NCERT Solutions Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Insight - the Duane–Hunt rule. A neat shortcut used by X-ray engineers: _min (in pm) ≈ 1240V (in kV). Plug V = 30 kV: _min ≈ 1240/30 = 41.3 pm. This matches our long calculation. Memorise the "1240" - it pops up everywhere in photon problems (it's hc in units of eV·nm).

Alternative method - work in eV directly. Since eV = 30 keV = 3× 10⁴ eV, and hc = 1240 eV·nm, _min = hceV = 1240 eV·nm3× 10⁴ eV ≈ 0.0413 nm. No need to chase 10⁻¹⁹s and 10⁻³⁴s.

Common mistake - confusing the spectrum. The continuous X-ray spectrum has a sharp short-wavelength cut-off (set by V) but no long-wavelength cut-off. The cut-off is on the shortest wavelength side because no photon can carry more than the electron's full kinetic energy. Don't write "_max" - there isn't one (it tails off into longer-wavelength soft X-rays smoothly).

Real-world - medical and crystallographic X-rays. A dental X-ray tube runs at about 60 kV, giving _min ≈ 21 pm. Chest X-rays use 80–120 kV (harder, more penetrating photons). Crystallography (Bragg diffraction) uses characteristic K-α lines of copper λ ≈ 154 pm rather than the bremsstrahlung continuum - the continuum is just a background.

Einstein's insight (1905, Nobel 1921). Classical wave theory predicted that brighter light = more energetic electrons. Experiment said NO - the maximum kinetic energy depends only on the frequency, not the intensity. Einstein's leap: light comes in indivisible packets (photons) of energy hν, and one photon ejects one electron in an all-or-nothing collision. Intensity controls how many photons per second arrive (i.e., the photocurrent), not how energetic each one is. This was the first concrete evidence for quantization of light - and it earned Einstein his Nobel Prize (not relativity!).

Alternative method - keep eV all the way. Convert ν to a "frequency in eV" using h in eV·s: h = 4.136× 10^-15 eV·s. Then hν = (4.136× 10^-15)(6× 10¹⁴) = 2.482 eV, K_max = 2.482 - 2.14 = 0.342 eV. Almost no powers of ten to chase.

Common mistake - eV vs joule mix-up. The work function is given in eV, but hν comes out in joules unless you're careful. Always convert both to the SAME unit before subtracting. A surprising fraction of exam slips happen here.

Real-world - photomultiplier tubes and solar cells. Caesium is used as the photoemissive coating in many photodetectors precisely because its low work function (2.14 eV) makes it sensitive even to visible light. Solar cells exploit the same effect in semiconductors: a photon kicks an electron out of the valence band into the conduction band, where it can flow as current.

Why this is a one-line problem. The defining relation K_max = eV₀ is the entire content of the problem. In eV the answer is literally "the same number" as the voltage. That's a recurring trick in atomic physics - keep energies in eV and voltages in volts, and conversions disappear.

Stopping potential's experimental meaning. The retarding voltage V₀ is the smallest reverse bias at which the photocurrent vanishes. Below V₀, some photoelectrons still squeak through to the collector; at V₀, even the most energetic one is turned back; above V₀, no electron makes it. Plotting current vs voltage gives a smooth I–V curve that hits zero at V₀.

Common mistake. Don't write K_max = V₀ and forget the factor of e. The voltage is in volts (V), the energy is in joules (J), and the factor that converts them is e (in coulombs).

Real-world - Millikan's photoelectric measurement of h. Robert Millikan measured stopping potentials at various frequencies in 1916, plotted V₀ vs ν, and the slope (he got h/e) gave Planck's constant to 0.5% accuracy. That experiment, intended to disprove Einstein's quantum hypothesis, ended up confirming it. Both got Nobel Prizes for the work.

Why a photon has momentum despite being massless. Classical mechanics says p = mv, but for relativistic massless particles, energy and momentum are connected by E = pc. Photons obey this - they carry momentum even though their rest mass is zero. Photon momentum is the basis of radiation pressure (which propels solar sails) and of the recoil that lets laser-cooling slow atoms to near absolute zero.

Alternative method - the "1240 eV·nm" shortcut. E (eV) = 1240λ (nm) = 1240632.8 ≈ 1.96 eV. Memorise this and you'll save a minute on every photon problem.

Common mistake - units of momentum. Sometimes students write p = E/c and get a number, but then can't tell what the units are. With E in J and c in m/s, p is in J·s/m = kg·m/s. Stick with SI and the units take care of themselves.

