Q.One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in

The key insight here is that a photon is both a wave and a particle. Its electromagnetic wavelength (the one we measure in optics) and its de Broglie wavelength (the one associated with its momentum) are the same thing. This is not a coincidence — it’s built into the definition.

Let’s walk through why.

1. Recall the de Broglie relation for any particle For any particle with momentum $p$, the de Broglie wavelength is

where $h$ is Planck’s constant. This applies to electrons, neutrons, baseballs — and photons.

2. What is the momentum of a photon? A photon has energy $E = h\nu$ (where $\nu$ is frequency) and also obeys $E = pc$ for massless particles. Equating:

Since $\nu = c / \lambda_{\text{EM}}$ (where $\lambda_{\text{EM}}$ is the electromagnetic wavelength), we get

3. Plug this into the de Broglie formula

So the de Broglie wavelength of a photon is identical to its electromagnetic wavelength.

4. Apply to the given number The problem states the electromagnetic wavelength is $4288\ \text{Å}$. Therefore, the de Broglie wavelength is also $4288\ \text{Å}$.

The Paschen series corresponds to transitions ending at $n = 3$. Its shortest wavelength (highest energy photon) is the series limit, from $n = \infty \to n = 3$:

A band gap of about $1.5\ \text{eV}$ is characteristic of a semiconductor (e.g. Si $\approx 1.1\ \text{eV}$, GaAs $\approx 1.4\ \text{eV}$). Conductors have essentially zero gap, insulators have gaps of several eV ($>3\ \text{eV}$), and a superconductor's gap is not defined by such a value.

Concept & Intuition

In a semiconductor, an electron-hole pair is created when an electron gains enough energy to jump from the valence band to the conduction band. The minimum energy required for one such pair is exactly the band gap energy $E_g$. If the band gap equals the energy of a photon of wavelength 620 nm, then $E_g$ is that photon’s energy. To generate multiple pairs, the total energy needed is simply the number of pairs times the band gap (assuming no losses and that each pair is created independently). The question asks for the minimum thermal energy — so we use the ideal, lossless case.

Step-by-step solution

1. Find the band gap energy from the given wavelength

   The energy of a photon is $E = \frac{hc}{\lambda}$, where

   $h = 4.135667 \times 10^{-15} \ \text{eV·s}$ (Planck’s constant in eV·s),

   $c = 3 \times 10^8 \ \text{m/s}$, and

   $\lambda = 620 \ \text{nm} = 620 \times 10^{-9} \ \text{m}$.

   Compute:

First, numerator: $4.1357 \times 10^{-15} \times 3 \times 10^8 = 1.2407 \times 10^{-6} \ \text{eV·m}$.

Then divide by $620 \times 10^{-9} = 6.2 \times 10^{-7} \ \text{m}$:

So the band gap $E_g \approx 2 \ \text{eV}$.

2. Energy needed per electron-hole pair

   The minimum energy to create one pair is exactly $E_g$ (the electron must gain at least this much to cross the gap). So one pair requires $2 \ \text{eV}$.

3. Total energy for 8 pairs

   If we need 8 independent pairs, the minimum total thermal energy is:

4. Match with options The options are: (A) 8 eV, (B) 4 eV, (C) 16 eV, (D) 2 eV. Our result is 16 eV, which corresponds to option (C).

The question tests your understanding of what a neutrino is and how it behaves. A neutrino is a fundamental particle with no electric charge and an extremely tiny mass (so small it was long thought to be zero). Its only interactions are through the weak nuclear force and gravity — and gravity is negligible at particle scales. The weak force has a very short range and a minuscule cross-section, meaning neutrinos almost never hit anything.

That property is the key: neutrinos can sail through ordinary matter as if it were nearly empty space. Let's examine each option.

1. Option (A): "Neutrino is emitted from the nucleus in the alpha decay process"

   Alpha decay emits an alpha particle (two protons and two neutrons). No neutrino appears in that process. Neutrinos are produced in beta decay, where a neutron turns into a proton and emits an electron and an antineutrino (or a proton turns into a neutron and emits a positron and a neutrino). So this statement is false.

