as the temperature rises.

Thus, option '2' is correct.

Solution:

To determine the Miller indices for a plane in a cubic lattice with intercepts at $a,\frac{a}{2}\text{ and }\frac{2a}{3}$ along the x, y and z-axes respectively, follow these steps:

Identify the intercepts:

x-intercepts: a

y-intercepts: $\frac{a}{2}$

z-intercept: $\frac{2a}{3}$

Calculate the reciprocals of the intercepts:

Reciprocal of a: $\frac{1}{a}$

Reciprocal of $\frac{a}{2}$:$\frac{2}{a}$

Reciprocal of $\frac{2a}{3}$: $\frac{3}{2a}$

Multiply each reciprocal by $a$ to eliminate the variable $a$:

$\frac{1}{a}\times a=1$

$\frac{2}{a}\times a=2$

$\frac{3}{2a}\times a=\frac{3}{2}$

then:

$1\times 2=2$, $2\times 2=4$ and $\frac{3}{2}\times 2=3$

The Miller indices are (2, 4, 3).

Thus, option '1' is correct.

Explanation:

Circuit Description:

• The circuit includes an AC voltage source $V_{in}$ with a peak voltage of $V_p=2V$  connected to two silicon diodes in opposite directions.
• Silicon diodes have a forward voltage drop of approximately 0.7V." id="MathJax-Element-283-Frame" role="presentation" style="position: relative;" tabindex="0">0.7V.0.7V." id="MathJax-Element-62-Frame" role="presentation" style="position: relative;" tabindex="0">0.7V.$\mathrm{0.7V.}$ 0.7V. 0.7V.

Waveform Analysis:

• When the input voltage exceeds +0.7V,  the diode connected in the forward direction conducts, clamping the output voltage $V_{PQ}$ to approximately +0.7V.
• Similarly, when the input voltage goes below -0.7V, the other diode conducts, clamping $V_{PQ}$ to approximately -0.7V.
• For any input voltage between -0.7V and +0.7V, neither diode conducts, and the output voltage follows the input voltage.

Output Waveform:

The output waveform will be clamped between +0.7V and -0.7V.

Thus, option '3' is correct.

Calculation:

 To calculate the Fermi energy of a metal, we use the formula:

$E_F=\frac{{\hslash }^2}{2m_e}{\left(3{\pi }^{2}n\right)}^{\frac{2}{3}}$

 Where:

$\hslash =\frac{h}{2\pi }$ is the reduced Planck's constant,

$m_e$  is the mass of the electron,

$n$ is the electron density,

$E_F$ is the Fermi energy.

Given:

$\begin{array}{rl}h& =6.626\times {10}^{-34}\text{J·s},\\ m_e& =9.11\times {10}^{-31}\text{kg},\\ n& =6.4\times {10}^{28}{\text{m}}^{-3},\\ 1\text{eV}& =1.6\times {10}^{-19}\text{J}.\end{array}$

Substitute the values into the formula:

$E_F=\frac{{\hslash }^2}{2m_e}{\left(3{\pi }^{2}n\right)}^{\frac{2}{3}}$ we get:

$E_F=\frac{(1.055\times {10}^{-34}{)}^2}{2\cdot 9.11\times {10}^{-31}}{(3{\pi }^2\cdot 6.4\times {10}^{28})}^{\frac{2}{3}}$

$E_F=\frac{9.34\times {10}^{-19}}{1.6\times {10}^{-19}}=5.8\text{eV}.$

Thus, the correct answer is: 5.8eV

Calculation:

Given:

• AC voltage: $V_{\text{rms}}=24{\text{V}}_{\text{rms}}$
• The time constant RC is much greater than the input signal period, implying that the capacitor provides excellent filtering and the output voltage will have minimal ripple. The diode is ideal.
• The diode is ideal.
• Resistance R is large.

Step 1: Peak value of the input AC voltage

The peak voltage $V_{\text{peak}}$ of an AC signal is related to its RMS value by:

$V_{\text{peak}}=\sqrt{2}\cdot V_{\text{rms}}$

Substituting the given value:

$V_{\text{peak}}=\sqrt{2}\cdot 24\approx 33.94\text{V}.$

Step 2: DC output voltage across R .

For a $\mathbf{\text{full-wave rectifier}}$  with a capacitor filter, the DC output voltage is approximately twice the peak voltage:

$V_{\text{dc}}\approx 2\cdot V_{\text{peak}}$

Substituting $V_{\text{peak}}$:

$V_{\text{dc}}\approx 2\cdot 33.94=67.88\text{V}$

The final answer is: 67.9V

Calculation:

The problem involves finding the angular positions of diffraction minima for light of two different wavelengths, where only one of the colors is observed at specific angles.