Real-world - laser pointers, optical tweezers. A 1 mW laser pointer emits roughly 3×10¹⁵ red photons per second. Each carries vanishingly small momentum, but a focused laser beam can apply enough radiation pressure to trap a microbe (Arthur Ashkin's Nobel-winning "optical tweezers"). That's the same physics as part (c): the photon momentum is real, just tiny.

Insight - Millikan's measurement. Robert Millikan performed exactly this experiment in 1916. By measuring stopping potentials at several frequencies, taking the slope, and multiplying by e (which he had already measured in his famous oil-drop experiment), he extracted h to about 0.5% accuracy. This was the first precision measurement of Planck's constant - and it confirmed Einstein's photon hypothesis, which Millikan had set out to disprove.

Alternative method - use the y-intercept. The y-intercept of the V₀ vs ν line is -₀/e; the x-intercept is the threshold frequency ₀ = ₀/h. So a single graph yields BOTH the work function and Planck's constant. This is the cleanest experimental signature of the photoelectric effect.

Common mistake - slope sign and units. The slope must be positive. Units: V₀ in volts, ν in Hz = 1/s; slope is V·s. Don't mix Hz with rad/s - they differ by 2π, and your "Planck's constant" will come out a factor of 2π off you'd be measuring = h/2π instead.

Real-world - every spectroscopist uses this. The slope-of-V₀-vs-ν method, with modern electronics, gives h to better than 1 ppm precision. Combined with the Josephson constant (2e/h) and the von Klitzing constant h/e², these define modern electrical metrology - the volt and the ohm are now built from h and e, not artefacts.

Insight - what the threshold frequency means physically. Every metal has a characteristic minimum energy ₀ (its work function) needed to pry one electron off the surface. Light of frequency below ₀ = ₀/h simply cannot do it - no matter how intense the light. This was the experimental observation that classical wave theory could not explain: a very intense low-frequency beam ejects ZERO electrons, while a very weak high-frequency beam ejects them immediately.

Alternative method - use the h/e trick directly. Notice that h/e ≈ 4.14× 10^-15 V·s is exactly the slope from Q 11.5. So V₀ = (4.14× 10^-15)Δν, with Δν in Hz. That single number condenses every photoelectric stopping-potential problem.

Common mistake - confusing intensity and frequency. Cranking up the brightness of light below ₀ does NOT eventually liberate electrons. The photoemission depends on individual photon energy, not total power. Many students lose marks writing "if the light is intense enough, electrons will be emitted regardless of frequency" - that's the classical (wrong) picture.

Real-world - solar cells and night-vision goggles. Choosing a photocathode material with low ₀ = low ₀ extends sensitivity to longer wavelengths. Caesium telluride sees UV; gallium arsenide sees visible; bialkali photocathodes are the workhorses of photomultiplier tubes used in particle physics detectors.

Insight - the threshold wavelength shortcut. A neat way to handle "yes/no" photoemission questions: compute ₀ = hc/₀. If the incident λ < ₀, emission happens; if λ > ₀, it does not. With hc = 1240 eV·nm, ₀ (nm) = 1240₀ (eV). For ₀ = 4.2 eV, ₀ ≈ 295 nm - that's deep UV. Anything redder fails. The given 330 nm is in the near-UV but still too red.

Alternative method - compare frequencies. ν = c/λ = 3× 10⁸/(330× 10^-9) = 9.09× 10¹⁴ Hz, ₀ = ₀/h ≈ 1.01× 10¹⁵ Hz. Since ν < ₀, no emission.

Common mistake - confusing wavelength and frequency comparisons. A SHORTER wavelength means a HIGHER frequency, hence a MORE energetic photon. Many students invert this. Always do a sanity check: shorter λ ↔ higher ν ↔ more energetic photon ↔ more likely to emit.

Real-world - UV sensors and ozone-layer protection. Most metals (work functions 4-5 eV) require UV light to eject electrons. Ozone absorbs UV-B (280-315 nm) and UV-C (100-280 nm) before they reach Earth's surface, protecting our skin AND making most outdoor metal surfaces immune to photoelectric loss. Tungsten ₀ ≈ 4.5 eV and copper ₀ ≈ 4.7 eV are basically photo-inert in sunlight.

Insight - what the threshold frequency tells us about the material. Once we know ₀, we know the work function: ₀ = h₀. Here ₀ ≈ 6.626× 10^-344.74× 10¹⁴ ≈ 3.14× 10^-19 J ≈ 1.96 eV. That's quite low - consistent with an alkali metal (sodium, potassium, caesium).

