Semiconductor Electronics – NEET UG PhysicsPractice Questions & PYQs

A body-centred cubic crystal structure features constituent particles located at each of the eight corners of the unit cell and a single particle situated entirely within the interior center. Each corner atom is shared among eight adjacent unit cells, contributing a fractional share of $\frac{1}{8}$ to the specific cell, which amounts to one full atom from the corners combined. Adding this to the single unshared atom located at the body center results in a total count of two atoms per unit cell.

The common emitter current gain is expressed as the ratio of collector current to base current, denoted by $\beta = \frac{I_{C}}{I_{B}}$. Since the emitter current is the sum of the base and collector currents, the base current is given by $I_{E} - I_{C}$. Substituting the reciprocal of the given collector-to-emitter ratio into the rearranged formula $\beta = \frac{I_{C}}{I_{E} - I_{C}}$ yields $\frac{0.96}{0.04}$, which evaluates to twenty-four.

A digital logic gate that produces a low output exclusively when all of its inputs are high implements the inverse operation of a conjunction. Examining the provided truth table reveals that the output remains high for all input combinations except when both inputs are simultaneously high, which results in a zero. This specific operational behavior corresponds to a circuit combining an intersection operation followed by a logical negation.

Diodes function as directional switches that block current when reverse biased and conduct current when forward biased. In this specific configuration, diode $D_1$ is reverse biased and blocks current flow through its respective branch, while diode $D_2$ is forward biased and allows current to pass. Consequently, the active loop includes the 30 ohm and 20 ohm resistors connected in series with the 5 volt source. Calculating the total current through this active path gives the formula $I = \frac{5}{30+20}$, which simplifies to the final equation $I = \frac{5}{50}$.

The current gain in a common emitter configuration ($\beta$) is related to the current gain in a common base configuration ($\alpha$) through the relation $\beta = \frac{\alpha}{1-\alpha}$. Given that the ratio of collector current to emitter current represents $\alpha$, substituting 0.98 into the formula yields $\beta = \frac{0.98}{1 - 0.98}$. Simplifying this expression results in a value of 49.

Ionic crystals are formed by the electrostatic attraction between positively charged metal cations and negatively charged non-metal anions. To maximize attractive forces and minimize repulsive interactions between like charges, these constituent ions occupy alternating lattice sites throughout the crystal structure. This periodic, alternating arrangement creates the characteristic three-dimensional repeating pattern seen in typical ionic solids.

In a body-centered cubic crystal lattice, the lattice parameter and atomic radius are related by the formula $a = \frac{4r}{\sqrt{3}}$. The distance between two nearest neighbor atoms in this structure equals twice the atomic radius, giving a radius of 1.85. Substituting this value into the equation yields $a = \frac{4(1.85)}{\sqrt{3}}$, which evaluates to a lattice parameter of 4.3 Å.

Transistor current gains are defined by the ratios of terminal currents, where $\alpha = \frac{I_C}{I_E}$ and $\beta = \frac{I_C}{I_B}$. Since the emitter current equals the sum of collector and base currents, the base current can be expressed as $I_B = I_E - I_C$. Substituting this into the definition of $\beta$ and dividing the numerator and denominator by $I_E$ directly yields the relation $\beta = \frac{\alpha}{1-\alpha}$.

A p-n junction diode enters a forward-biased state when its p-type terminal is held at a positive potential with respect to its n-type terminal. By inspecting the terminal polarities across the choices, this specific condition is uniquely fulfilled in circuit A. This proper voltage alignment narrows the depletion region and permits easy current flow.

The distance between the body-centered atom and a corner atom in a body-centered cubic lattice corresponds to half of the body diagonal, determined by the formula $\frac{\sqrt{3}a}{2}$. Multiplying the square root of three by the given edge length of 4.225 Å and dividing the product by two yields the final calculated spacing. This arithmetic operation evaluates to a distance of 3.66 Å.