2. Option (B): "Neutrino interacts very strongly with matter"

   This is the opposite of the truth. Neutrinos interact extremely weakly — that's their defining feature. A typical neutrino can pass through billions of kilometres of lead without a single interaction. "Strongly" would describe particles like protons or neutrons, which feel the strong nuclear force. So this is false.

3. Option (C): "Neutrino can penetrate through the earth without being absorbed"

   Exactly right. Because neutrinos interact so weakly, the vast majority of neutrinos passing through the Earth never interact at all. In fact, trillions of solar neutrinos pass through your body every second without you noticing. The Earth is essentially transparent to them. This is the correct statement.

4. Option (D): "The mass of neutrino is equal to the mass of neutron"

   A neutron has a mass of about $939.6 \text{ MeV}/c^2$. A neutrino's mass is less than $0.1 \text{ eV}/c^2$ — that's at least ten million times smaller. They are not even close. (Neutrino masses are nonzero, as shown by oscillation experiments, but they are minuscule.) So this is false.

The key is to recall the electromagnetic spectrum in order of decreasing wavelength (or increasing frequency). Microwaves have the longest wavelengths among the given options, followed by infrared, then ultraviolet, then X‑rays, and finally gamma rays with the shortest. Each wavelength range in the right column corresponds to one of these bands. Let’s match them step by step.

1. Microwaves (A) – These have wavelengths from about 0.1 m down to 1 mm. That matches IV (0.1 m to 1 mm).

   Reasoning: Microwaves are longer than infrared but shorter than radio waves; the given range fits perfectly.

2. Infrared (B) – Infrared spans from 1 mm down to about 700 nm (the edge of visible red). That is I (1 mm to 700 nm).

   Reasoning: IR sits between microwaves and visible light; 700 nm is the red end of visible, so the range just above it is IR.

3. Ultraviolet (D) – Ultraviolet goes from about 400 nm (violet end of visible) down to 1 nm. That is II (400 nm to 1 nm).

   Reasoning: UV is shorter than visible light but longer than X‑rays; the given range matches.

4. X‑ray (C) – X‑rays range from about 1 nm down to 10⁻³ nm. That is V (1 nm to 10⁻³ nm).

   Reasoning: X‑rays are shorter than UV but longer than gamma rays; the range fits.

5. Gamma rays (E) – Gamma rays have wavelengths less than about 10⁻³ nm. That is III (< 10⁻³ nm).

   Reasoning: Gamma rays are the shortest of all; the “less than” range is correct.

Now assemble the matches in order A→B→C→D→E:

A → IV, B → I, C → V, D → II, E → III.

That sequence is IV I V II III, which corresponds to option (D).

The key idea is that a p-n junction photodiode works by absorbing photons and creating electron-hole pairs. For that to happen, the photon’s energy must be at least equal to the semiconductor’s band gap energy $E_g$. If the photon energy is smaller than $E_g$, the photon passes through without being absorbed — the device simply cannot “see” it.

So the question becomes: which of the given wavelengths corresponds to a photon energy less than 2.8 eV? That wavelength will be the one the detector misses.

1. Find the cutoff wavelength. The relation between photon energy $E$ and wavelength $\lambda$ is

where $h = 6 \times 10^{-34} \ \mathrm{J \cdot s}$ and $c = 3 \times 10^8 \ \mathrm{m/s}$.

At the threshold, $E = E_g = 2.8 \ \mathrm{eV}$. First convert eV to joules:

so

2. Solve for $\lambda$:

Compute step by step:

That is about $402 \ \mathrm{nm}$ — but wait, let’s do it more carefully with the given numbers.

Actually, using $h = 6 \times 10^{-34}$ exactly:

But a more common shortcut uses $hc \approx 1240 \ \mathrm{eV \cdot nm}$:

The slight difference comes from using $h = 6.63 \times 10^{-34}$ in the standard constant; with the given $h = 6 \times 10^{-34}$, the 402 nm result is the one to trust for this problem.

