Question

Silicon is doped with which of the following to obtain p-type semiconductor ?

• A. Phosphorus
• B. Arsenic
• C. Boron
• D. Antimony
• A. $5 \times 10^{22} \text{ m}^{-3}$
• B. $1.5 \times 10^6 \text{ m}^{-3}$
• C. $9 \times 10^8 \text{ m}^{-3}$
• D. $4.5 \times 10^9 \text{ m}^{-3}$
• A. current is mainly due to drifting of majority carriers.
• B. current is mainly due to drifting of minority carriers.
• C. diffusion and drift currents are equal.
• D. current is of the order of 1 A.
• A. 0.2 V
• B. 0.5 V
• C. 0.7 V
• D. 1.5 V
• A. $0.01$ eV
• B. $0.05$ eV
• C. $0.1$ eV
• D. $0.15$ eV

Answer 1

The Correct Option is C

Concept:
• Pure elemental semiconductors like Silicon (Si) and Germanium (Ge) belong structurally to Group 14 of the periodic table and essentially have exactly 4 valence electrons.

• Doping is the intentional introduction of specific impurities into an intrinsic semiconductor for the purpose of intensely modulating its electrical properties.

• A p-type (positive-type) semiconductor is strategically created by heavily doping the pure crystal with trivalent impurities (atoms possessing only 3 valence electrons).

• A trivalent impurity atom forms exactly three covalent bonds with neighboring Si atoms, leaving a sharp vacancy or "hole" in the fourth bond, which practically acts as a positive charge carrier.

Step 1: Analyze the fundamental requirements
To successfully obtain a p-type semiconductor, we strictly need a trivalent dopant element from Group 13 of the periodic table.
Commonly utilized Group 13 elements include Boron (B), Aluminum (Al), Gallium (Ga), and Indium (In).

Step 2: Evaluate the chemical nature of the given options
We carefully check the periodic group properties of all four options provided:
(A) Phosphorus (P): This is a prominent Group 15 element. It has precisely 5 valence electrons (pentavalent). Doping with it yields an n-type semiconductor.
(B) Arsenic (As): This is also a Group 15 pentavalent element. It similarly produces an n-type semiconductor.
(C) Boron (B): This is a defining Group 13 element. It has exactly 3 valence electrons (trivalent). When heavily substituted into a silicon lattice, it successfully generates holes, creating a p-type semiconductor.
(D) Antimony (Sb): This is yet another Group 15 pentavalent element. It unequivocally produces an n-type semiconductor.

Step 3: Conclusion
Based on the thorough chemical classification, Boron is the sole trivalent impurity critically listed in the options.
Therefore, heavily doping silicon with Boron will reliably produce the desired p-type semiconductor.
This makes option (C) correct.

Answer 2

Concept:
• Under conditions of absolute thermal equilibrium, regardless of the level of chemical doping, a foundational relationship exists between the concentrations of electrons and holes.

• This strict relationship is universally known as the Mass Action Law for semiconductors.

• The highly significant mathematical formulation states that the product of majority and minority carrier concentrations is perfectly equal to the square of the intrinsic carrier concentration.

• Formula: $n_e \cdot n_h = n_i^2$, where $n_e$ is electron concentration, $n_h$ is hole concentration, and $n_i$ is intrinsic carrier concentration.

Step 1: Identify the provided physical parameters
The intrinsic carrier concentration parameter is given carefully as:
$n_i = 1.5 \times 10^{16} \text{ m}^{-3}$
The given steady-state electron concentration is heavily defined as:
$n_e = 5 \times 10^{22} \text{ m}^{-3}$
Notice importantly that $n_e \gg n_i$, clearly indicating this is a strongly n-type extrinsic semiconductor.

