Energy bands - valence band and conduction band: Exercises

If any symbol here is unfamiliar, revisit the parent note first — this page assumes only what is listed above.

Level 1 — Recognition

L1·Q1 — Name the three regions

A vertical energy diagram shows three horizontal strips stacked top to bottom. The bottom strip is fully shaded (all states occupied), the middle strip is blank with no states drawn in it, and the top strip is empty (states drawn but unoccupied). Name each strip and the quantity separating the bottom two occupied/empty edges.

Figure — Energy bands - valence band and conduction band

Recall Solution

Read the picture from bottom to top (mnemonic "Very Good Cars"):

Bottom, fully shaded → the valence band (VB). Its top edge is called $E_{v}$ .
Middle, no states at all → the band gap, a forbidden zone. No electron may have an energy here.
Top, empty → the conduction band (CB). Its bottom edge is $E_{c}$ .
The separation between the two edges is the band gap $E_{g} = E_{c} - E_{v}$ .

What it looks like: the shaded rectangle is the "full parking floor," the blank middle is the "floor with no ramp," the empty top is the "open floor with room to drive."

L1·Q2 — Classify by gap size

Sort these by band picture into metal / semiconductor / insulator: (a) $E_{g} = 0$ with a partly-filled top band, (b) $E_{g} = 1.12 eV$ , (c) $E_{g} = 5.5 eV$ .

Recall Solution

(a) No effective gap / partly-filled band → metal. Empty states sit immediately above filled ones, so any tiny field moves electrons.
(b) $1.12 eV$ is in the $0.5$ – $2 eV$ window → semiconductor (this is silicon).
(c) $5.5 eV > 5 eV$ → insulator (this is diamond). The jump is far too big for room-temperature heat to lift electrons across.

Level 2 — Application

L2·Q1 — Photon absorption cutoff of silicon

What is the longest wavelength $λ_{m a x}$ of light that silicon ( $E_{g} = 1.12 eV$ ) can absorb by lifting a valence electron into the conduction band? Is $850 nm$ infrared absorbed?

Recall Solution

WHY this tool: a photon can only excite an electron across the gap if it carries at least the gap energy, $E_{photon} \geq E_{g}$ . The photon energy comes from $E = h c / λ$ , so the largest $λ$ (lowest energy) that still works is exactly where $E_{photon} = E_{g}$ .

Set them equal and solve for $λ$ : $E_{g} = \frac{h c}{λ _{m a x}} \Rightarrow λ_{m a x} = \frac{h c}{E _{g}} = \frac{1240 eV\cdotnm}{1.12 eV} \approx 1107 nm .$

What we did: rearranged $E = h c / λ$ for $λ$ .
Why: longer wavelength = lower energy; the threshold sits where energy just equals $E_{g}$ .

Check $850 nm$ : $E = 1240/850 = 1.46 eV > 1.12 eV$ → yes, absorbed. Any $λ < 1107 nm$ is absorbed; anything longer passes through (Si is transparent to deep infrared).

L2·Q2 — Gap needed for a green LED

A light-emitting diode emits green light at $λ = 520 nm$ when a conduction electron drops across the gap and releases one photon. What band gap $E_{g}$ does the emitting material need?

Recall Solution

WHY: in emission the electron falls from $E_{c}$ to $E_{v}$ , releasing exactly $E_{g}$ as one photon. So $E_{photon} = E_{g}$ here (equality, not inequality, because emission gives back precisely the gap). $E_{g} = \frac{h c}{λ} = \frac{1240}{520} \approx 2.38 eV .$ A material with $E_{g} \approx 2.4 eV$ (e.g. GaP-based alloys) emits green.

Level 3 — Analysis

L3·Q1 — Carrier ratio between two temperatures

For silicon, by what factor does the exponential part of $n_{i}$ grow when temperature rises from $300 K$ to $400 K$ ? At $300 K$ , $k_{B} T = 0.02585 eV$ ; scale $k_{B} T$ linearly with $T$ .

