Q.In an unbiased p-n junction, holes diffuse from the p-region to n-region because

1. What happens the instant a p-n junction is formed

The n-side has a large concentration of free electrons; the p-side has a large concentration of holes. Because of this huge concentration gradient:

• Electrons diffuse from n $\rightarrow$ p.
• Holes diffuse from p $\rightarrow$ n.

2. Why the region becomes empty of carriers

A carrier that crosses the junction meets the opposite type of carrier in overwhelming numbers and recombines. So near the junction:

• On the n-side, electrons leave behind immobile positive donor ions ($\mathrm{N_D^+}$).
• On the p-side, holes leave behind immobile negative acceptor ions ($\mathrm{N_A^-}$).

These ions are locked in the crystal lattice — they are not free charge carriers.

3. Self-limiting nature

The exposed ions set up an internal electric field (n $\rightarrow$ p) and hence a barrier potential $V_B$ (≈ 0.7 V for Si, 0.3 V for Ge). This field opposes further diffusion, so the layer stops growing at a definite width — typically a fraction of a micrometre. …

The key idea is that Ohm’s law is a local relation, not just a macroscopic one for a whole wire. Inside a conductor, the current density $\vec{j}$ (current per unit area) is directly proportional to the electric field $\vec{E}$ that drives it. The proportionality constant is the conductivity $\sigma$, so $\vec{j} = \sigma \vec{E}$. Since resistivity $\rho$ is the reciprocal of conductivity ($\rho = 1/\sigma$), you can also write $\vec{E} = \rho \vec{j}$.

Let’s see why the other options are wrong and why (C) is correct.

1. Start with the microscopic form of Ohm’s law.

   For a conducting material, the drift velocity of charge carriers is proportional to the applied electric field. This gives $\vec{j} = \sigma \vec{E}$. Here $\sigma$ is the conductivity — a material property that tells you how easily current flows. A high $\sigma$ means a small field produces a large current density.

2. Rewrite using resistivity.

   Resistivity $\rho$ is defined as $\rho = 1/\sigma$. So from $\vec{j} = \sigma \vec{E}$, multiply both sides by $\rho$:

Hence $\vec{E} = \rho \vec{j}$. This is exactly option (C).

3. Check the other options.
   • (A) $\vec{E} = \sigma \vec{j}$ would mean the field equals conductivity times current density. But $\sigma$ is large for good conductors, so this would imply a huge field for a modest current — opposite to reality. Also, units don’t match: $\sigma$ has units $(\Omega \cdot \text{m})^{-1}$, so $\sigma \vec{j}$ has units of $\text{A/m}^2 \cdot (\Omega \cdot \text{m})^{-1} = \text{V/m}^2$, not V/m. …

The key idea here is that for a p-n junction, the depletion region behaves approximately like a parallel-plate capacitor with a uniform electric field. The potential difference across the depletion layer is simply the product of the electric field and the thickness, provided the field is constant. That’s a standard approximation for an abrupt junction under low bias.

1. Identify the given quantities

   • Thickness of depletion layer: $d = 1\,\mu\text{m} = 1 \times 10^{-6}\,\text{m}$
   • Potential across it: $V = 0.05\,\text{V}$

2. Recall the relation between electric field, potential, and distance

   For a uniform field, the potential difference is

This is because the electric field is the negative gradient of potential, and for constant field, the change in potential over a distance $d$ is simply $E d$.

3. Solve for the electric field

Since $1\,\text{V/m} = 1\,\text{N/C}$, the electric field is $5 \times 10^{4}\,\text{N/C}$.

4. Match with the options …

Step 1 — What a wire-bound resistor has to do.

A resistor is a standard: its value must stay put. Two properties are demanded of the wire:

1. High resistivity $\rho$, so that a usefully large $R=\rho L/A$ fits in a small coil;
2. A very small temperature coefficient of resistivity $\alpha$, where

Current through the resistor dissipates $I^2R$ of heat and warms it; if $\alpha$ were large the resistance would drift with the current, and the resistor would be useless as a fixed component.

Step 2 — Which materials qualify.

