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Electron, Photon, Photoelectric Effect and X-rays

NEET > Physics > Dual Nature of Matter and Radiation

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Chapter Snapshot - Electron, Photon, Photoelectric Effect and X-rays

This chapter covers the dual nature of radiation and matter - from the discovery of cathode rays and J.J. Thomson's e/m measurement, through Einstein's photoelectric equation (hf = W + KE_max), to de Broglie matter waves and X-ray production. NEET frequently tests stopping potential, threshold frequency, de Broglie wavelength of accelerated particles, minimum wavelength of X-rays, and Moseley's law. The Compton effect and characteristic vs continuous X-ray spectra complete the picture of photon-matter interactions.

✓ Use This To Plan Your First 2–3 Hours
Expected Questions (Typical)
Q
2-3
NEET regularly asks 2-3 questions from this chapter. Photoelectric effect (Einstein equation, stopping potential, threshold frequency graphs) is the most tested area. de Broglie wavelength of charged particles and minimum wavelength of X-rays are other favourites.
Time Required (Practical)
⏱
8-10 hrs
Theory and derivations ~3 hrs; photoelectric effect numericals ~2 hrs; de Broglie wavelength problems ~1.5 hrs; X-ray spectrum and Moseley's law ~1.5 hrs; MCQ revision ~1-2 hrs.
Difficulty Level
⚡
Moderate
The photoelectric equation is straightforward but graph interpretation (V0 vs frequency, photocurrent vs intensity) trips many students. de Broglie wavelength formulae require careful substitution. Moseley's law needs attention to the screening constant b.
Most Asked Style: Numerical - calculate stopping potential given wavelength and work function; find de Broglie wavelength of an electron accelerated through V volts; determine minimum X-ray wavelength from tube voltage. Conceptual MCQs on graphs of photoelectric effect and intensity vs wavelength of X-ray spectra.Biggest Trap: Confusing threshold frequency (property of the metal) with frequency of incident light. Students often set stopping potential proportional to intensity instead of frequency. Remember: intensity changes photocurrent, frequency changes stopping potential.Fast Win: Memorise three golden formulae: (i) eV0 = h(f minus f0); (ii) de Broglie wavelength of electron = 12.27/sqrt(V) angstrom; (iii) minimum X-ray wavelength = 12375/V angstrom. These three alone cover about 60% of NEET marks from this chapter.Revision-Friendly: Draw the V0 vs frequency graph (straight line, x-intercept = threshold frequency, slope = h/e) and the I vs wavelength X-ray spectrum (continuous curve with sharp characteristic peaks). These two diagrams anchor the entire chapter.

Subtopics - Electron, Photon, Photoelectric Effect and X-rays (NEET)

Dual nature of matter and radiation - from cathode rays to X-ray spectra, covering the particle and wave aspects of electrons and photons

Revision tip: Build your revision around four anchors: (1) Thomson's e/m = E squared / (2VB squared); (2) Einstein's photoelectric equation hf = W + eV0; (3) de Broglie wavelength = h/p = h/sqrt(2mE); (4) X-ray cutoff wavelength = hc/eV. Master these, and every numerical in this chapter becomes a substitution exercise.
NCERT LinesMCQsQuick Test

1) Cathode Rays, Canal Rays and e/m Determination

Discovery of cathode rays in gas discharge tubes, properties of cathode rays, J.J. Thomson's crossed-field experiment to measure e/m, Millikan's oil drop experiment for charge quantisation, positive (canal) rays and their properties, Thomson's and Bainbridge mass spectrographs.

