TECHNIQUE: Modern-Physics Option Elimination (capstone)

The four-screen reflex is the professional's first response to any claimed result. Referees run dimension and magnitude checks before reading a derivation; experimentalists sanity-check a detector reading against the order of magnitude it should have before trusting the electronics; theorists plug limiting cases into a new formula before publishing it. Fermi-style estimation is a taught skill in every serious physics programme precisely because the ability to bound an answer fast — and to reject a wrong one faster — separates a physicist from a calculator operator. The habit this programme drilled across its modern-physics core, structure and magnitude first, computation second, is not exam technique that happens to be physics; it is physics technique that happens to win exams — the one skill you will still be using long after the last paper is marked.

Option elimination is not a trick; it is what a working physicist does to any claimed result before believing it, compressed to exam speed. The four screens below are ordered by cost — cheapest first — so that a distractor killable by a ten-second units check never survives to waste a minute on symmetry. What makes this technique so lethal in modern physics specifically is the density of checkable structure: every formula on the core syllabus pins a quantity to a unit, a magnitude band, a conservation ledger and a limiting value all at once, so almost every wrong option violates something you can see. The screens are universal; the ammunition is chapter-specific. The mistake students make is treating elimination as a mood ("let me look for the obviously silly one") rather than a protocol ("units, then magnitude, then conservation, then limits — and I know exactly what each chapter feeds each screen"). This module is that ammunition, sorted.

Ninety percent of modern-physics unit checks are settled by two conversion bridges. Cache them as facts; reconstruct them when a question challenges you.

The photon bridge, $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$. A photon's energy and wavelength are locked by $E = hc/\lambda$. Feeding $h$ and $c$ in SI and converting to electron-volts and nanometres gives a single number worth more than any three formulas:

where: $E$ is the photon energy in electron-volts and $\lambda$ its wavelength in nanometres. The physical reading: a 500 nm green photon carries $1240/500 = 2.48$ eV; a violet 400 nm photon carries $3.1$ eV; a deep-UV 248 nm photon carries $5.0$ eV. Visible light lives in the narrow $1.6$–$3.1$ eV window, which is exactly why most metals — with work functions of $2$–$5$ eV — need violet or ultraviolet to eject electrons at all. Any option offering a "visible photon of 50 eV" or "an X-ray of 2 eV" has misplaced the energy by orders of magnitude and dies at Screen 1 before Screen 2 even wakes.

The electron bridge, $\lambda = 1.226/\sqrt{V}\ \text{nm}$. An electron accelerated through $V$ volts gains kinetic energy $eV$, hence momentum $p = \sqrt{2m_e eV}$, hence de Broglie wavelength $\lambda = h/p$. Collapsing the constants:

where: $V$ is the accelerating potential in volts and $\lambda$ the electron's wavelength in nanometres. The physical reading: a 100 V electron has $\lambda = 1.226/\sqrt{100} = 0.123$ nm; a 150 V electron has $\lambda = 1.226/\sqrt{150} = 0.100$ nm — an ångström, comparable to atomic spacings, which is precisely why electrons diffract off crystals (Davisson–Germer) while a cricket ball never does. The bridge derives cleanly, and we do it below, because a formula you can rebuild is a formula no distractor can spoof.

Formula options die at Screen 1; numerical options die against cached magnitudes. Modern physics is unusually kind here because its quantities occupy tight, memorable bands separated by clean powers of ten. Fermi's whole method was to place a quantity on this ladder before computing it — and a distractor that lands on the wrong rung is dead without a single line of work.

