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?

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Pick out the correct statement from the following; (A) The number density of free electrons in the valance band decides the strength of the electric current (B) Valance band is always completely filled, while conduction band is always partially filled (C) The maximum energy required to shift an electron from the conduction band to valance band is called energy band gap (D) In a semiconductor no free electrons are found in the conduction band at 0 K

   ›Reveal solutionSolution

   The key idea is understanding the definitions of valence band, conduction band, band gap, and the behavior of electrons at absolute zero. The correct statement is that in a semiconductor at 0 K, the conduction band contains no free electrons.

   The question tests your grasp of basic semiconductor physics — specifically, the structure of energy bands and how electrons populate them. Let’s clarify each concept before evaluating the options.

   Concept & Intuition:

   In solids, electrons occupy energy bands: the valence band (filled with electrons in their ground state) and the conduction band (empty or partially filled, where electrons can move freely). The band gap is the energy difference between the top of the valence band and the bottom of the conduction band. At absolute zero (0 K), all electrons are in their lowest energy states — in a semiconductor, this means the valence band is completely full and the conduction band is completely empty. No thermal energy is available to excite electrons across the gap.

   Now, evaluate each statement:

   1. Option (A): “The number density of free electrons in the valence band decides the strength of the electric current.”

      • Reasoning: Free electrons in the conduction band (not valence band) are the primary charge carriers for current. In the valence band, electrons are bound and cannot move freely unless they leave holes. So this statement is false.

   2. Option (B): “Valence band is always completely filled, while conduction band is always partially filled.”

      • Reasoning: This is not always true. In metals, the conduction band may be partially filled and the valence band may be partially filled or overlapping. In insulators and semiconductors at 0 K, the valence band is full and conduction band empty. The word “always” makes this incorrect.

   3. Option (C): “The maximum energy required to shift an electron from the conduction band to valence band is called energy band gap.”

      • Reasoning: The band gap is the minimum energy needed to move an electron from the valence band to the conduction band (not the maximum, and not from conduction to valence). The direction is reversed and the word “maximum” is wrong. So this is false.

   4. Option (D): “In a semiconductor no free electrons are found in the conduction band at 0 K.”

      • Reasoning: At 0 K, all electrons occupy the lowest available energy states. In a pure semiconductor, the valence band is completely filled and the conduction band is completely empty — no thermal energy exists to promote electrons. Thus, there are indeed no free electrons in the conduction band. This statement is correct.

   Watch out

   A common mistake is confusing the direction of electron transition for the band gap. The band gap is the minimum energy to go from valence to conduction, not the maximum or the reverse.

   Tip

   Remember: At 0 K, semiconductors behave like insulators — no free carriers. This is a defining property that distinguishes them from conductors.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. COMEDK 2026Set 2026-A1 markMCQ
   Q.What is the minimum wavelength of radiation required to detect a p-n junction diode made of a semiconductor having band gap 3.3 eV . [Planck's constant $h=6.6 \times 10^{34} \mathrm{~J} . \mathrm{s}$ ] (A) $3300^{\circ} \mathrm{A}$ (B) $4800^{\circ} \mathrm{A}$ (C) $3750^{\circ} \mathrm{A}$ (D) $7500^{\circ} \mathrm{A}$

   ›Reveal solutionSolution

   The minimum detectable wavelength corresponds to the photon energy exactly matching the band gap. Using $E = \frac{hc}{\lambda}$, the wavelength is $\lambda = \frac{hc}{E} \approx 3750 \, \text{Å}$, so option (C) is correct.

   The key idea is that a p-n junction diode can detect radiation only if the incoming photons have enough energy to excite electrons across the band gap. The minimum photon energy needed equals the band gap energy $E_g = 3.3 \, \text{eV}$. The corresponding wavelength is the maximum wavelength that can be detected — but the question asks for the minimum wavelength of radiation required, which is actually the same threshold: any photon with energy at least $E_g$ works, so the shortest wavelength that still has enough energy is given by $E = hc/\lambda$.

   1. Convert the band gap to joules Since $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$,

   $E_g = 3.3 \, \text{eV} = 3.3 \times 1.6 \times 10^{-19} = 5.28 \times 10^{-19} \, \text{J}.$

   2. Use the photon energy–wavelength relation The energy of a photon is $E = \frac{hc}{\lambda}$, so

   $\lambda = \frac{hc}{E}.$

   Here $h = 6.6 \times 10^{-34} \, \text{J·s}$ and $c = 3 \times 10^8 \, \text{m/s}$.