Cathode Rayse/m = 1.77 x 10^11 C/kgMillikan's Oil DropCanal RaysBainbridge Spectrograph
›
Gas Discharge and Cathode RaysSystematic study of electric discharge through gases at decreasing pressures - from sparks at 10 mm Hg through positive column, Faraday's dark space, Crooke's dark space, striations, to cathode ray emission at 0.01 mm Hg. Properties: travel in straight lines, produce fluorescence, deflected by E and B fields.
›
J.J. Thomson's e/m ExperimentCrossed electric and magnetic fields balance on an electron beam: eE = evB gives v = E/B. Combined with eV = (1/2)mv squared, yields e/m = E squared / (2VB squared). Thomson measured e/m = 1.77 x 10^11 C/kg. Specific charge decreases at relativistic speeds due to mass increase.
›
Millikan's Oil Drop ExperimentMeasurement of electronic charge by balancing gravitational, electric and viscous forces on charged oil droplets. Charge always found to be an integral multiple of e = 1.602 x 10^(minus 19) C, establishing charge quantisation.
›
Canal Rays and Mass SpectrographsPositive ions formed by electron-atom collisions pass through perforated cathode as canal rays. q/m depends on gas (unlike cathode rays). Thomson's mass spectrograph: parabolic traces z squared = k(q/m)y. Bainbridge spectrograph: velocity selector (v = E/B) followed by magnetic deflection r = mv/(qB'), separating isotopes by mass.

2) Matter Waves and Wave-Particle Duality

de Broglie hypothesis that every moving particle has an associated wavelength, expressions for de Broglie wavelength of charged and uncharged particles, Davisson-Germer experimental verification, characteristics of matter waves, and Heisenberg's uncertainty principle.

lambda = h/mvlambda_e = 12.27/sqrt(V) angstromDavisson-GermerHeisenberg: dx.dp >= h/4piMatter Waves
›
de Broglie WavelengthWavelength of a moving particle: lambda = h/p = h/mv = h/sqrt(2mE). For charged particle accelerated through V: lambda = h/sqrt(2mqV). Numerical shortcuts: electron = 12.27/sqrt(V) angstrom, proton = 0.286/sqrt(V) angstrom, alpha-particle = 0.101/sqrt(V) angstrom.
›
Davisson and Germer ExperimentElectron beam scattered from nickel crystal shows intensity maximum at scattering angle 50 degrees for 54 V accelerating potential. Bragg's formula 2d sin(theta) = n lambda gives lambda = 1.65 angstrom, matching de Broglie prediction of 12.27/sqrt(54) = 1.67 angstrom. This experimentally verified the wave nature of electrons.
›
Characteristics of Matter WavesMatter waves represent probability of finding a particle, are not electromagnetic, and are associated with all moving particles regardless of charge. Phase velocity can exceed c. Number of de Broglie waves in nth Bohr orbit equals n, linking to Bohr's quantisation condition 2 pi r = n lambda.
›
Heisenberg's Uncertainty PrincipleSimultaneous measurement of position and momentum is fundamentally limited: dx times dp >= h/(4 pi). Also applies to energy-time: dE times dt >= h/(4 pi). Explains non-existence of electrons in the nucleus and finite width of spectral lines.

3) Photon and Photoelectric Effect

Einstein's photon model of light, photon energy-mass-momentum relations, photoelectric effect - work function, threshold frequency, stopping potential, Einstein's photoelectric equation, effect of intensity and frequency on photocurrent, and the Compton effect as evidence for photon momentum.

E = hf = hc/lambdap = h/lambdahf = W + eV0V0 independent of intensityCompton shift
›
Photon PropertiesEnergy E = hf = hc/lambda; in eV: E = 12375/lambda(angstrom). Rest mass zero but effective mass m = hf/c squared. Momentum p = E/c = h/lambda. Number of photons per second from source of power P: n = P lambda/(hc). Intensity I = P/(4 pi r squared) for point source.
›
Photoelectric Effect and Einstein's EquationPhoton absorbed by surface electron: hf = W0 + KE_max. Work function W0 = hf0 = hc/lambda0. Stopping potential: eV0 = h(f minus f0). If f < f0, no emission regardless of intensity. Photo-effect is instantaneous with no time delay, confirming particle nature of light.
›
Effect of Intensity and FrequencyIncreasing intensity (same frequency) raises photocurrent but stopping potential V0 stays unchanged. Increasing frequency (same intensity) raises stopping potential V0 but saturated photocurrent stays the same. V0 vs f graph: straight line with slope h/e, x-intercept at threshold frequency f0.
›
Compton EffectScattering of a photon by a free electron. Compton shift: delta lambda = (h/m0c)(1 minus cos phi). At phi = 90 degrees, delta lambda = h/(m0c) = 0.024 angstrom. At phi = 180 degrees, delta lambda = 2h/(m0c) = 0.048 angstrom. Shift depends only on scattering angle, not on incident wavelength or material.