The ladder in words, so you can recite it: energies run from thermal $0.025$ eV at the bottom, through the semiconductor gap ($\sim 1$ eV), the visible photon and work function ($1.6$–$5$ eV), hydrogen's ionisation ($13.6$ eV), X-rays (keV), up to binding energy per nucleon ($\approx 8$ MeV) and gamma rays (MeV). Lengths run from the visible wavelength (hundreds of nm), down through the atom ($\sim 0.5$ Å), the X-ray and electron wavelength ($\sim 0.1$ nm), to the nucleus (a few fm, with radius $R = 1.2\,A^{1/3}$ fm). Memorise the rungs, not the digits — because an MCQ that offers a nuclear binding energy of "8 keV" or an atomic radius of "5 nm" has slipped a rung, and Screen 2 kills it the instant you place it.

Where the first two screens read structure, the third reads bookkeeping — and bookkeeping is where setters hide their most tempting distractors, because a wrong number that looks computed feels trustworthy. Three ledgers cover the core syllabus.

Energy, in the photoelectric effect. Einstein's equation $KE_{\max} = hf - \phi$ is a ledger: the photon brings $hf$, the metal charges an entry fee $\phi$ (the work function), and what remains is the electron's kinetic energy. Two hard bounds follow with no computation. The ejected electron's energy can never exceed the photon's energy $hf$ — so an option with $KE_{\max} > hf$ is dead. And $KE_{\max}$ can never be negative — so if $hf < \phi$, there is simply no emission (not "negative kinetic energy"), and an option claiming photocurrent below threshold is dead. The stopping potential obeys $eV_0 = KE_{\max}$, so a claimed $V_0$ exceeding $hf/e$ is likewise impossible.

Charge and nucleon number, in nuclear reactions. Every nuclear reaction or decay conserves the total charge $Z$ (proton number) and the total nucleon number $A$ (mass number), written as subscript and superscript. In $\alpha$-decay, $A$ drops by 4 and $Z$ by 2; in $\beta^-$-decay, $A$ is unchanged and $Z$ rises by 1 (a neutron becomes a proton). An option whose products' $A$ and $Z$ do not sum to the reactant's is dead on arithmetic alone — you never needed the energetics. This is the same conservation habit the particle chapter drilled, applied to the core-tested nuclear reactions.

Energy in equals energy out, everywhere. The $Q$-value of a reaction is the mass-energy released, $Q = (\Delta m)\,c^2$, with $1\,\text{u} = 931.5$ MeV. A fusion or fission that releases energy must move toward more tightly bound nuclei (higher binding energy per nucleon), so an option claiming energy release from a process that lowers binding-per-nucleon is dead — the binding-energy curve is itself a conservation screen.

The last screen is the deepest and the slowest, so it runs last. You plug a kind number — usually an extreme or a ground state — into every option and demand the physically obvious result.

The Bohr ladder. Hydrogen's energy levels are $E_n = -13.6/n^2$ eV. Plug the kind values: $n=1$ gives $-13.6$ eV (the ground state, the most tightly bound), $n=2$ gives $-3.4$ eV, and $n\to\infty$ gives $0$ (the electron just free — the ionisation edge). An option for a level formula that does not give $-13.6$ eV at $n=1$, or does not approach $0$ from below as $n$ grows, is dead. The ionisation energy from the ground state is therefore exactly $13.6$ eV, and from $n=2$ it is $3.4$ eV — two anchor numbers an option cannot fake.

The photoelectric threshold. Emission switches on only when $\lambda \le \lambda_0 = hc/\phi$. For a metal with $\phi = 2.5$ eV, $\lambda_0 = 1240/2.5 = 496$ nm — green light is right at the edge, red is dead, blue works. Any option predicting emission by long-wavelength light below threshold, however intense, fails this limit (intensity sets how many electrons, never whether — a Screen-3/Screen-4 favourite).

The correspondence limit. At large $n$ the Bohr level spacing shrinks and the quantum ladder blurs into the smooth classical orbit — the sanity check that quantum results must reduce to classical ones where classical physics works. An option that stays wildly quantum in the classical regime, or classical in the quantum one, has failed correspondence.

The screens live or die on knowing what "obvious" looks like at the extremes. This table is the kind-number crib — the value each core formula must return when you plug the limit, and the distractor it therefore kills.