   3. Plug in the numbers

   $\lambda = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{5.28 \times 10^{-19}} = \frac{1.98 \times 10^{-25}}{5.28 \times 10^{-19}} \approx 3.75 \times 10^{-7} \, \text{m}.$

   4. Convert to angstroms $1 \, \text{Å} = 10^{-10} \, \text{m}$, so

   $\lambda = 3.75 \times 10^{-7} \, \text{m} = 3750 \, \text{Å}.$

   Tip

   A common shortcut: $\lambda (\text{in Å}) = \frac{12400}{E (\text{in eV})}$. Here $12400 / 3.3 \approx 3758 \, \text{Å}$, close enough to 3750 Å given the approximations in constants.

   Watch out

   A classic mistake is to confuse "minimum wavelength" with "maximum wavelength." Since energy is inversely proportional to wavelength, the minimum wavelength corresponds to the maximum energy — but here the threshold energy is fixed, so the minimum detectable wavelength is simply the one whose photon energy equals the band gap. Any shorter wavelength (higher energy) also works, but the question asks for the minimum required, which is this boundary value.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. COMEDK 2025Set 2025-A1 markMCQ
   Q.Diamond is considered as an insulator because (A) Diamond has very large forbidden energy gap. (B) Valance band and conduction band of diamond are overlapping X . (C) In the diamond the carbon atoms are held by weak covalent bonds (D) Refractive index of diamond is very low

   ›Reveal solutionSolution

   Diamond is an insulator because its electrons cannot jump from the valence band to the conduction band — the forbidden energy gap is very large (~5.5 eV), so no conduction occurs at ordinary temperatures.

   Concept & Intuition

   In solid-state physics, materials are classified by their band structure — the arrangement of energy levels that electrons can occupy.

   • Conductors have overlapping valence and conduction bands, so electrons move freely.
   • Insulators have a large energy gap between the valence band (full of electrons) and the conduction band (empty).
   • Semiconductors have a small gap that electrons can cross with a little energy (heat, light).

   Diamond is the classic example of an insulator: its carbon atoms are held by strong covalent bonds, and those bonds lock electrons into a full valence band. To conduct electricity, an electron must gain enough energy to jump the gap — but diamond’s gap is huge (~5.5 eV), far larger than thermal energy at room temperature (~0.025 eV). So essentially no electrons make the jump, and diamond does not conduct.

   Now let’s examine each option.

   1. Option (A): Diamond has very large forbidden energy gap.

      This is correct. The forbidden gap (band gap) in diamond is about 5.5 eV. For comparison, silicon (a semiconductor) has a gap of 1.1 eV, and insulators typically have gaps > 3 eV. A large gap means that at ordinary temperatures, virtually no electrons are thermally excited into the conduction band — hence no electrical conduction. This is the fundamental reason diamond is an insulator.

   2. Option (B): Valence band and conduction band of diamond are overlapping.

      This is false. Overlapping bands are characteristic of conductors (metals), where electrons can move freely. In diamond, the bands are well separated by the large gap. Overlap would make diamond a conductor, which it is not.

   3. Option (C): In diamond, the carbon atoms are held by weak covalent bonds.

      This is false. Diamond’s covalent bonds are extremely strong (each carbon is sp³ hybridized, bonded to four others in a tetrahedral lattice). The strength of these bonds is what gives diamond its hardness and high melting point, but it is the band gap that determines electrical insulation, not bond strength directly. Weak bonds would actually make conduction easier, not harder.

   4. Option (D): Refractive index of diamond is very low.

      This is false. Diamond has a very high refractive index (~2.42), which is why it sparkles. Refractive index is an optical property, not directly related to electrical conductivity. Even if it were low, that wouldn’t make diamond an insulator.

   Watch out

   A common mistake is to confuse “strong covalent bonds” with “insulating behavior.” While strong bonds contribute to a large band gap, the direct cause of insulation is the size of the forbidden energy gap, not bond strength itself. Option (C) is tempting but incorrect.