4) X-ray Production, Spectra and Moseley's Law

Production of X-rays in a Coolidge tube, continuous and characteristic X-ray spectra, minimum wavelength (cutoff), properties of X-rays, absorption, Moseley's law relating characteristic frequency to atomic number, and applications of X-rays.

lambda_min = 12375/V angstromContinuous (Bremsstrahlung)Characteristic K, L, M seriesMoseley: sqrt(f) = a(Z minus b)Hard vs Soft X-rays
›
Coolidge Tube and X-ray ProductionTungsten filament cathode emits electrons accelerated through 10-80 kV onto a high-Z, high-melting-point target. Only about 2% of electron energy converts to X-rays; rest becomes heat dissipated through copper block and cooling fins. Intensity controlled by filament current; penetrating power (quality) controlled by accelerating voltage.
›
Continuous X-ray SpectrumProduced by deceleration of electrons near target nuclei (Bremsstrahlung). When an electron loses all energy in one collision: eV = hc/lambda_min, giving cutoff wavelength lambda_min = 12375/V angstrom. Intensity-wavelength curve starts at lambda_min, peaks, then drops. The peak shifts to shorter wavelengths at higher voltage.
›
Characteristic X-ray Spectrum and Moseley's LawInner-shell electron knocked out; outer electron fills vacancy, emitting X-ray photon of definite wavelength. K-series: transitions to K-shell (K-alpha from L to K, K-beta from M to K). L-series: transitions to L-shell. Moseley's law: sqrt(f) = a(Z minus b), where b = 1 for K-series, 7.4 for L-series. For K-alpha: f = (3Rc/4)(Z minus 1) squared.
›
Properties and Applications of X-raysX-rays are EM waves with wavelength 0.1 to 100 angstrom. Hard X-rays: higher frequency (~10^19 Hz), shorter wavelength (0.1-4 angstrom), greater penetration. Soft X-rays: lower frequency (~10^16 Hz), longer wavelength (4-100 angstrom). Absorption: I = I0 exp(minus mu x), with mu proportional to lambda cubed and Z to the fourth power. Applications include crystallography, medical imaging, and radiotherapy.

Electron, Photon, Photoelectric Effect and X-rays Download Notes & Weightage Plan

For each topic in the Electron, Photon, Photoelectric Effect and X-rays chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.

2 Downloads

Cathode Rays, Canal Rays and e/m Determination

Discovery of cathode rays in gas discharge tubes, properties of cathode rays, J.J. Thomson's crossed-field experiment to measure e/m, Millikan's oil drop experiment for charge quantisation, positive (canal) rays and their properties, Thomson's and Bainbridge mass spectrographs.

Cathode Rayse/m = 1.77 x 10^11 C/kgMillikan's Oil DropCanal RaysBainbridge Spectrograph

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Focus on Thomson's crossed-field logic: balance eE = evB to get v = E/B, then use eV = (1/2)mv squared to derive e/m = E squared/(2VB squared). For Millikan's experiment, remember charge is always an integral multiple of e. Canal rays: q/m is NOT universal (depends on gas), unlike cathode rays. Bainbridge spectrograph separates isotopes using velocity selector followed by magnetic deflection.
Download NotesPrintable PDF
★
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Draw the J.J. Thomson apparatus diagram with E and B fields perpendicular to the beam. Write the two-step derivation for e/m. Tabulate differences between cathode rays and canal rays. Practise one numerical on Bainbridge spectrograph (r = mE/(qBB')).