Board level — the photon-energy screen (NEET pattern, about 30 seconds)

A metal has work function 2.5 eV. Which of these can eject photoelectrons? (a) red light, 700 nm (b) green light, 550 nm (c) blue light, 450 nm (d) infrared, 1200 nm.

Step 1. Screen 3 (energy ledger): emission needs photon energy $E = 1240/\lambda \ge \phi = 2.5$ eV. No computation of $KE$ needed — just the threshold comparison. The threshold wavelength is $\lambda_0 = 1240/2.5 = 496$ nm; anything longer is dead.

Step 2. Place each option against $496$ nm: (d) 1200 nm and (a) 700 nm and (b) 550 nm all exceed 496 nm — too long, too little energy per photon, no emission at any intensity. Only (c) 450 nm is below threshold: $1240/450 = 2.76$ eV $> 2.5$ eV, leaving $KE_{\max} = 0.26$ eV. Answer (c). Three options killed by one threshold, thirty seconds, no photoelectric equation solved in full.

Board level — the Bohr ladder (NEET/AP pattern, about 25 seconds)

The energy required to ionise a hydrogen atom already in its first excited state ($n=2$) is: (a) 13.6 eV (b) 10.2 eV (c) 3.4 eV (d) 1.5 eV.

Step 1. Screen 4 (special values): the Bohr ladder is $E_n = -13.6/n^2$ eV, and ionisation means climbing from the current level to $E_\infty = 0$. From $n=2$, $E_2 = -13.6/4 = -3.4$ eV, so the energy to reach zero is exactly $+3.4$ eV.

Step 2. Read the distractors as the traps they are: (a) 13.6 eV is ionisation from the ground state — the setter's bait for "ionisation energy of hydrogen"; (b) 10.2 eV is the $n=1\to 2$ excitation energy, not ionisation; (d) 1.5 eV is $|E_3|$, the wrong level. Answer (c). The whole item is a single plug into $-13.6/n^2$.

Competitive level — the de Broglie units trap (JEE-Main pattern, about 40 seconds)

Two quick screens, one option set each. (i) An electron accelerated through 150 V has de Broglie wavelength closest to: (a) 0.01 nm (b) 0.1 nm (c) 1 nm (d) 10 nm. (ii) A proposed formula for that wavelength is quoted; which scaling with accelerating voltage $V$ is correct? (a) $\lambda \propto V$ (b) $\lambda \propto \sqrt{V}$ (c) $\lambda \propto 1/\sqrt{V}$ (d) $\lambda \propto 1/V$.

Step 1. (i) Screen 2, the length ladder: an electron at ~100 V sits on the "about 0.1 nm" rung by the bridge $\lambda = 1.226/\sqrt{V}$. Compute only enough to confirm the rung: $1.226/\sqrt{150} = 0.100$ nm. Options (a), (c), (d) are one or two rungs off. Answer (b) — placement, not precision.

Step 2. (ii) Screen 1, the scaling audit: from the bridge, $\lambda = 1.226/\sqrt{V}$, so $\lambda \propto 1/\sqrt{V}$ outright. Answer (c). Options (a) and (b) get the direction backwards (faster electrons have shorter waves), and (d) has the wrong power — quadrupling $V$ halves $\lambda$, it does not quarter it. Timing note: 40 seconds for the pair, and notice that neither item was "solved" so much as placed and scaled. The one arithmetic step in (i) existed only to confirm which rung; (ii) needed no arithmetic at all. This is the capstone's product — the structure of every core formula familiar enough to audit at a glance.

Competitive level — the full protocol on a nuclear reaction (JEE-Advanced flavour, about 80 seconds)

In the reaction ${}^{14}_{7}\text{N} + {}^{4}_{2}\text{He} \rightarrow X + {}^{1}_{1}\text{H}$, and given that the process absorbs energy, which statement survives all four screens? (a) $X$ is ${}^{17}_{8}\text{O}$ and the products are more tightly bound than the reactants. (b) $X$ is ${}^{17}_{8}\text{O}$ and energy must be supplied for the reaction to proceed. (c) $X$ is ${}^{16}_{8}\text{O}$ and the nucleon number is conserved. (d) $X$ is ${}^{18}_{9}\text{F}$ with a $Q$-value of about 8 keV.