   Tip

   A quick way to remember: Large gap = insulator, small gap = semiconductor, no gap = conductor. Diamond’s gap is the largest among common materials, so it’s the ultimate insulator.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. COMEDK 2025Set 2025-A1 markMCQ
   Q.What is the maximum wave length of EM radiation required to move an electron from the valance band to conduction band of a semiconductor? [Given :Energy gap $\mathrm{E}_{\mathrm{g}}=1.98 \times 10^{-19} \mathrm{~J}$; Planck's constant $h=6.6 \times 10^{-34} \mathrm{Js}$ ] (A) $10^{-9} \mathrm{~m}$ (B) $10^{-6} \mathrm{~m}$ (C) $10^{-10} \mathrm{~m}$ (D) $10^{-12} \mathrm{~m}$

   ›Reveal solutionSolution

   The maximum wavelength corresponds to the minimum photon energy needed to bridge the band gap. Using $E = hc/\lambda$, we find $\lambda \approx 10^{-6}\,\text{m}$, so option (B) is correct.

   Concept & Intuition

   To promote an electron from the valence band to the conduction band, a photon must supply at least the band-gap energy $E_g$. The maximum wavelength that can do this is the one whose photon energy exactly equals $E_g$ — any longer wavelength would have lower energy and be insufficient. So we simply set $E_{\text{photon}} = E_g$ and solve for $\lambda$.

   Step-by-step solution

   1. Relate photon energy to wavelength The energy of a photon is

   $E = \frac{hc}{\lambda}$

   where $h = 6.6 \times 10^{-34}\,\text{Js}$, $c = 3.0 \times 10^8\,\text{m/s}$, and $\lambda$ is the wavelength.

   2. Set photon energy equal to the band gap For the threshold case,

   $\frac{hc}{\lambda_{\text{max}}} = E_g$

   so

   $\lambda_{\text{max}} = \frac{hc}{E_g}$

   3. Plug in the numbers

   $\lambda_{\text{max}} = \frac{(6.6 \times 10^{-34})(3.0 \times 10^8)}{1.98 \times 10^{-19}}$

   First compute numerator:

   $6.6 \times 3.0 = 19.8 \quad\Rightarrow\quad 19.8 \times 10^{-26} = 1.98 \times 10^{-25}$

   Then divide:

   $\lambda_{\text{max}} = \frac{1.98 \times 10^{-25}}{1.98 \times 10^{-19}} = 10^{-6}\,\text{m}$

   4. Match with the options $10^{-6}\,\text{m}$ corresponds to option (B).

   Watch out

   A common mistake is to forget that maximum wavelength means minimum energy. Using a larger $E_g$ than given, or misplacing the decimal in $hc$, can lead to answers like $10^{-9}$ or $10^{-10}$ m — but careful calculation gives exactly $10^{-6}$ m.

   Tip

   Notice that $hc \approx 2 \times 10^{-25}\,\text{J·m}$ and $E_g \approx 2 \times 10^{-19}\,\text{J}$; their ratio is neatly $10^{-6}$ m. This quick mental check confirms the answer without a calculator.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

5. COMEDK 2025Set 2025-M1 markMCQ
   Q.The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $1.24 \mu \mathrm{~m}$ is incident on it. The band gap (in eV ) for the semiconductor is (A) 1 eV (B) 1.1 eV (C) 2.48 eV (D) 0.7 eV

   ›Reveal solutionSolution

   The band gap energy equals the photon energy at the threshold wavelength; using $E = \frac{hc}{\lambda}$ with $\lambda = 1.24\ \mu\text{m}$ gives exactly 1.00 eV, so the correct option is (A).

   The key idea is that a semiconductor becomes conductive when incident photons have enough energy to excite electrons across the band gap. The threshold wavelength tells us the minimum photon energy that can do this — that energy is the band gap energy. So we just need to convert the given wavelength into energy using the photon energy formula.

   1. Recall the photon energy–wavelength relation.

      The energy of a photon is $E = \frac{hc}{\lambda}$, where $h$ is Planck’s constant, $c$ is the speed of light, and $\lambda$ is the wavelength. In electron-volts and micrometres, there’s a handy constant: $hc \approx 1240\ \text{eV·nm}$. Since $1\ \mu\text{m} = 1000\ \text{nm}$, we can write $hc \approx 1.24\ \text{eV·μm}$.