Step 1: Diffraction condition for minima

he condition for the minima in a single-slit diffraction pattern is given by:

$a\sin \theta =m\lambda ,$

where;

$\theta$ is the angle at which the minimum occurs,

a is the width of the slit, 

m is the order of the minimum $m=1,2,3,\dots$

$\lambda$ is the wavelength of light.

Step 2: Finding the angular positions

For the smallest angle where only orange light is observed, the blue light minima must not overlap. This happens when $m_{\text{orange}}$ and $m_{\text{blue}}$ correspond to different orders. 

For orange light:

$a\sin {\theta }_1=m_{\text{orange}}{\lambda }_{\text{orange}},$

where;

 ${\lambda }_{\text{orange}}=600\text{nm}=600\times {10}^{-9}\text{m}$ and $a=18\mu \text{m}=18\times {10}^{-6}\text{m}$

For blue light:

$a\sin {\theta }_2=m_{\text{blue}}{\lambda }_{\text{blue}},$

where;

${\lambda }_{\text{blue}}=450\text{nm}=450\times {10}^{-9}\text{m}$

At these specific angles, we require:

$m_{\text{blue}}\ne m_{\text{orange}}.$

Let us calculate for $m=1$ (first-order minima):

Angular position for orange light $m_{\text{orange}}=1$ :

$\sin {\theta }_1=\frac{m_{\text{orange}}{\lambda }_{\text{orange}}}{a}=\frac{1\times 600\times {10}^{-9}}{18\times {10}^{-6}}=0.03333.$

${\theta }_1=\arcsin (0.03333)\approx {1.91}^{\circ }.$

Angular position for blue light $\text{blue}=1$:

$\sin {\theta }_2=\frac{m_{\text{blue}}{\lambda }_{\text{blue}}}{a}=\frac{1\times 450\times {10}^{-9}}{18\times {10}^{-6}}=0.025.$

${\theta }_2=\arcsin (0.025)\approx {1.43}^{\circ }.$

Step 3: Angular difference

The angular difference between the minima is: 

${\theta }_1-{\theta }_2=-{1.43}^{\circ }+{1.91}^{\circ }\approx {0.48}^{\circ }.$

Calculation:

To find the angle at which the first-order maximum occurs, we use Bragg's Law: 

$n\lambda =2d\sin \theta$

• n=1 is the order of diffraction.
• $\lambda =0.141nm$the wavelength of the X-ray.
• d is the interplanar spacing,
• $\theta$ is the diffraction angle.

For the NaCl crystal, the interplanar spacing d is related to the lattice constant a by:

$d=\frac{a}{\sqrt{h^2+k^2+l^2}}$

 For the (111) plane, $h=1,k=1,l=1$. Thus:

$d=\frac{a}{\sqrt{1^2+1^2+1^2}}=\frac{a}{\sqrt{3}}$

Substituting the lattice constant $a=0.563\text{nm}$.

$d=\frac{0.563}{\sqrt{3}}\approx 0.325\text{nm}$

Now, substituting into Bragg's Law:

$\sin \theta =\frac{n\lambda }{2d}$

Substituting $n=1,\lambda =0.141\text{nm },\text{and }d=0.325\text{nm}$:

$\sin \theta =\frac{1\times 0.141}{2\times 0.325}=\frac{0.141}{0.65}\approx 0.2169$

To find $\theta$ take the inverse sine:

$\theta =\arcsin (0.2169)$

$\theta \approx {12.53}^{\circ }$

The angle at which the first-order maximum occurs is:$\theta \approx {12.53}^{\circ }$

Explanation:

Let's analyze each option:

1. Body Centered Cubic (BCC) – 8:

In a Body Centered Cubic structure, the central atom is surrounded by 8 nearest neighbors. So, this statement is correct.

2. Face Centered Cubic (FCC) – 6:

In a Face Centered Cubic structure, each atom has 12 nearest neighbors (not 6). So, this statement is incorrect.

3. Diamond – 4:

In a diamond structure, each carbon atom forms four covalent bonds in a tetrahedral arrangement, so each atom has 4 nearest neighbors. This statement is correct.

Hexagonal Closed Packed (HCP) – 12:

In the Hexagonal Closed Packed structure, each atom has 12 nearest neighbors. So, this statement is correct.

Thus, Correct options: 1, 3, and 4.