Q.Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to $(E_g)_C$, $(E_g)_{Si}$ and $(E_g)_{Ge}$. Which of the following statements is true?

Concept & Intuition

In an intrinsic semiconductor, electrons and holes are created in equal numbers: $n_i = p_i$. When we dope the material, we disturb this balance — but in thermal equilibrium, the product $n \cdot p$ remains constant and equals $n_i^2$. This is the law of mass action for semiconductors, analogous to the equilibrium constant in chemistry. So even after doping, if we know the new electron concentration, we can directly find the hole concentration using $p = n_i^2 / n$.

1. Identify the intrinsic carrier concentration

   The problem gives $n_i = 6 \times 10^{15} \ \mathrm{m}^{-3}$ for the intrinsic (undoped) semiconductor.

2. State the law of mass action

   In thermal equilibrium, for a non-degenerate semiconductor:

This holds regardless of doping.

3. Plug in the doped electron concentration After doping, $n = 4 \times 10^{22} \ \mathrm{m}^{-3}$. So:

4. Calculate step by step First, square the intrinsic concentration:

Then divide:

5. Match with the options The result $9 \times 10^{8} \ \mathrm{m}^{-3}$ corresponds exactly to option (C).

The key concept here is charge neutrality in a semiconductor. Even though we call it "n-type" because electrons are the majority carriers, the material as a whole must remain electrically neutral. Doping introduces extra electrons, but it also introduces an equal number of positive charges from the ionized donor atoms. The net charge is zero.

Let’s walk through the reasoning step by step:

1. Understand what doping does. In an n-type semiconductor, we add donor atoms (like phosphorus in silicon). Each donor atom has five valence electrons; four bond with neighboring silicon atoms, and the fifth becomes a free electron. This leaves behind a positively charged donor ion (a fixed positive charge in the crystal lattice).

2. Count the charges. For every free electron added by doping, there is exactly one positive donor ion created. So the number of free electrons (negative charges) equals the number of ionized donors (positive charges). The holes (minority carriers) are negligible in number compared to the electrons and donors.

3. Apply charge neutrality. The total charge density in a semiconductor is given by:

where $q$ is the elementary charge, $p$ is hole concentration, $n$ is electron concentration, $N_D^+$ is ionized donor concentration, and $N_A^-$ is ionized acceptor concentration. For an n-type semiconductor with no acceptors ($N_A^- = 0$), and at typical temperatures where all donors are ionized, we have $n \approx N_D^+$ and $p \ll n$. Thus:

The small number of holes ($p$) is balanced by a tiny excess of electrons, but overall the net charge is zero.

4. Why the common pitfall? Many students think "n-type means extra electrons, so it must be negative." But they forget that the extra electrons come from donor atoms that become positively charged. The material is like a neutral atom: equal numbers of protons and electrons. The dopant atoms are part of the crystal, not external charges.

The key idea here is that the total current in a semiconductor is the sum of electron and hole currents. Each current depends on the concentration of the charge carrier, its charge, and its drift velocity. By writing the ratio of these currents and substituting the given ratios, we can solve for the unknown concentration ratio.

Why this works:

In a semiconductor under an electric field, the current due to electrons is $I_e = n e v_e A$ and due to holes is $I_h = p e v_h A$, where $n$ and $p$ are concentrations, $e$ is the elementary charge, $v_e$ and $v_h$ are drift velocities, and $A$ is the cross-sectional area. Since the area and charge cancel when taking the ratio, we get a simple relation between current ratio, concentration ratio, and drift velocity ratio.

1. Write the ratio of currents: Given $\dfrac{I_e}{I_h} = \dfrac{7}{4}$. Using $I_e = n e v_e A$ and $I_h = p e v_h A$, we have:

2. Substitute the known ratio of drift velocities: Given $\dfrac{v_e}{v_h} = \dfrac{5}{4}$, so:

3. Solve for the concentration ratio: Multiply both sides by $\dfrac{4}{5}$:

Thus, the ratio of concentrations of electrons to holes is $7:5$.

Concept & Intuition

In an intrinsic (pure) semiconductor, the number of electrons ($n$) equals the number of holes ($p$), both denoted $n_i$ (intrinsic carrier concentration). When we add donor impurities (like phosphorus in silicon), each donor atom contributes an extra electron to the conduction band. This directly raises the electron concentration $n$. But the product $n \cdot p$ is a constant at a given temperature (the law of mass action: $n p = n_i^2$). So if $n$ goes up, $p$ must go down to keep the product constant. The result: more electrons, fewer holes.

Step-by-step reasoning

1. Recall the law of mass action In thermal equilibrium, for a non-degenerate semiconductor:

where $n_i$ depends only on temperature and the material’s band gap. This relation holds regardless of doping.