   2. Apply the threshold condition.

      The problem says conductivity increases for wavelengths shorter than $1.24\ \mu\text{m}$. That means the threshold wavelength is $\lambda_{\text{max}} = 1.24\ \mu\text{m}$. At this wavelength, the photon energy equals the band gap $E_g$:

   $E_g = \frac{hc}{\lambda_{\text{max}}} = \frac{1.24\ \text{eV·μm}}{1.24\ \mu\text{m}} = 1.00\ \text{eV}.$

   3. Interpret the result. Photons with wavelength shorter than $1.24\ \mu\text{m}$ have energy greater than 1 eV, so they can excite electrons across the gap. Photons with longer wavelength have less energy and cannot. Thus the band gap is exactly 1 eV.

   Tip

   The product $hc = 1240\ \text{eV·nm}$ is a constant worth memorizing — it turns wavelength-in-nm problems into simple division. Here, because the wavelength in μm is numerically equal to 1.24, the energy comes out as exactly 1 eV.

   Watch out

   A common mistake is to forget that “shorter than” means the threshold is the longest wavelength that still works. Using $1.24\ \mu\text{m}$ as the cutoff is correct; using a shorter wavelength would give a larger (wrong) energy.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

6. COMEDK 2025Set 2025-M1 markMCQ
   Q.In the energy band diagram of a material shown below, open circles and filled circles denote holes and electrons respectively. The material is a (A) metal (B) insulator (C) n-type semiconductor (D) p-type semiconductor

   ›Reveal solutionSolution

   The material has a finite band gap with electrons in the conduction band and holes in the valence band. Counting carriers, the holes in the valence band outnumber the electrons in the conduction band, so holes are the majority carriers — the hallmark of a p-type semiconductor. The correct option is (D).

   ---

   The key to this question is reading the energy band diagram — a picture that tells you where the electrons and holes are sitting. In solid-state physics, the conduction band (upper band) is where free electrons roam and conduct current; the valence band (lower band) is where electrons are bound in covalent bonds, and a missing electron leaves a hole (a positive charge carrier). The gap between them, labelled $E_g$, is the band gap.

   Why this approach works:

   If you see filled circles (electrons) in the conduction band, that means some electrons have been excited across the gap. If you see open circles (holes) in the valence band, that means some electrons have left behind vacancies. The relative numbers of electrons and holes tell you whether the material is intrinsic (equal numbers), n-type (more electrons), or p-type (more holes). Metals have no band gap (bands overlap), and insulators have a very large gap with no carriers at room temperature.

   ---

   1. Identify the bands and carriers

      • The upper rectangle is the conduction band. It contains 2 filled circles → these are electrons.
      • The lower rectangle is the valence band. It contains 4 open circles (holes) and 2 filled circles (electrons that remain in the valence band).
      • So the material has 2 free electrons in the conduction band and 4 holes in the valence band.

   2. Interpret the band gap

      • The vertical double-headed arrow labelled $E_g$ shows a finite, non-zero band gap. This rules out a metal (which has overlapping bands, i.e., $E_g = 0$).
      • The gap is not huge (no numerical value given, but the diagram shows a moderate gap — typical of a semiconductor). An insulator would have a very large $E_g$ and essentially no carriers in the conduction band at ordinary temperatures.

   3. Compare electron and hole counts

      • Electrons in conduction band: 2
      • Holes in valence band: 4
      • This is not equal — there are more holes than electrons. In an intrinsic (pure) semiconductor, the number of electrons in the conduction band equals the number of holes in the valence band (because each excited electron leaves one hole). Here, the numbers are unequal, so the material is doped.

   4. Determine the doping type

      • In an n-type semiconductor, donor impurities add extra electrons, so the electron concentration $n$ is greater than the hole concentration $p$. But wait — the diagram shows more holes (4) than electrons (2). That seems backward at first glance.
      • However, remember: the diagram is a snapshot of carriers at some temperature. In an n-type material, the majority carriers are electrons, but there are still some holes (minority carriers) from thermal generation. The diagram could represent a situation where the electron count in the conduction band is smaller than the hole count in the valence band if the doping is light and temperature is moderate? Actually, no — in n-type, $n > p$ always at equilibrium.
      • Let’s re-examine: The valence band has 2 filled circles (electrons) and 4 open circles (holes). That means the valence band originally had 6 electron states (since each hole is a missing electron). So the total number of electrons in the valence band is 2, and holes = 4. The conduction band has 2 electrons.
      • Total electrons in the system = 2 (conduction) + 2 (valence) = 4. Total holes = 4. So n = 2, p = 4 → $p > n$. That is characteristic of a p-type semiconductor, where acceptor impurities create extra holes.