• Manganin (Cu–Mn–Ni), constantan (Cu–Ni) and nichrome (Ni–Cr) all have high $\rho$ and $\alpha$ of the order of $10^{-5}\ \mathrm{K^{-1}}$ or less — essentially flat with temperature. These are exactly the alloys NCERT names for wire-bound resistors, standard resistance coils and metre-bridge/potentiometer wires.

Step 3 — Reject the others.

• (A) Ge, Au, Ga: germanium is a semiconductor — its resistance falls sharply with temperature (large negative $\alpha$), the worst possible behaviour for a standard resistor; gold is also far too expensive. …

The question asks what physical principle Kirchhoff’s junction rule reflects. The rule itself states: at any junction in an electrical circuit, the sum of currents entering the junction equals the sum of currents leaving the junction. That is, $\sum I_{\text{in}} = \sum I_{\text{out}}$.

Why does this hold? Current is the flow of charge per unit time. If more charge flowed into a junction than flowed out, charge would pile up there. In a steady-state circuit (the kind we analyse with Kirchhoff’s rules), no charge accumulates anywhere — the charge density at every point is constant in time. So the net rate at which charge enters a junction must be zero, which is exactly the junction rule.

This is a direct statement of conservation of charge, not of energy, momentum, or current density. Let’s see why the other options are wrong.

1. Conservation of current density vector — Current density $\vec{J}$ is a vector field, but the junction rule deals with scalar currents (the flow along wires). There is no separate “conservation of current density” law; current density itself is not a conserved quantity. The rule is about the integral of current density over cross-sections, which reduces to charge conservation.

2. Conservation of energy — That is the basis of Kirchhoff’s loop rule (the sum of potential differences around a closed loop is zero). The junction rule has nothing to do with energy; it applies even in circuits with no energy sources.

3. Conservation of momentum — Momentum conservation is relevant in mechanics and particle collisions, not in the steady flow of charge through wires. There is no momentum principle behind the junction rule. …

Step 1 — The governing relation.

For a conductor of cross-section $A$ carrying free electrons of number density $n$ drifting with speed $v_d$,

This comes from counting the charge crossing a cross-section in time $\Delta t$: the electrons within a cylinder of length $v_d\Delta t$ get through, and that cylinder holds $n A v_d \Delta t$ electrons, each carrying charge $e$.

Step 2 — Put in the actual magnitudes.

• $e=1.6\times10^{-19}\ \text{C}$ — extremely small.
• $v_d\sim 10^{-4}\ \text{m s}^{-1}$ (a fraction of a mm per second) — extremely small.
• $n\sim 10^{29}\ \text{m}^{-3}$ for a typical metal like copper (about one free electron per atom) — enormous.

Step 3 — Multiply.

For a wire of area $A\sim10^{-6}\ \text{m}^2$:

an everyday current. The $10^{29}$ is what defeats the $10^{-19}$ and the $10^{-4}$.

Step 4 — Why the other options fail. …

Step 1 — The defining relation.

For a transistor in common-emitter configuration the dc current gain is

This relation must hold whatever the biasing, so it is the strongest test we can apply to the options.

Step 2 — Test each option against $I_C = 50 I_B$.

• (A) $I_B = 10\ \mu\mathrm{A} = 10\times10^{-6}$ A $\Rightarrow 50 I_B = 500\times10^{-6}\ \mathrm{A} = 0.5\ \mathrm{mA}$. This matches the quoted $I_C = 0.5$ mA. ✓
• (B) $I_B = 0.5\ \mu$A $\Rightarrow 50I_B = 25\ \mu\mathrm{A} = 0.025$ mA, not 10 mA. ✗
• (C) $I_B = 5\ \mu$A $\Rightarrow 50I_B = 250\ \mu\mathrm{A} = 0.25$ mA, not 1 mA. ✗
• (D) $I_B = 1\ \mu$A $\Rightarrow 50I_B = 50\ \mu\mathrm{A} = 0.05$ mA, not 0.5 mA. ✗

Only option (A) is internally consistent with $\beta = 50$.

Step 3 — Cross-check with the output loop (load line).

The collector–emitter loop gives

$V_{CC} = V_{CE} + I_C R_C .$ …