Step 1. Screen 3 (conservation): balance the ledgers. Nucleon number: $14 + 4 = 18$, and the ${}^{1}_{1}\text{H}$ carries 1 away, so $X$ has $A = 17$. Charge: $7 + 2 = 9$, minus the proton's 1, so $X$ has $Z = 8$ — oxygen. Thus $X = {}^{17}_{8}\text{O}$. That instantly kills (c) — its ${}^{16}_{8}\text{O}$ has $A = 16$, so nucleons do not balance — and (d), whose ${}^{18}_{9}\text{F}$ has the wrong $A$ and $Z$ both.

Step 2. Screen 3 again, on energetics: the problem states the reaction absorbs energy (it is endoergic, historically Rutherford's first artificial transmutation). Energy in equals energy out means the products are less tightly bound and energy must be supplied. That kills (a), which claims the products are more tightly bound — the sign is backwards. Only (b) states $X = {}^{17}_{8}\text{O}$ and that energy must be supplied.

Step 3. Screen 2 as a final guard: even the surviving option's cousin (d) fails the magnitude ladder — a nuclear $Q$-value of "8 keV" is a rung too low; nuclear energetics live at MeV, not keV. Answer (b), eighty seconds, and the only arithmetic was two integer sums. Timing note: the stretch here is that nothing announced "conserve $A$ and $Z$ first". The screens fire on structure regardless of costume — two integer balances killed two options, one sign argument killed a third, and the magnitude ladder stood guard over the rest. This is the last worked example of the programme; the drill ladder below is yours.

Under 45 seconds each; the screen to fire is named after the em-dash. Answers stated — mark honestly.

Drill 1 (M1). A photon of wavelength 620 nm is said to carry 20 eV — Screen 1/2: $1240/620 = 2.0$ eV, a visible photon on the eV rung; "20 eV" is a rung too high. Dead.

Drill 2 (M1). Doubling the light intensity is claimed to double the maximum photoelectron energy — Screen 4: intensity sets the number of electrons, never $KE_{\max} = hf - \phi$. Dead.

Drill 3 (M1). An electron at 100 V is given de Broglie wavelength 12 nm — Screen 1/2: $1.226/\sqrt{100} = 0.123$ nm, an ångström-scale wave; "12 nm" is two rungs off. Dead.

Drill 4 (M2). The ground-state energy of hydrogen is quoted as $-3.4$ eV — Screen 4: $-13.6/n^2$ at $n=1$ is $-13.6$ eV; $-3.4$ eV is the $n=2$ level. Dead.

Drill 5 (M2). An option gives the $n=3\to 2$ (H-alpha) photon as 1.89 keV — Screen 2: $E_3 - E_2 = -1.51-(-3.4) = 1.89$ eV, a visible line at 656 nm; "keV" is X-ray territory. Dead.

Drill 6 (M2). As $n$ grows, hydrogen levels are said to spread further apart — Screen 4: $-13.6/n^2$ crowds toward 0 as $n$ rises; the spacing shrinks. Dead.

Drill 7 (M3). A nuclear binding energy per nucleon is quoted as 8 keV — Screen 2: the curve peaks near 8.8 MeV (iron); binding-per-nucleon is MeV-scale, never keV. Dead.

Drill 8 (M3). A $\beta^-$-decay is written to raise the nucleon number $A$ by one — Screen 3: in $\beta^-$-decay $A$ is unchanged (a neutron becomes a proton), only $Z$ rises by one. Dead.

Drill 9 (M4). The band gap of silicon is quoted as 11 eV — Screen 2: semiconductor gaps are about 1 eV (Si 1.1, Ge 0.7); 11 eV is insulator territory. Dead.