   5. Check the options

      • (A) Metal: No — band gap present.
      • (B) Insulator: No — there are carriers in the conduction band.
      • (C) n-type: No — holes outnumber electrons.
      • (D) p-type: Yes — holes are majority carriers.

   Watch out

   A common mistake is to count only the open circles as holes and forget that the filled circles in the valence band are electrons, not holes. The valence band’s total states are the sum of filled circles (electrons) and open circles (holes). Here, 2 filled + 4 open = 6 states; holes = 4, electrons in valence = 2. So holes dominate.

   Tip

   In any band diagram, remember: Conduction band electrons = filled circles in the upper band; Valence band holes = open circles in the lower band. The majority carrier type is the one with the larger count.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

7. COMEDK 2024Set 2024-M1 markMCQ
   Q.Though $\mathrm{Sn}$ and $\mathrm{Si}$ are $4^{\text {th }}$ group elements, $\mathrm{Sn}$ is a metal while $\mathrm{Si}$ is a semiconductor because (A) $\mathrm{Sn}$ has more electrons than $\mathrm{Si}$ (B) The energy gap of $\mathrm{Sn}$ is zero volt while that of $\mathrm{Si}$ is $0.07 \mathrm{~V}$ (C) The energy gap of $\mathrm{Sn}$ is $1.1 \mathrm{~eV}$ volt while that of $\mathrm{Si}$ is $0.07 \mathrm{~V}$ (D) $\mathrm{Sn}$ has more holes than $\mathrm{Si}$

   ›Reveal solutionSolution

   The key idea is that the electrical behavior of an element (metal vs. semiconductor) depends on its band gap energy. Tin (Sn) has a zero band gap, making it a metal, while silicon (Si) has a finite band gap (~1.1 eV), making it a semiconductor. The correct option is (B).

   ---

   The question asks why Sn and Si, both in Group 14 of the periodic table, behave so differently electrically — Sn is a metal, Si is a semiconductor. The answer lies not in the number of electrons or holes, but in the energy band structure of the solid. In particular, the band gap — the energy difference between the valence band (filled with electrons) and the conduction band (where electrons can move freely) — determines whether a material conducts like a metal, a semiconductor, or an insulator.

   • Metals have overlapping bands or a zero band gap, so electrons can move into the conduction band with minimal energy.
   • Semiconductors have a small but nonzero band gap (typically 0.1–2 eV), so they conduct only when enough energy (heat, light) is supplied.
   • Insulators have a large band gap (>3 eV), so they barely conduct.

   Now, let’s examine each option carefully.

   1. Option (A): "Sn has more electrons than Si"

      While it’s true that Sn (atomic number 50) has more electrons than Si (atomic number 14), the number of valence electrons is the same (4 each) because both are in Group 14. The total number of electrons does not determine metallic vs. semiconducting behavior — many elements with many electrons are insulators (e.g., lead is a metal, but bismuth has more electrons and is a semimetal). So this is not the reason.

   2. Option (B): "The energy gap of Sn is zero volt while that of Si is 0.07 V"

      This is almost correct, but the numbers need checking. Actually, the band gap of Si is about 1.1 eV (not 0.07 V). The statement says "0.07 V" for Si, which is wrong. However, the idea that Sn has a zero band gap (or overlapping bands) is correct. In fact, gray tin (the stable form at low temperature) is a semimetal with a very small band gap, but white tin (the common metallic form) has overlapping bands — effectively a zero band gap. So the concept in (B) is right, but the number for Si is incorrect. Still, among the options, this is the only one that points to the band gap difference.

   3. Option (C): "The energy gap of Sn is 1.1 eV while that of Si is 0.07 V"

      This reverses the actual values. Sn does not have a 1.1 eV band gap (it’s essentially zero), and Si’s band gap is ~1.1 eV, not 0.07 eV. So this is completely wrong.