Recall Solution

WHY only the exponential: $n_{i} = N_{c} N_{v} e^{- E_{g} / (2 k_{B} T)}$ . The prefactor changes slowly (a power of $T$ ), but the exponential changes dramatically, so it dominates the ratio.

Step 1 — get $k_{B} T$ at $400 K$ by scaling: $k_{B} T (400) = 0.02585 \times \frac{400}{300} = 0.03447 eV$ .

Step 2 — form the ratio (the $N_{c} N_{v}$ cancels): $\frac{e ^{- E_{g} / (2 k_{B} T_{2})}}{e ^{- E_{g} / (2 k_{B} T_{1})}} = exp [\frac{E _{g}}{2} (\frac{1}{k _{B} T _{1}} - \frac{1}{k _{B} T _{2}})] .$

What we did: used $e^{a} / e^{b} = e^{a - b}$ so the two exponents subtract.

Step 3 — plug in $E_{g} = 1.12$ : $exp [\frac{1.12}{2} (\frac{1}{0.02585} - \frac{1}{0.03447})] = exp [0.56 (38.69 - 29.01)] = e^{5.42} \approx 226.$ So the exponential factor grows by roughly $\sim 2 \times 1 0^{2}$ — a $100 K$ rise multiplies intrinsic carriers by hundreds. This is why leakage current in devices climbs sharply as they heat up.

L3·Q2 — Ge vs Si intrinsic ratio (redo from the exponent)

Confirm the parent note's claim that Ge has $\sim 7 \times 1 0^{3}$ times more intrinsic carriers than Si at $300 K$ , using only the exponential factor.

Recall Solution

WHY: smaller gap → less energy to climb → exponentially more carriers. The gaps differ, and the gap lives in the exponent. $\frac{n _{i} ( Ge )}{n _{i} ( Si )} \approx \frac{e ^{- 0.66/ (2 \cdot 0.02585)}}{e ^{- 1.12/ (2 \cdot 0.02585)}} = e^{(1.12 - 0.66) / (0.0517)} = e^{0.46/0.0517} = e^{8.90} \approx 7.3 \times 1 0^{3} .$ Note the larger gap sits on top after subtraction, giving a positive exponent → Ge wins by thousands. Matches the parent note.

Level 4 — Synthesis

L4·Q1 — Absolute $n_{i}$ for silicon from scratch

Given $N_{c} N_{v} = 1.5 \times 1 0^{38} cm^{- 6}$ for Si, compute the intrinsic carrier density $n_{i}$ at $300 K$ ( $E_{g} = 1.12 eV$ , $k_{B} T = 0.02585 eV$ ). Comment on "tiny exponential × huge prefactor."

Recall Solution

Step 1 — prefactor: $N_{c} N_{v} = 1.5 \times 1 0^{38} = 1.22 \times 1 0^{19} cm^{- 3}$ .

Step 2 — exponent: $\frac{E _{g}}{2 k _{B} T} = \frac{1.12}{2 \times 0.02585} = \frac{1.12}{0.0517} = 21.66$ , so $e^{- 21.66} = 3.9 \times 1 0^{- 10}$ .

Step 3 — multiply: $n_{i} = 1.22 \times 1 0^{19} \times 3.9 \times 1 0^{- 10} \approx 4.8 \times 1 0^{9} cm^{- 3} .$ Interpretation: the exponential ( $\sim 1 0^{- 10}$ ) says crossing is astronomically rare per state, but there are $\sim 1 0^{19}$ states pushing. Their product lands near the textbook $n_{i} \sim 1 0^{10} cm^{- 3}$ for Si — a small but very real carrier population. (The commonly quoted $1.5 \times 1 0^{10}$ arises from slightly different $N_{c} N_{v}$ / temperature-dependence data.)

L4·Q2 — Design a photodetector cutoff

You want a detector that responds to light out to $1550 nm$ (fiber-optic wavelength) but is transparent to anything longer. What is the maximum allowed band gap $E_{g}$ ? Would Si work?