Step 2: Apply the Mass Action Law equation
We aim to precisely determine the unknown hole concentration ($n_h$).
Rearranging the Mass Action Law heavily for $n_h$:
\[ n_h = \frac{n_i^2}{n_e} \]
Substitute the provided heavy numerical values deeply into the numerator and denominator:
\[ n_h = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} \]

Step 3: Perform the mathematical calculation carefully
First, meticulously compute the square of the complex intrinsic concentration in the numerator:
\[ (1.5 \times 10^{16})^2 = (1.5)^2 \times (10^{16})^2 \]
\[ = 2.25 \times 10^{32} \text{ m}^{-6} \]
Now, divide this heavily by the given steady electron concentration:
\[ n_h = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} \]
Isolate the numerical fraction from the deep powers of ten:
\[ n_h = \left(\frac{2.25}{5}\right) \times 10^{(32 - 22)} \]
\[ n_h = 0.45 \times 10^{10} \]
Convert this raw result fully into proper standard scientific notation for final matching:
\[ n_h = 4.5 \times 10^{-1} \times 10^{10} = 4.5 \times 10^9 \text{ m}^{-3} \]

Step 4: Conclusion
The heavily calculated minority hole concentration is exactly $4.5 \times 10^9 \text{ m}^{-3}$.
This strictly correlates with option (D).

Answer 3

Concept:
• A deeply fundamental p-n junction features two vastly competing charge transport mechanisms: diffusion and drift.

• Diffusion current is strictly driven by the extreme concentration gradient across the junction. Majority carriers (holes from p-side, electrons from n-side) diffuse rapidly into the opposite region.

• Drift current is forcefully driven by the internally built-in electric field situated securely within the depletion region. It sweeps newly generated minority carriers across the junction.

• During forward bias, an external voltage severely opposes the built-in potential, heavily shrinking the depletion region and strongly lowering the potential barrier height.

Step 1: Analyze the internal physics of forward bias
When a p-n diode is robustly forward-biased, the external electric field aggressively pushes majority carriers heavily towards the junction.
Because the barrier height is heavily reduced, a massive number of majority carriers easily acquire sufficient energy to cross over the junction.
This results in an exponentially massive surge in the diffusion current.
Simultaneously, the heavily opposing external field slightly decreases the already minuscule drift of minority carriers.
Thus, the total forward current is overwhelmingly dominated by the massive motion of majority carriers forcefully crossing the junction.

Step 2: Evaluate the phrasing of the provided options
Let us meticulously scrutinize the given options based on semiconductor physics:
(C) suggests diffusion and drift are perfectly equal. This is purely the condition for an unbiased, open-circuit diode in thermal equilibrium, not forward bias.
(D) suggests current is strictly around 1 A. Forward currents in typical small-signal semiconductor diodes are securely in the heavily lower milliampere (mA) range, so this is generally false.
(B) suggests current is mainly due to drifting of minority carriers. This is heavily true only for reverse bias, where diffusion is completely stopped and only leakage drift occurs.
(A) claims current is mainly due to drifting of majority carriers. Physically speaking, the dominant mechanism for majority carriers crossing the lowered barrier is strictly termed "diffusion", not "drift". Drift is firmly associated with the electric field forcefully moving carriers.
However, in heavily simplified or poorly translated contexts (like translating the Hindi term 'अपवाह' loosely), "drifting" might inappropriately be used interchangeably as general "motion" or "flow".
Despite the heavily flawed usage of the strict word "drifting", the undeniable core concept intended by the examiner is that forward current is overwhelmingly driven by the massive flow of majority carriers.
Therefore, option (A) is definitely the most conceptually aligned intended answer among the highly flawed choices.

Step 3: Conclusion
While strictly speaking forward current is a "diffusion" current of majority carriers, option (A) correctly identifies that the current is heavily dominated by "majority carriers".
We securely select option (A) as the intended correct response.

Answer 4

Concept:
• The threshold voltage (also widely known as the knee voltage, cut-in voltage, or strongly built-in potential barrier) of a p-n junction diode is a highly critical parameter.

• It firmly represents the absolute minimum forward bias voltage that must be externally applied to massively overcome the internal potential barrier of the heavily depleted region.

• Below this specific threshold voltage, the heavily suppressed forward current is practically negligible.

• Once the external voltage firmly exceeds this threshold, the diffusion current rises massively and exponentially.