   4. Option (D): "Sn has more holes than Si"

      Holes are missing electrons in the valence band. In a pure (intrinsic) semiconductor, the number of holes equals the number of electrons in the conduction band. Sn, being a metal, has no band gap, so the concept of "holes" in the usual sense doesn’t apply — its conduction is due to free electrons, not holes. This option is misleading and incorrect.

   Watch out

   A common mistake is to think that having more electrons automatically makes an element metallic. But electron count alone doesn’t determine conductivity — it’s the arrangement of energy bands in the solid that matters. Also, note that the band gap of Si is ~1.1 eV, not 0.07 eV; the latter is a typical voltage drop across a diode, not a band gap.

   Tip

   A neat way to remember: In Group 14, as you go down (C → Si → Ge → Sn → Pb), the band gap decreases. Carbon (diamond) is an insulator (~5.5 eV), Si and Ge are semiconductors (~1.1 eV and ~0.67 eV), and Sn and Pb are metals (zero band gap). So the trend is clear: heavier elements have smaller band gaps, eventually becoming metallic.

   Thus, the only option that correctly identifies the reason (band gap difference) is (B), even though the numerical value for Si is misstated. In multiple-choice questions, we select the best available answer — and here, (B) is the only one that points to the correct physical cause.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. COMEDK 2023Set 2023-E1 markMCQ
   Q.The energy gap between valance band and the conduction band for a given material is $6 \mathrm{~eV}$, then the material is : (A) A semiconductor (B) A metal (C) An insulator (D) A superconductor

   ›Reveal solutionSolution

   An energy gap of 6 eV is far too large for thermal excitation, so the material is an insulator. (Superconductivity is not defined by a band-gap of this kind, so (D) is not the classification asked for.)

   Concept: classification of solids by the forbidden-energy-gap Eg between valence and conduction bands.

   • Metals (conductors): the bands overlap, Eg is effectively zero.
   • Semiconductors: small gap, Eg of the order of ~1 eV (Si 1.1 eV, Ge 0.7 eV), so thermal energy can promote some electrons.
   • Insulators: large gap, Eg roughly > 3 eV (e.g. diamond ~ 6 eV); at ordinary temperatures essentially no electron can cross it.

   An energy gap of 6 eV is far too large for thermal excitation, so the material is an insulator.

   (Superconductivity is not defined by a band-gap of this kind, so (D) is not the classification asked for.)

   ✓Final answer

   The correct option is (C) — An insulator

   ANSWER: C

9. COMEDK 2023Set 2023-M1 markMCQ
   Q.An LED is constructed from a $p$-$n$ junction diode using GaAsP. The energy gap is $1.9 \mathrm{~eV}$. The wavelength of the light emitted will be equal to (A) $10.4 \times 10^{-26} \mathrm{~m}$ (B) $654 \mathrm{~nm}$ (C) $654 ~\mathop A\limits^o$ (D) $654 \times 10^{-11} \mathrm{~m}$

   ›Reveal solutionSolution

   Using $\lambda = hc/E$ with $hc=1240\,$eV·nm and $E=1.9\,$eV gives $\lambda\approx654\,$nm (visible red).

   The emitted photon energy equals the band gap, $E=1.9\,$eV. Its wavelength:

   $\lambda = \frac{hc}{E} = \frac{1240\ \text{eV·nm}}{1.9\ \text{eV}} \approx 653\text{–}654\ \text{nm}.$

   Checking the alternatives: $654\,\text{Å}=65.4\,$nm and $654\times10^{-11}\,$m$=6.54\,$nm are far too short (and would not be visible light), while $10.4\times10^{-26}\,$m is nonsensical. So the answer is $654\,$nm.

   ✓Final answer

   The correct option is (B) — $654 \mathrm{~nm}$

10. KCET 2022Set B-31 markMCQ
    Q.The forbidden energy gap for ‘Ge’ crystal at ‘0’ K is (A) 1.2 eV (B) 6.57 eV (C) 0.071 eV (D) 0.71 eV

    ›Reveal solutionSolution

    The forbidden energy gap of a germanium crystal at 0 K is a well-known material constant. The correct value is 0.71 eV, which corresponds to option (D).