Recall Solution

WHY: to absorb $1550 nm$ the gap must be small enough that this photon still carries $\geq E_{g}$ . The largest usable gap is where equality holds at $1550 nm$ . $E_{g}^{m a x} = \frac{h c}{λ} = \frac{1240}{1550} = 0.80 eV .$ So you need $E_{g} \leq 0.80 eV$ . Silicon fails ( $E_{g} = 1.12 eV > 0.80 eV$ ): a $1550 nm$ photon carries only $0.80 eV < 1.12 eV$ , not enough to cross Si's gap, so Si is transparent there. You would use a narrow-gap material like Ge ( $0.66 eV$ ) or InGaAs instead.

Level 5 — Mastery

L5·Q1 — Find the temperature that matches two materials' carrier crossings

At what temperature $T$ would silicon's exponential factor $e^{- E_{g} / (2 k_{B} T)}$ equal germanium's value at $300 K$ ? (Use $E_{g}^{Si} = 1.12$ , $E_{g}^{Ge} = 0.66$ , and $k_{B} T = 0.02585 eV$ at $300 K$ ; scale $k_{B} T$ linearly with $T$ .)

Recall Solution

WHY set exponents equal: matching the two exponential factors means the two exponents are equal (the exponential is one-to-one).

Ge at $300 K$ : exponent $= \frac{0.66}{2 \times 0.02585} = 12.77$ .

For Si at temperature $T$ , write $k_{B} T = 0.02585 \cdot \frac{T}{300}$ . Its exponent must equal $12.77$ : $\frac{1.12}{2 \times 0.02585 \times ( T /300 )} = 12.77.$ Solve for $T$ : $T = 300 \times \frac{1.12}{2 \times 0.02585 \times 12.77} = 300 \times \frac{1.12}{0.660} = 300 \times 1.697 \approx 509 K .$ Interpretation: you must heat silicon to $\approx 509 K$ ( $\approx 236^{\circ} C$ ) before its thermal carrier-crossing probability matches germanium's at room temperature — a quantitative statement of why Ge is "leakier" and why Si tolerates hotter operation. This ties directly to Fermi level and Fermi-Dirac distribution and Intrinsic and extrinsic semiconductors.

L5·Q2 — Full chain: colour of an emitted photon after a temperature-driven gap change

Many semiconductors' gaps shrink with temperature. Suppose a material has $E_{g} = 1.42 eV$ (GaAs) at $300 K$ and its emission wavelength is set by $λ = h c / E_{g}$ . (a) Find its emission wavelength at $300 K$ . (b) If heating shrinks the gap to $1.35 eV$ , does the emitted light shift toward red or blue, and what is the new $λ$ ?

Recall Solution

Part (a): emission gives back the gap exactly, so $λ_{300} = \frac{h c}{E _{g}} = \frac{1240}{1.42} \approx 873 nm (near-infrared) .$ Part (b): smaller gap → lower photon energy → longer wavelength → shift toward red: $λ_{hot} = \frac{1240}{1.35} \approx 919 nm .$ WHY the direction: $λ$ and $E_{g}$ are inversely related ( $λ = h c / E_{g}$ ). Shrink $E_{g}$ , and $λ$ grows — the emission "red-shifts." This is why LED colour drifts as the chip warms. Whether the gap is direct or indirect controls how efficiently this photon is emitted — see Direct vs indirect band gap.

Recall Self-test checklist

By now you should be able to, without notes: Read a band diagram and name VB, gap, CB ::: L1 Convert between $λ$ and $E_{g}$ using $h c = 1240 eV\cdotnm$ ::: L2 Take ratios of exponential factors by subtracting exponents ::: L3 Combine prefactor and exponential to get an absolute $n_{i}$ ::: L4 Solve for the temperature or gap that matches a target crossing ::: L5

Connections

Parent: Energy bands (VB & CB) — the concepts these problems drill.
Fermi level and Fermi-Dirac distribution — the exact occupation rule behind the Boltzmann approximation.
Density of states — where $N_{c}$ , $N_{v}$ (and $N_{c} N_{v}$ ) come from.
Intrinsic and extrinsic semiconductors — how doping changes $n$ without touching $E_{g}$ .
Holes as charge carriers — the partner created every time an electron crosses.
Direct vs indirect band gap — controls emission efficiency in L5·Q2.