2. Effect of donor doping on electron concentration Donor atoms are ionized at room temperature, releasing electrons. For moderate doping, nearly all donors are ionized, so:

(where $N_D$ is the donor concentration), assuming $N_D \gg n_i$. Thus $n$ increases significantly above $n_i$.

3. Consequence for hole concentration From the mass-action law:

Since $n$ has increased, $p$ must decrease below $n_i$. The more donors added, the smaller $p$ becomes.

4. Compare with the options
   • (A) hole concentration decreases, electron concentration increases — matches our reasoning.
   • (B) hole increases, electron decreases — opposite of donor doping (that would be acceptor doping).
   • (C) both increase — impossible because $n p$ would exceed $n_i^2$.
   • (D) both decrease — also impossible; at least one carrier type must increase to maintain the product.

Concept. In the photoelectric effect, $eV_0 = h\nu - \phi$: the stopping potential $V_0$ is fixed by the radiation frequency and the work function $\phi$ of the material, and is completely independent of intensity. Intensity controls only the saturation current.

Reading the graph.

• Curves 1 and 2 start from the same (more negative) stopping-potential intercept $\Rightarrow$ same material/work function, with curve 1 at higher intensity (higher saturation current) than curve 2.
• Curves 3 and 4 start from the same (less negative) intercept $\Rightarrow$ they form the second same-material pair, again at two intensities.

Matching to the options. The same-material pairs are (1, 2) and (3, 4). Of the offered choices, only "3 and 4" is such a pair; "1 and 3", "1 and 4", "2 and 3" mix the two different stopping potentials, i.e. different materials.

ANSWER: D

The question tests three independent ideas from current electricity: the temperature dependence of resistivity, the effect of stretching a wire on its resistance, and the definition of drift velocity. Each statement must be judged on its own merit.

Let’s examine them one by one.

1. Statement I: “Specific resistance depends on nature of material and independent of temperature of the material.”

   Specific resistance is another name for resistivity ($\rho$). Resistivity does depend on the material’s nature (e.g., copper vs. nichrome), but it is not independent of temperature. For most conductors, $\rho$ increases with temperature; for semiconductors, it decreases. The statement is therefore false.

2. Statement II: “A wire of resistance $6~\Omega$ is drawn out so that its new length is four times its original length. The resistance of the new wire is $48~\Omega$.”

   When a wire is stretched, its volume remains constant. If length becomes $4$ times, then cross-sectional area becomes $1/4$ times (since $V = A L$ constant). Resistance $R = \rho \frac{L}{A}$.

   Original: $R_0 = \rho \frac{L_0}{A_0} = 6~\Omega$.

   New: $L = 4L_0$, $A = A_0/4$.

   So $R = \rho \frac{4L_0}{A_0/4} = \rho \frac{16 L_0}{A_0} = 16 \times 6 = 96~\Omega$, not $48~\Omega$.

   The statement is false.

3. Statement III: “Drift velocity is the average constant velocity acquired by free electrons inside a metal by the application of an electric field which results in current.”

   This is the standard definition. Electrons in a metal move randomly at high speeds, but an applied electric field gives them a small net drift in the opposite direction. The average of this net velocity is the drift velocity, which is constant for a steady field and current. The statement is true.

Thus, only Statement III is correct.

The key idea here is that for an intrinsic semiconductor, the conductivity depends on both electrons and holes. Since the material is pure, the electron concentration equals the hole concentration, both equal to $n_i$. The total conductivity is then $\sigma = n_i e (\mu_n + \mu_e)$, and resistivity is just its reciprocal.

Let’s work through it step by step.

1. Identify the formula for conductivity in an intrinsic semiconductor. In an intrinsic semiconductor, $n = p = n_i$. The conductivity is

where $e = 1.6 \times 10^{-19}\ \text{C}$ is the electronic charge.

2. Plug in the given values. $n_i = 2.5 \times 10^{19}\ \text{m}^{-3}$, $\mu_n = 0.15\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$, $\mu_e = 0.35\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$ (note: $\mu_e$ here is the hole mobility, often denoted $\mu_p$).

3. Resistivity is the reciprocal of conductivity.

4. Convert to the units given in the options. Options are in $\Omega\ \text{cm}$, $\Omega\ \text{mm}$, $\Omega\ \text{m}$. $0.5\ \Omega\text{m} = 50\ \Omega\text{cm}$ (since $1\ \text{m} = 100\ \text{cm}$, so $0.5\ \Omega\text{m} = 0.5 \times 100 = 50\ \Omega\text{cm}$). Also $0.5\ \Omega\text{m} = 500\ \Omega\text{mm}$ (since $1\ \text{m} = 1000\ \text{mm}$), but that’s not among the options. So the matching option is $50\ \Omega\text{cm}$.