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1NEET occasionally asks a conceptual question on cathode ray properties or charge quantisation. Direct formula-based questions from this topic are rare.
Time Required2 hrsTheory ~1 hr (discharge stages, cathode ray properties, canal ray properties). Thomson and Millikan derivations ~0.5 hr. Mass spectrograph concepts ~0.5 hr.
DifficultyEasyMostly factual. The derivation of e/m is the only moderately challenging piece. Mass spectrograph parabola equation appears intimidating but is rarely tested.
  • Scoring Focus: Know properties of cathode rays (deflected by E and B, travel in straight lines, produce fluorescence) and canal rays (q/m depends on gas, deflection smaller). Thomson's e/m value = 1.77 x 10^11 C/kg and Millikan's e = 1.6 x 10^(minus 19) C are must-remember constants.
  • High-risk Area: Confusing cathode rays with canal rays - cathode rays have constant e/m regardless of gas; canal rays do not. Also, forgetting that cathode ray direction is independent of anode position.
  • Best Practice Style: Theory + factual recall
Priority rule: Cover this topic first as it provides historical context. Spend minimal time on derivations; focus on properties and key values.

Matter Waves and Wave-Particle Duality

de Broglie hypothesis that every moving particle has an associated wavelength, expressions for de Broglie wavelength of charged and uncharged particles, Davisson-Germer experimental verification, characteristics of matter waves, and Heisenberg's uncertainty principle.

lambda = h/mvlambda_e = 12.27/sqrt(V) angstromDavisson-GermerHeisenberg: dx.dp >= h/4piMatter Waves

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Master four formulae: lambda = h/mv = h/sqrt(2mE) = h/sqrt(2mqV). Charged particle shortcuts: electron 12.27/sqrt(V), proton 0.286/sqrt(V), alpha 0.101/sqrt(V) - all in angstrom. Thermal neutron: lambda = 25.17/sqrt(T) angstrom. Davisson-Germer gave lambda = 1.65 angstrom from diffraction, matching 1.67 angstrom from de Broglie formula. Bohr quantisation follows from 2 pi r = n lambda.
Download NotesPrintable PDF
★
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write the de Broglie formula tree: start from lambda = h/p, branch into lambda = h/mv, h/sqrt(2mE), h/sqrt(2mqV). Add numerical shortcuts for each particle. Sketch the Davisson-Germer setup and note the key numbers: 54 V, 50 degree scattering, Ni crystal.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1NEET asks 1 question almost every year - typically a numerical on de Broglie wavelength of an electron or proton accelerated through a given potential difference.
Time Required2 hrsde Broglie formulae and derivations ~1 hr. Davisson-Germer + matter wave characteristics ~0.5 hr. Uncertainty principle ~0.5 hr.
DifficultyEasy-ModerateThe formulae are direct substitution. The main challenge is remembering which shortcut applies to which particle and handling unit conversions between eV and joules.
  • Scoring Focus: lambda = 12.27/sqrt(V) for electrons is the single most-tested formula. Also know the ratio lambda_photon/lambda_electron for same energy. Uncertainty principle: dx.dp >= h/(4 pi).
  • High-risk Area: Using the electron shortcut formula (12.27/sqrt(V)) for protons or alpha particles. Each particle has a different numerical constant because mass differs.
  • Best Practice Style: Formula application + numerical drill
Priority rule: Study this immediately after Topic 1. Practise at least 5 numerical problems on de Broglie wavelength with different particles and energy units.

Photon and Photoelectric Effect

Einstein's photon model of light, photon energy-mass-momentum relations, photoelectric effect - work function, threshold frequency, stopping potential, Einstein's photoelectric equation, effect of intensity and frequency on photocurrent, and the Compton effect as evidence for photon momentum.

