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Dual Nature of Radiation & Matter: Error Notes

Aamirah Fathima · NCERT Class 12 Physics, Chapter 11 · 80 questions · +4 / −1 / 0

Score147 / 320
Correct43
Wrong25
Skipped12
Accuracy63%
Target 290: missed by 143 marks. Danger line 260: below it by 113.
Target allows 6 wrong, 0 blank: 25 wrong and 12 blank. The wrong answers cost 125 marks; the blanks left 48 marks unclaimed.
Quickest gain: the Davisson–Germer block (Q77–80) is pure NCERT recall worth 16 marks, and Q8, Q10 and Q37 are one-line questions.
Biggest fix: intensity → number of electrons (current); frequency → energy (stopping potential). Three mistakes (Q4, Q36, Q45) come from this one idea.
Strong areas: Photons: energy, momentum & intensity (10/12) and Einstein's equation (10/16).

Section-wise marks

SectionQs✔✘–Marks
Electron emission & photoelectric effectQ1–103438/40
Einstein's photoelectric equationQ11–26104236/64
Threshold frequency & stopping potentialQ27–373626/44
Reading photoelectric graphsQ38–4663021/36
Photons: energy, momentum & intensityQ47–58102038/48
de Broglie waves of particlesQ59–6955115/44
de Broglie λ from accelerating potentialQ70–7661023/28
Davisson–Germer & evidence for matter wavesQ77–800040/16

Answer grid

1C2A→C3D4A→C5C→D6B7C→A8–→B9–→D10–→A11B→D12D13B14D15–→C16A17C18C19A20C21A→B22B→D23B24D25C→A26–→C27C→B28C29–→A30B→D31B→D32C→A33A34C35B→D36A→B37–→D38A39A40D→B41B42D43D→C44C45B→D46C47D48B→D49A50C51B52B53D54D→B55B56A57C58C59D→A60D61D62A→B63B→A64B→C65A66A→D67A68A69–→C70C71B72C73B74B75B→A76A77–→B78–→A79–→B80–→B
Correct (her answer)Wrong (hers → key)Skipped (– → key)Click a red or grey box to jump to its note.

Error patterns

PatternQuestions
Intensity vs frequency mixed upQ4, Q45 (stopping potential), Q36 (A–R), Q5 (below threshold)
Power-of-ten / unit slipsQ21 (eV given for J), Q30, Q32, Q59
Formula used incompletelyQ22 (K ∝ ν assumed), Q25 (φ not subtracted), Q27, Q31 (h/2e), Q48 (no 4πr²), Q75 (missed a factor)
Ratios invertedQ62, Q63, Q64 (heavier particle ⇒ shorter λ)
Graph readingQ40, Q43 (intercept tells φ, slope tells h/e), Q54
Recall / wordingQ2 (thermionic), Q7, Q11, Q35 (calculate, not recall), Q66
SkippedQ8, Q9, Q10, Q15, Q26, Q29, Q37, Q69 and the whole Davisson–Germer block Q77–80

Error notes (25 wrong + 12 skipped)

Q2Electron emissionWrong · −1

A tungsten filament inside an evacuated tube is raised to 2500 K and electrons leave its surface. This process is called

  1. (A)photoelectric emissionyour answer
  2. (B)field emission
  3. (C)thermionic emissioncorrect
  4. (D)secondary emission
What went wrong: Nothing in the question mentions light. The energy comes from heating.

Solution

  1. The escape energy is supplied by heat (2500 K).
  2. Emission caused by heat = thermionic emission (electron guns, old valves).
Thermionicheat Field~10⁸ V m⁻¹ Photoelectricphotons Secondaryfast particles Named after the source of the escape energy (≥ φ).
Remember: Name the source of energy: heat → thermionic, strong field → field, light → photoelectric, fast particles → secondary.
Q4Effect of intensityWrong · −1

In Lenard's photoelectric experiment the intensity of the incident light is increased while its frequency is held fixed. Then

  1. (A)the stopping potential increases and the saturation current is unchangedyour answer
  2. (B)both the saturation current and the stopping potential increase
  3. (C)the saturation current increases and the stopping potential is unchangedcorrect
  4. (D)neither the saturation current nor the stopping potential changes
What went wrong: You swapped the two effects. Intensity controls the number of electrons; frequency controls their energy.