Drill 10 (cross). A photoelectron is claimed to leave with more kinetic energy than the incident photon carried — Screen 3: $KE_{\max} = hf - \phi \le hf$ always; you cannot get out more than came in. Dead. Ten kills, almost zero solving. If any drill cost more than its screen's ten-to-twenty seconds, the chapter behind it is what to re-read — the screens are only as fast as the ammunition is familiar.

Screens in cost order. Units (10 s) then magnitude (10 s) then conservation (20 s) then limits (deepest). Firing a limit argument on a units-killable option wastes its depth; reach for the cheap screen first, always.

Compute, then audit, for numerical-entry. In integer/numerical-answer questions (JEE-Main and Advanced both use them) there are no options to eliminate — the screens run in audit mode: solve honestly, then check your answer sits on the right rung and obeys the ledgers before committing.

This capstone drills the CORE, not the frontier. The syllabus-tested modern physics on JEE-Main, JEE-Advanced, NEET and AP Physics 2 is the dual nature of radiation and matter, atoms (Bohr), nuclei (binding energy, radioactivity), and semiconductors. Special relativity, particle physics, stellar physics and cosmology — the rest of this chapter — are enrichment beyond the Indian core boards (they appear in IB-HL and some A-Level options only); they are not on your competitive paper, and this technique deliberately stays inside the core.

Intensity versus energy is the most-tested photoelectric trap. Raising the light's brightness lifts the number of photoelectrons (the current), never their maximum kinetic energy — that is set by frequency alone through $hf - \phi$. Setters manufacture this distractor on almost every photoelectric MCQ.

Ionisation from the ground state is 13.6 eV; from $n=2$ it is 3.4 eV. The commonest atomic-physics bait swaps the excitation energy (10.2 eV for $n=1\to 2$) for the ionisation energy, or quotes ground-state ionisation when the atom starts excited. Read which level the question names before you plug.

Marking-scheme honesty. JEE and NEET use negative marking, so a two-survivor guess pays only when the scheme's arithmetic is in your favour — check your paper's exact scheme (Advanced's partial-credit multi-correct differs from Main's single-correct, and AP-2's differs again) before you gamble on a survivor.

The core physics this module drills lives in its prerequisites, which are all live: the photoelectric effect (hs2-photoelectric-effect, the source of $KE_{\max} = hf - \phi$ and the intensity-versus-energy trap), the Bohr model (hs2-bohr-model, the $-13.6/n^2$ ladder and the ionisation anchors), and mass defect and binding energy (hs2-mass-defect-binding-energy, the $1\,\text{u} = 931.5$ MeV bridge and the binding-energy curve as a conservation screen). The elimination protocol itself — the four screens in cost order and the guessing arithmetic — is the technique this capstone packages for the core modern-physics core. Beyond the core, this chapter's own enrichment modules await for IB-HL and A-Level readers, and the deepest layer of matter — quarks, leptons and the Standard Model (hs2-particle-physics-standard-model) — is where the conservation-law screen returns with baryon and lepton number added to charge. That is the natural next step: the same bookkeeping habit, one deep dive further. Provenance: fresh technique module and capstone of the HS physics programme; the drills are composed for this article and every kill is verified against the honest physics of its home module. Recompute list: $hc = 1240\ \text{eV}\,\cdot\,\text{nm}$; photon energies at 400, 450, 500, 550, 620, 700 nm; de Broglie $1.226/\sqrt{V}$ at 100 and 150 V; Bohr levels $-13.6/n^2$ and the ionisation/excitation anchors; the H-alpha and Lyman-alpha lines; threshold $\lambda_0 = hc/\phi$; nuclear radius $1.2\,A^{1/3}$; binding energy per nucleon; the ${}^{14}$N($\alpha$,p)${}^{17}$O balance; the cricket-ball de Broglie wavelength.