E = hf = hc/lambdap = h/lambdahf = W + eV0V0 independent of intensityCompton shift

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Photon: E = hf, m = hf/c squared, p = h/lambda. Photoelectric equation: hf = W0 + eV0. Work function W0 = hf0 = hc/lambda0. Stopping potential V0 = (h/e)(f minus f0) is independent of intensity; photocurrent is proportional to intensity. V0 vs f graph: slope = h/e, intercept on frequency axis = f0. Compton shift delta lambda = (h/m0c)(1 minus cos phi) depends only on scattering angle.
Download NotesPrintable PDF
★
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Draw V0 vs frequency and photocurrent vs collector potential graphs side by side. Write Einstein's equation in all four equivalent forms. Tabulate work functions of common metals (Cs = 1.9 eV, Na = 2.7 eV, Cu = 4.7 eV). Solve 5 numericals on stopping potential and maximum velocity.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2Photoelectric effect is the highest-yield subtopic. NEET asks 1-2 questions every year - numericals on stopping potential, threshold wavelength, or graph-based conceptual questions.
Time Required3 hrsEinstein's equation and photon theory ~1 hr. Graphs and conceptual analysis ~1 hr. Numericals and Compton effect ~1 hr.
DifficultyModerateThe equation itself is simple. Difficulty arises in graph interpretation - reading off work function from intercepts, distinguishing intensity effects from frequency effects, and applying Compton shift formula.
  • Scoring Focus: Einstein's photoelectric equation in all forms. V0 vs f graph: slope = h/e, y-intercept = minus W0/e, x-intercept = f0. Stopping potential is independent of intensity. Photocurrent saturates at higher intensity. No emission below threshold frequency regardless of intensity.
  • High-risk Area: Assuming stopping potential increases with intensity (it does not - only frequency changes V0). Forgetting that maximum KE corresponds to surface electrons; deeper electrons lose energy escaping. Mixing up Compton wavelength (0.024 angstrom) with Compton shift at 180 degrees (0.048 angstrom).
  • Best Practice Style: Graph analysis + numerical drill
Priority rule: This is the highest-weightage topic. Study it thoroughly before X-rays. Spend maximum time on graph interpretation and numerical practice.

X-ray Production, Spectra and Moseley's Law

Production of X-rays in a Coolidge tube, continuous and characteristic X-ray spectra, minimum wavelength (cutoff), properties of X-rays, absorption, Moseley's law relating characteristic frequency to atomic number, and applications of X-rays.

lambda_min = 12375/V angstromContinuous (Bremsstrahlung)Characteristic K, L, M seriesMoseley: sqrt(f) = a(Z minus b)Hard vs Soft X-rays

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Coolidge tube: filament current controls intensity, accelerating voltage controls quality. Continuous spectrum from Bremsstrahlung; cutoff lambda_min = hc/eV = 12375/V angstrom. Characteristic spectrum from inner-shell transitions: K-alpha (L to K), K-beta (M to K). Moseley's law: sqrt(f) = a(Z minus b), b = 1 for K-series. For K-alpha: E = 10.2(Z minus 1) squared eV. Absorption coefficient mu is proportional to lambda cubed and Z to the fourth.
Download NotesPrintable PDF
★
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Draw the X-ray intensity vs wavelength graph showing continuous background with characteristic peaks superimposed. Mark lambda_min. Write Moseley's law with b values for K, L, M series. Sketch the energy level diagram for K-alpha, K-beta, L-alpha transitions.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1NEET asks about 1 question - usually lambda_min calculation from tube voltage, or a conceptual question on continuous vs characteristic spectra, or Moseley's law application.
Time Required2.5 hrsCoolidge tube and X-ray production ~0.5 hr. Continuous and characteristic spectra ~1 hr. Moseley's law and properties ~1 hr.
DifficultyModerateThe cutoff wavelength formula is straightforward. Moseley's law requires remembering screening constants. Distinguishing continuous from characteristic spectra is conceptual. The K-series energy level diagram can be confusing.
  • Scoring Focus: lambda_min = 12375/V angstrom (same formula as photon energy, applied inversely). Know that characteristic wavelength depends on Z (target material), not on accelerating voltage. Moseley's sqrt(f) = a(Z minus 1) for K-alpha. X-ray production is the reverse of the photoelectric effect.
  • High-risk Area: Thinking characteristic X-ray wavelength changes with accelerating voltage (it does not - it depends only on atomic number Z). Confusing hard and soft X-rays - hard means higher frequency, shorter wavelength, greater penetration. Forgetting screening constant b = 1 for K-series in Moseley's law.
  • Best Practice Style: Conceptual clarity + formula application
Priority rule: Study after photoelectric effect - X-ray production is its reverse process. Focus on lambda_min formula and Moseley's law. Properties of X-rays are factual - read once and revise from a table.