Step 1: Material dependency of threshold voltage
The intrinsic value of the cut-in voltage is heavily dependent on the fundamental energy bandgap of the specific semiconductor material used to fabricate the diode.
Materials with a heavily larger forbidden energy bandgap strictly require a much higher forward voltage to enable carriers to successfully cross the junction barrier.

Step 2: Comparing standard semiconductor materials
Germanium (Ge) strongly possesses a relatively small bandgap of approximately $0.66 \text{ eV}$ at room temperature.
Because of this physically smaller energy requirement, the threshold voltage for a standard Germanium diode is securely around $0.2 \text{ V}$ to $0.3 \text{ V}$.
Silicon (Si) strongly possesses a significantly larger bandgap of approximately $1.12 \text{ eV}$ at room temperature.
Consequently, establishing a sufficient forward current in Silicon heavily requires a substantially higher applied external potential.
The universally accepted, standard threshold voltage for a typical Silicon diode firmly resides at approximately $0.7 \text{ V}$.

Step 3: Evaluating the options
Looking closely at the rigidly provided options:
(A) 0.2 V roughly corresponds to heavily doped Germanium or some Schottky diodes.
(C) 0.7 V precisely matches the universally accepted physical standard for Silicon p-n junction diodes.

Step 4: Conclusion
The heavily demanded threshold voltage for a standard Silicon diode is universally recognized as approximately $0.7 \text{ V}$.
This strictly makes option (C) the firmly correct choice.

Answer 5

Concept:
• When a pentavalent elemental impurity (like Phosphorus or Arsenic) heavily dopes a Germanium (Ge) or Silicon (Si) crystal lattice, it forcefully replaces a host atom.

• Four of the five valence electrons from the impurity atom form incredibly strong covalent bonds with the four immediately adjacent host atoms.

• The critical fifth electron is left entirely unbound to any specific covalent bond. It continues to orbit the heavily localized positive ion core of the impurity atom.

• This highly specific physical setup can be modeled extremely well using a heavily modified version of the Bohr model of the Hydrogen atom.

Step 1: Understanding the modified Bohr model for semiconductors
In a pure vacuum, the ionization energy of a pristine Hydrogen atom is heavily established as $13.6 \text{ eV}$.
However, the weakly bound fifth electron in a doped semiconductor operates inside a dense crystalline medium.
This heavy crystalline environment drastically alters the physics in two major ways:
1. The electrostatic Coulomb force is heavily reduced by the large relative dielectric constant ($\epsilon_r$) of the host semiconductor crystal.
2. The electron's dynamic motion is heavily altered, so it operates with an effective mass ($m^*$) that is typically much lighter than the true free electron mass ($m_e$).
The severely modified ionization energy $E_d$ for this localized donor electron is given by the formula:
\[ E_d = 13.6 \text{ eV} \times \left( \frac{m^*}{m_e} \right) \times \left( \frac{1}{\epsilon_r^2} \right) \]

Step 2: Comparing the physical values for Silicon and Germanium
Silicon (Si) has a relative dielectric constant of roughly $\epsilon_r \approx 11.7$.
Due to this moderate dielectric screening, the calculated ionization energy required to firmly free the fifth electron in Silicon is experimentally found to be approximately $0.05 \text{ eV}$.
Germanium (Ge) operates with a significantly larger relative dielectric constant of roughly $\epsilon_r \approx 16$.
Because the dielectric constant is heavily squared in the denominator of our modified formula, the physically higher $\epsilon_r$ of Ge causes a massive reduction in the binding energy.
Additionally, the effective mass of an electron in Germanium is also structurally different.
As a direct result of these heavy crystalline factors, the ionization energy for the weakly bound fifth electron in heavily doped Germanium plunges to approximately $0.01 \text{ eV}$.

Step 3: Conclusion
The heavily researched ionization energy for liberating the fifth electron in a pentavalent-doped Germanium crystal firmly sits at about $0.01 \text{ eV}$.
This strictly matches the numerical value provided in option (A).