3. Interpret the cutoff.

   The photodiode detects photons with $\lambda \leq \lambda_{\text{cutoff}}$ (higher energy). It cannot detect photons with $\lambda > \lambda_{\text{cutoff}}$ (lower energy).

   Our computed cutoff is about 402 nm. Now compare the options:

   • 100 nm — much shorter, easily detected.
   • 200 nm — shorter, detected.
   • 400 nm — just at the edge; actually 400 nm < 402 nm, so it can detect it (barely).
   • 600 nm — longer than 402 nm, so it cannot detect it.

   Therefore the wavelength it cannot detect is 600 nm.

The key idea is that a thermopile bolometer is a thermal detector — it responds to the heating caused by radiation, not to the radiation’s frequency directly. Any radiation that gets absorbed and heats the detector can, in principle, be sensed. But in practice, the design and materials of a thermopile bolometer are optimized for a specific range: infrared (IR) radiation.

Why infrared? Because IR radiation is strongly absorbed by most materials and produces a significant temperature rise. Ultraviolet, X-rays, and gamma rays are much more energetic, but they tend to pass through or damage the detector rather than heat it uniformly. Moreover, thermopile bolometers are commonly used in IR thermography, motion sensors, and spectroscopy — not for high-energy photon detection.

Let’s walk through the reasoning step by step.

1. What a thermopile bolometer does

   A bolometer measures radiation by the change in temperature of an absorbing material. A thermopile is a series of thermocouples connected in series. When one set of junctions (the “hot” junctions) is heated by absorbed radiation and the other set (the “cold” junctions) is kept at a reference temperature, a voltage is generated due to the Seebeck effect. That voltage is proportional to the incident radiation power.

2. Which radiations cause significant heating?

   • Infrared radiation (wavelengths roughly 0.7 µm to 1 mm) is readily absorbed by most solids and liquids, converting its energy into heat. This makes it ideal for thermal detectors.
   • Ultraviolet radiation (10 nm to 400 nm) is more energetic but often causes photoelectric effects or chemical changes rather than simple heating. It also tends to be absorbed in the very surface layer, potentially damaging the detector.
   • X-rays and gamma rays (wavelengths below 10 nm) are highly penetrating. They pass through thin absorbing layers without depositing much heat, and they require specialized detectors (like scintillators or Geiger counters) that rely on ionization, not thermal effects.

3. Practical application confirms the choice

   Thermopile bolometers are standard tools in infrared astronomy, thermal imaging, and non-contact temperature measurement. They are not used for X-ray or gamma-ray detection because those radiations would mostly go right through the thin absorbing film, producing negligible heating.

4. Eliminating the other options
   • (A) Ultraviolet radiation: UV is typically detected by photodiodes or photomultipliers, not by thermal effects.
   • (B) X-rays: Detected via ionization in gas-filled tubes or solid-state detectors.
   • (C) Gamma radiation: Similar to X-rays, requires high-Z materials for absorption and scintillation or semiconductor detection.
   • (D) Infrared radiation: Matches the operating principle of a thermopile bolometer perfectly.

The core idea is simple: to create an electron-hole pair in a semiconductor, a photon must supply at least the band gap energy. If the photon’s energy is less than the band gap, it cannot promote an electron from the valence band to the conduction band — it just passes through. The maximum wavelength corresponds to the minimum photon energy that still works, which is exactly the band gap energy.

So we set the photon energy equal to the band gap and solve for wavelength.

1. Write the relation between photon energy and wavelength.

   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.

2. Plug in the given values.

   Band gap $E_g = 0.7$ eV, $h = 4.136 \times 10^{-15}$ eV·s, $c = 3 \times 10^8$ m/s.

   We want $\lambda$ such that $E = E_g$:

3. Calculate $hc$ first.

This product ($hc \approx 1.24 \times 10^{-6}$ eV·m) is a handy constant to remember — it saves time in such problems.

4. Divide by the band gap.

5. Convert to the form given in the options. The options are in $10^{-8}$ m or $10^{-9}$ m. $1.7726 \times 10^{-6}$ m = $1772.6 \times 10^{-9}$ m, which rounds to $1773 \times 10^{-9}$ m.