Electron, Photon, Photoelectric Effect and X-rays Chapter NEET Traps & Common Mistakes (Topic-Wise)

Each subtopic below is of the Electron, Photon, Photoelectric Effect and X-rays chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.

! Avoid Easy Negatives
Stopping Potential and Intensity
Photoelectric EffectStopping PotentialIntensity

Mistake Snapshot (What Students Do Wrong)

  • Intensity changes V0: Students assume that increasing light intensity increases stopping potential. Stopping potential V0 depends only on frequency of incident light, not on intensity. Higher intensity means more photons per second, hence more photoelectrons, hence higher photocurrent - but V0 stays fixed.
  • Emission below threshold: Some students believe that sufficiently high intensity can cause emission even below threshold frequency. This is incorrect. If f is less than f0, no single photon carries enough energy to liberate an electron, regardless of how many photons arrive per second.
2–3 Line Example (Typical Error)

A sodium surface (W0 = 2.3 eV) is illuminated by light of wavelength 400 nm. Photon energy = 12375/4000 = 3.09 eV. KE_max = 3.09 minus 2.3 = 0.79 eV. Stopping potential V0 = 0.79 V. Doubling the intensity doubles the photocurrent but V0 remains 0.79 V.

How NEET Frames The Trap

NEET sets questions where two light sources have different intensities but same frequency, or same intensity but different frequencies, then asks which parameter changes.

NEET-Style Trap Question Format

Q. When the intensity of incident light on a photosensitive surface is doubled (frequency unchanged), which quantity doubles?
A. Stopping potential   B. Maximum kinetic energy of photoelectrons   C. Number of photoelectrons emitted per second   D. Work function of the metal  
Trick: Stopping potential and KE_max depend on frequency, not intensity. Work function is a material property. Only the number of photoelectrons (and hence photocurrent) doubles when intensity doubles at constant frequency.

Quick rule: Intensity controls COUNT of photoelectrons; frequency controls ENERGY of each photoelectron.
de Broglie Wavelength Shortcuts
de BroglieWavelengthCharged Particles

Mistake Snapshot (What Students Do Wrong)

  • Using electron shortcut for all particles: The formula lambda = 12.27/sqrt(V) angstrom applies only to electrons. Protons use 0.286/sqrt(V) and alpha-particles use 0.101/sqrt(V). Each constant depends on the particle mass. Using 12.27 for a proton gives a wavelength that is too large by a factor of about 43.
  • Forgetting charge in qV: For alpha-particles, kinetic energy = qV = 2eV (charge is 2e, not e). Substituting q = e instead of q = 2e in lambda = h/sqrt(2mqV) gives the wrong answer.
2–3 Line Example (Typical Error)

An alpha-particle (m = 4 amu, q = 2e) is accelerated through 100 V. lambda = h/sqrt(2 x 4 x 1.67 x 10^(minus 27) x 2 x 1.6 x 10^(minus 19) x 100) = 0.101/sqrt(100) = 0.0101 angstrom. Using the electron formula gives 12.27/sqrt(100) = 1.227 angstrom - over 100 times too large.

How NEET Frames The Trap

NEET may give the same accelerating voltage for different particles and ask for the ratio of their de Broglie wavelengths. Students who memorise only the electron shortcut get trapped.

NEET-Style Trap Question Format

Q. An electron and a proton are accelerated through the same potential difference V. The ratio of their de Broglie wavelengths (lambda_e / lambda_p) is:
A. 1   B. sqrt(m_p / m_e)   C. m_p / m_e   D. sqrt(m_e / m_p)  
Trick: lambda = h/sqrt(2mqV). For same q and V, lambda is proportional to 1/sqrt(m). Therefore lambda_e/lambda_p = sqrt(m_p/m_e), which is approximately 43. Option (b) is correct.