Solution

  1. More intensity → more photons per second → more photoelectrons → saturation current ↑.
  2. Each photon still has energy hν → K_max unchanged → stopping potential unchanged.
2I (doubled)I −V₀collector potential → Same ν → same −V₀; intensity changes only the saturation level.
Remember: Intensity → current. Frequency → stopping potential.
Q5Threshold frequencyWrong · −1

Light whose frequency is below the threshold frequency falls on a clean metal surface. Its intensity is then made one thousand times as large. The number of photoelectrons emitted per second

  1. (A)increases in proportion to the square root of the intensity
  2. (B)becomes one thousand times as large
  3. (C)increases, but only after a measurable delayyour answer
  4. (D)remains zerocorrect
What went wrong: Option C is the wave-theory prediction, and the photon picture rejects it. Below threshold, no emission happens however long or bright the light.

Solution

  1. Each photon has hν < φ, so no single photon can free an electron.
  2. An electron absorbs one photon at a time; energy does not build up.
  3. More photons with too little energy each still give zero emission.
Remember: Below ν₀: no emission at any intensity, and no waiting time helps.
Q7Assertion–ReasonWrong · −1

A: However intense the incident light, no photoelectron leaves the metal if the frequency of the light is below the threshold frequency of that metal.
R: An electron takes up the energy of one whole photon at a time, so the energy available to it is fixed by the frequency alone and not by how many photons arrive each second.

  1. (A)Both A and R are true and R is the correct explanation of Acorrect
  2. (B)Both A and R are true but R is not the correct explanation of A
  3. (C)A is true but R is falseyour answer
  4. (D)A is false but R is true
What went wrong: You put crosses on R, but R is a correct statement of the one-photon-one-electron rule, and it is exactly why A is true.

Solution

  1. A is the observed threshold law, so it is true.
  2. R: absorption is quantum (one photon → one electron), and the energy per photon is hν, which is true.
  3. R explains A: if hν < φ, no single photon is enough, and intensity can't add photons together → option (A).
Remember: Ask: “Does R tell me why A happens?” Here it does.
Q8Work functionsSkipped · 0

Of caesium, sodium, calcium and copper, the metal that responds to the longest wavelengths, because its work function is the smallest, is

  1. (A)sodium
  2. (B)caesiumcorrect
  3. (C)calcium
  4. (D)copper
What went wrong: Left blank. Remember the NCERT work-function table order.

Solution

  1. Work functions (NCERT): Cs 2.14 < K 2.30 < Na 2.75 < Ca 3.20 < … < Cu 4.65 eV.
  2. Smallest φ → largest threshold wavelength λ₀ = hc/φ → caesium (it responds even to visible light).
Remember: Alkali metals (Cs, K, Na) have the lowest φ and Cs is the lowest of all, which is why it is used in photocells.
Q9Saturation currentSkipped · 0

The photocurrent reaches a saturation value once the collector is made sufficiently positive, because at that stage

  1. (A)the collector potential has exceeded the stopping potential
  2. (B)the cathode has run out of free electrons
  3. (C)all the incident photons have been used up
  4. (D)every electron emitted from the cathode is collectedcorrect
What went wrong: Left blank.

Solution

  1. Increasing positive collector potential pulls in more of the emitted electrons.
  2. Once all emitted electrons reach the collector, the current can't rise further, so it saturates.
  3. The number emitted is fixed by intensity, so the saturation level depends on intensity.
2I (doubled)I −V₀collector potential → Same ν → same −V₀; intensity changes only the saturation level.
Remember: Saturation = 100% collection, not 100% emission.
Q10Emission types (match)Skipped · 0

Match each kind of electron emission with the agent that supplies the energy:
A. Thermionic · B. Field · C. Photoelectric · D. Secondary
I. Incident photons · II. Heating the emitter · III. Bombardment by fast particles · IV. A very strong external electric field

  1. (A)A–II, B–IV, C–I, D–IIIcorrect
  2. (B)A–II, B–I, C–IV, D–III
  3. (C)A–III, B–IV, C–II, D–I
  4. (D)A–I, B–II, C–III, D–IV
What went wrong: Left blank. This is a direct NCERT match, worth 4 easy marks.