Alternative method - solve in eV throughout. K_max = 1.64× 10^-19/(1.6× 10^-19) = 1.025 eV; hν = (4.14× 10^-15 eV·s)(7.21× 10¹⁴) = 2.985 eV; ₀ = hν - K_max = 1.96 eV; ₀ = ₀/h = 1.96/(4.14× 10^-15) = 4.73× 10¹⁴ Hz.

Common mistake - forgetting to halve. The kinetic energy is 12mv², not mv². It's an easy slip when racing through arithmetic. A doubled K_max would lead to a much smaller, often negative, ₀ - a clear sign of error.

Real-world - photoelectric work functions of common metals. Caesium (2.14 eV), potassium (2.30 eV), sodium (2.75 eV), zinc (4.3 eV), copper (4.7 eV), platinum (6.35 eV). Lower work function = easier to extract electrons = good for photocathodes. Caesium is the photodetector poster child.

Insight - recognising the material. A work function of 2.16 eV is very close to caesium (2.14 eV) - so the emitter in this experiment was likely caesium or a caesium-based alloy. Recall that in Q 11.2 we were also given caesium with ₀ = 2.14 eV; the slight discrepancy is consistent with experimental uncertainty.

Alternative method - work in joules throughout. E = hc/λ = (1.988× 10^-25)/(488× 10^-9) = 4.073× 10^-19 J; K_max = eV₀ = (1.6× 10^-19)(0.38) = 6.08× 10^-20 J; ₀ = 4.073× 10^-19 - 0.608× 10^-19 = 3.465× 10^-19 J. Same answer, more arithmetic.

Common mistake - eV vs J units. Always make sure photon energy and work function are in the SAME unit before subtracting. The "1240 eV·nm" shortcut keeps everything in eV and prevents this mistake.

Real-world - argon-ion laser eye surgery. The 488 nm and 514 nm lines of the argon-ion laser are absorbed strongly by haemoglobin and the retinal pigment epithelium. Ophthalmologists use them to weld retinal tears (photocoagulation) - the same blue-green photons that liberate caesium photoelectrons here. The energy budget is identical; just the target material changes.

Insight - why classical mechanics works for everyday objects. The de Broglie wavelength of a baseball is 10³² times smaller than the Bohr radius. No matter how cleverly you arrange apertures, you cannot create a diffraction pattern from a baseball - there's no slit narrow enough. This is why classical mechanics is recovered as the macroscopic limit of quantum mechanics. Quantum effects exist for everything, but they are vanishingly small for big things.

Alternative method - kinetic-energy form. When kinetic energy K is given instead of velocity: p = √2mK λ = h√2mK. For an electron accelerated through V volts: λ = 1.227/√V nm (with V in volts) - a JEE classic.

Common mistake - units of momentum. Use SI: kg, m, s. If you slip a "km/s" or a "gram" in, your wavelength is off by orders of magnitude. Convert before plugging.

Real-world - when does matter behave as a wave? Electrons mass 9.11× 10^-31 kg at 100 eV have λ ≈ 0.12 nm - comparable to atomic spacings. That's why electron diffraction (Davisson–Germer, 1927) confirmed de Broglie's hypothesis and why electron microscopes resolve ∼ 0.1 nm features. Neutron diffraction (used in materials research) similarly works because neutrons at thermal energies have λ ∼ 0.1 nm. Even fullerenes (C₆₀ molecules) have shown matter-wave interference in lab experiments.

Insight - why de Broglie went looking for matter waves. De Broglie noticed exactly this consistency: for photons, the "wavelength" and "h/p" coincide. He then asked, "If a known wave (light) has λ = h/p, why shouldn't a known particle (an electron) also have a wave with λ = h/p?" That is the symmetry that led him to predict matter waves in his 1924 PhD thesis. Davisson and Germer (1927) confirmed his prediction by diffracting electrons off a nickel crystal.

Alternative method - start from the photon's effective mass. A photon of energy E has relativistic "mass-equivalent" m_rel = E/c². Its momentum is then p = m_rel c = E/c. Substituting E = hν = hc/λ gives p = h/λ - same result, different route.

Common mistake - using p = mc with rest mass. The photon's rest mass is zero, so p ≠ m_rest c = 0 (that would be nonsensical). The correct formula is the massless limit of relativistic energy-momentum: E = pc. Always start from E² = (pc)² + mc²², then set m = 0.

Real-world - unification of wave and particle. This identity is at the heart of quantum field theory: every "particle" is a quantum of an underlying field, and the field's wavelength is the particle's de Broglie wavelength. For photons, the EM field; for electrons, the Dirac field; for the Higgs boson, the Higgs field. The hypothesis you just proved for light extends to everything in the Standard Model.