    The forbidden energy gap, or band gap, is the energy difference between the top of the valence band and the bottom of the conduction band in a semiconductor. At absolute zero (0 K), the crystal is in its purest, most ordered state — no thermal energy is available to excite electrons across the gap. For germanium, this intrinsic property is a fixed number, determined by its atomic structure and bonding.

    Why does this matter? The band gap dictates whether a material behaves as an insulator, semiconductor, or conductor. For semiconductors like germanium, the gap is small enough that at room temperature, some electrons can jump across, making the material conduct. But at 0 K, the gap is the "true" value, unaltered by thermal expansion or lattice vibrations.

    1. Recall the known band gaps for common semiconductors.

       Silicon (Si) has a band gap of about 1.1 eV at 0 K. Germanium (Ge) has a smaller gap — around 0.67 eV at room temperature, but slightly larger at 0 K due to lattice contraction. The standard accepted value for Ge at 0 K is 0.71 eV.

    2. Eliminate the distractors.

       • Option (A) 1.2 eV is close to silicon’s band gap (1.1 eV), not germanium’s.
       • Option (B) 6.57 eV is far too large — that’s in the range of insulators like diamond (5.5 eV).
       • Option (C) 0.071 eV is too small; that would make germanium act almost like a conductor at low temperatures, which it doesn’t.
       • Option (D) 0.71 eV matches the known value.

    3. Confirm with a consistency check.

       At room temperature (300 K), germanium’s band gap shrinks to about 0.67 eV due to thermal expansion. The 0 K value being slightly higher (0.71 eV) is physically reasonable — cooling the crystal reduces atomic spacing, which slightly increases the gap.

    Watch out

    A common mistake is confusing germanium’s room-temperature gap (≈0.67 eV) with its 0 K value. Always check the temperature condition in the question — here it’s explicitly 0 K, so the correct value is 0.71 eV, not 0.67 eV.

    ✓Final answer

    The correct option is (D), with the forbidden energy gap of germanium at 0 K being 0.71 eV.

11. COMEDK 2022Set 20221 markMCQ
    Q.Choose the incorrect statements. (A) Silicon is used in the fabrication of LED. (B) LED works on the principle of electroluminescence. (C) LED is a power efficient device. (D) LED is fabricated with direct band gap semiconductor.

    ›Reveal solutionSolution

    (B) LEDs do work by electroluminescence (radiative recombination of electrons and holes at a forward-biased p-n junction) - CORRECT. (C) LEDs are highly power-efficient (long life, low operating voltage/power) - CORRECT. (D) LEDs are made from DIRECT band-gap semiconductors (GaAs, GaAsP, GaP...) so that recombination gives a photon rather than a phonon - CORRECT. (A) Silicon is an INDIRECT band-gap semiconductor; recombination in Si releases energy mostly as heat (phonons), not light. Silicon is therefore NOT used to fabricate LEDs - this statement is INCORRECT.

    Concept: LED physics.

    (B) LEDs do work by electroluminescence (radiative recombination of electrons and holes at a forward-biased p-n junction) - CORRECT.

    (C) LEDs are highly power-efficient (long life, low operating voltage/power) - CORRECT.

    (D) LEDs are made from DIRECT band-gap semiconductors (GaAs, GaAsP, GaP...) so that recombination gives a photon rather than a phonon - CORRECT.

    (A) Silicon is an INDIRECT band-gap semiconductor; recombination in Si releases energy mostly as heat (phonons), not light. Silicon is therefore NOT used to fabricate LEDs - this statement is INCORRECT.

    ✓Final answer

    The correct option is (A) — Silicon is used in the fabrication of LED.

    ANSWER: A

12. COMEDK 2021Set 20211 markMCQ
    Q.What is the minimum band-gap of the LED diode? (A) 1.5 eV (B) 1.7 eV (C) 1.8 eV (D) 0.8 eV

    ›Reveal solutionSolution

    Minimum band gap of an LED = 1.8 eV.

    Concept: LED band gap.

    For an LED emitting visible light the semiconductor band gap must be at least about the photon energy of red light. NCERT states that LEDs are made of semiconductors with band gap of about 1.8 eV or more (Eg ~ 1.8 eV corresponds to about 700 nm, red).

    Minimum band gap of an LED = 1.8 eV.

    ✓Final answer

    The correct option is (C) — 1.8 eV

    ANSWER: C