Solution

  1. Thermionic ← heat (II); Field ← strong field ≈ 10⁸ V m⁻¹ (IV); Photoelectric ← photons (I); Secondary ← fast particles (III).
Thermionicheat Field~10⁸ V m⁻¹ Photoelectricphotons Secondaryfast particles Named after the source of the escape energy (≥ φ).
Remember: Attempt every match question: getting even two pairs right usually narrows it to one option.
Q11Einstein's equationWrong · −1

Light of wavelength 400 nm falls on a metal whose work function is 2.0 eV. The wavelength is then changed to 300 nm, the metal being unchanged. The maximum kinetic energy of the photoelectrons increases by

  1. (A)3.11 eV
  2. (B)1.11 eVyour answer
  3. (C)2.07 eV
  4. (D)1.04 eVcorrect
What went wrong: The increase in K_max equals the increase in photon energy. φ cancels and you don't need it.

Solution

  1. E = 1243/λ(nm) eV: E₄₀₀ = 3.11 eV, E₃₀₀ = 4.14 eV.
  2. ΔK = ΔE = 4.14 − 3.11 = 1.04 eV.
Remember: Same metal, two wavelengths: ΔK_max = Δ(hc/λ).
Q15Work function (symbolic)Skipped · 0

Light of wavelength λ₁ gives photoelectrons of maximum kinetic energy K₁, and light of wavelength λ₂ gives maximum kinetic energy K₂, from the same surface. The work function of the surface is

  1. (A)(K₂ − K₁)/(λ₁K₁ − λ₂K₂)
  2. (B)(λ₁K₂ − λ₂K₁)/(λ₁ − λ₂)
  3. (C)(λ₁K₁ − λ₂K₂)/(λ₂ − λ₁)correct
  4. (D)(λ₂ − λ₁)/(λ₂K₁ − λ₁K₂)
What went wrong: Left blank (you underlined it). Eliminate hc first.

Solution

  1. hc = λ₁(K₁ + φ) = λ₂(K₂ + φ).
  2. λ₁K₁ + λ₁φ = λ₂K₂ + λ₂φ → φ(λ₁ − λ₂) = λ₂K₂ − λ₁K₁.
  3. φ = (λ₁K₁ − λ₂K₂)/(λ₂ − λ₁).
  4. Dimension check: (A) and (D) are not energy, so they're out immediately.
Remember: For symbolic options, check dimensions first; it often removes two options.
Q21Units: eV vs JWrong · −1

The threshold wavelength of a metal is 650 nm. Its work function, in joules, is about

  1. (A)1.91 × 10⁻¹⁹ Jyour answer
  2. (B)3.06 × 10⁻¹⁹ Jcorrect
  3. (C)4.9 × 10⁻¹⁹ J
  4. (D)2.04 × 10⁻¹⁹ J
What went wrong: 1.91 is the answer in eV. The question asks for joules.

Solution

  1. φ = hc/λ₀ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(650 × 10⁻⁹) = 3.06 × 10⁻¹⁹ J.
  2. (= 1.91 eV; × 1.6 × 10⁻¹⁹ → 3.06 × 10⁻¹⁹ J.)
Remember: Underline the unit asked for. The eV number is a planted trap in the J options.
Q22Doubling frequencyWrong · −1

Light of frequency ν gives photoelectrons of maximum kinetic energy K from a metal of work function φ. If the frequency is doubled, the new maximum kinetic energy is

  1. (A)4K
  2. (B)2Kyour answer
  3. (C)K + φ
  4. (D)2K + φcorrect
What went wrong: K is not proportional to ν; it is hν minus φ.