Quick rule: For same V and same charge: lambda ratio = inverse square root of mass ratio.
Minimum Wavelength of X-rays
X-raysCutoff WavelengthContinuous Spectrum

Mistake Snapshot (What Students Do Wrong)

  • Confusing continuous and characteristic wavelengths: The minimum wavelength (cutoff) belongs to the continuous spectrum and depends on accelerating voltage. Characteristic wavelengths depend on the target material (atomic number Z) and are independent of accelerating voltage. Students often mix these up.
  • Wrong formula application: lambda_min = 12375/V(in volts) angstrom is often confused with photon energy E = 12375/lambda angstrom. The formulas are the same rearranged, but careless substitution (using eV instead of V, or nm instead of angstrom) leads to errors.
2–3 Line Example (Typical Error)

A Coolidge tube operates at 50 kV. lambda_min = 12375/50000 = 0.2475 angstrom. This cutoff wavelength shifts to 0.124 angstrom if voltage is doubled to 100 kV. The characteristic K-alpha line of the target does NOT shift - it depends only on Z.

How NEET Frames The Trap

NEET asks: if accelerating voltage is increased, what happens to the X-ray spectrum? Students must identify that lambda_min decreases (continuous spectrum shifts) but characteristic lines stay at the same wavelength.

NEET-Style Trap Question Format

Q. In a Coolidge tube, if the accelerating voltage is doubled, which statement is correct?
A. Both lambda_min and characteristic wavelengths are halved   B. lambda_min is halved; characteristic wavelengths remain unchanged   C. lambda_min remains unchanged; characteristic wavelengths are halved   D. Both lambda_min and characteristic wavelengths remain unchanged  
Trick: lambda_min = hc/eV, so doubling V halves lambda_min. Characteristic wavelengths depend on atomic number Z (Moseley's law), not on voltage. Option (b) is correct.

Quick rule: Voltage controls the continuous cutoff; atomic number controls the characteristic lines.
Moseley's Law Screening Constant
Moseley's LawCharacteristic X-raysAtomic Number

Mistake Snapshot (What Students Do Wrong)

  • Using b = 1 for all series: The screening constant b = 1 applies only to K-series. For L-series b = 7.4 and for M-series b = 19.2. Using b = 1 universally gives incorrect characteristic frequencies for L and M series.
  • Forgetting (Z minus b) squared dependence: Moseley's law states sqrt(f) = a(Z minus b), meaning f is proportional to (Z minus b) squared. Students who write f proportional to Z miss the screening correction and the squared relationship.
2–3 Line Example (Typical Error)

For K-alpha line of copper (Z = 29): f = (3Rc/4)(Z minus 1) squared = (3Rc/4)(28) squared. For L-alpha of the same element the screening constant changes to b = 7.4, giving a completely different frequency. Applying b = 1 to the L-series overestimates the frequency significantly.

How NEET Frames The Trap

NEET may give two elements and ask the ratio of K-alpha frequencies. Students must use (Z1 minus 1) squared / (Z2 minus 1) squared, not Z1 squared / Z2 squared.

NEET-Style Trap Question Format

Q. The ratio of frequencies of K-alpha X-rays of two elements with atomic numbers 31 and 21 is approximately:
A. 31/21   B. (31/21)^2   C. (30/20)^2   D. 30/20  
Trick: For K-alpha, b = 1. Frequency is proportional to (Z minus 1) squared. Ratio = (30/20) squared = 9/4 = 2.25. Using Z squared instead of (Z minus 1) squared gives (31/21) squared = 2.18, which is close but wrong.

Quick rule: K-alpha frequency ratio = (Z1 minus 1) squared / (Z2 minus 1) squared. Never drop the screening constant.

Topics

Electric Discharge Through Gases

Cathode Rays

J.J. Thomson's Experiment

Millikan's Oil Drop Experiment

Positive Rays or Canal Rays

Bainbridge Mass Spectrograph

Matter Waves (de-Broglie Waves)

Thomson's Mass Spectrograph

Heisenberg's Uncertainty Principle

Photon

Photo-Electric Effect

Compton Effect

X-Rays

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