Solution

  1. hν = K + φ.
  2. K' = 2hν − φ = 2(K + φ) − φ = 2K + φ (more than double).
ν₀Aν₀B AB −φ/eνV₀ slope = h/e (same for all) V₀ = (h/e)ν − φ/e: x-intercept ν₀ = φ/h, y-intercept −φ/e
Remember: K_max vs ν is a straight line that does not pass through the origin, so doubling ν more than doubles K.
Q25Max speedWrong · −1

Light of wavelength 350 nm falls on a metal of work function 2.0 eV. The maximum speed of the photoelectrons is about

  1. (A)7.4 × 10⁵ m s⁻¹correct
  2. (B)5.2 × 10⁵ m s⁻¹
  3. (C)1.1 × 10⁶ m s⁻¹your answer
  4. (D)3.7 × 10⁵ m s⁻¹
What went wrong: 1.1 × 10⁶ m s⁻¹ is the speed you get if all 3.55 eV goes into KE, i.e. φ was not subtracted.

Solution

  1. E = 1243/350 = 3.55 eV → K = 3.55 − 2.0 = 1.55 eV = 2.48 × 10⁻¹⁹ J.
  2. v = √(2K/m) = √(2 × 2.48 × 10⁻¹⁹/9.1 × 10⁻³¹) = 7.4 × 10⁵ m s⁻¹.
Remember: Subtract φ before converting to speed.
Q26Photoelectrons in a B-fieldSkipped · 0

Photons of energy 3 eV fall on a metal of work function 1 eV. The fastest photoelectrons then enter a uniform magnetic field of 2 × 10⁻⁴ T at right angles to their velocity. The radius of their circular path is about

  1. (A)4.8 cm
  2. (B)1.2 cm
  3. (C)2.4 cmcorrect
  4. (D)0.6 cm
What went wrong: Left blank. It's a two-chapter link: Einstein's equation, then r = p/eB.

Solution

  1. K = 3 − 1 = 2 eV = 3.2 × 10⁻¹⁹ J.
  2. p = √(2mK) = √(2 × 9.1 × 10⁻³¹ × 3.2 × 10⁻¹⁹) = 7.63 × 10⁻²⁵ kg m s⁻¹.
  3. r = p/(eB) = 7.63 × 10⁻²⁵/(1.6 × 10⁻¹⁹ × 2 × 10⁻⁴) = 0.024 m = 2.4 cm.
rv B into the page ⊗ r = mv/(eB) = √(2mK)/(eB)
Remember: r = √(2mK)/(eB). Keep it ready for mixed photoelectric + magnetism questions.
Q27Halving λWrong · −1

Light of wavelength 300 nm falling on a metal of work function 2.0 eV gives a stopping potential of 2.14 V. If the wavelength is halved, the stopping potential becomes

  1. (A)4.28 V
  2. (B)6.29 Vcorrect
  3. (C)1.07 Vyour answer
  4. (D)8.29 V
What went wrong: Halving λ doubles the photon energy; it doesn't halve V₀. And V₀ isn't proportional to E anyway.

Solution

  1. E₃₀₀ = 2.14 + 2.0 = 4.14 eV → E₁₅₀ = 8.29 eV.
  2. V₀' = 8.29 − 2.0 = 6.29 V.
Remember: Change the photon energy first, then subtract φ.
Q29Two stopping potentialsSkipped · 0

A metal surface illuminated by light of wavelength λ gives a stopping potential V₀. When illuminated by light of wavelength 2λ, the stopping potential falls to V₀/3. The threshold wavelength of the surface is

  1. (A)4λcorrect
  2. (B)2λ
  3. (C)3λ
  4. (D)6λ
What went wrong: Left blank. This is the same structure as optics-paper style two-equation questions.

Solution

  1. eV₀ = hc/λ − φ … (1); eV₀/3 = hc/2λ − φ … (2).
  2. 3 × (2): eV₀ = 3hc/2λ − 3φ. Equate with (1): hc/λ − φ = 3hc/2λ − 3φ → 2φ = hc/2λ.
  3. φ = hc/4λ → λ₀ = hc/φ = 4λ.
Remember: Multiply one equation so the V₀ terms match, then subtract.
Q30Threshold frequencyWrong · −1

The threshold wavelength of a metal is 500 nm. Its threshold frequency is

  1. (A)3.0 × 10¹⁴ Hz
  2. (B)1.5 × 10¹⁴ Hzyour answer
  3. (C)6.0 × 10¹⁵ Hz
  4. (D)6.0 × 10¹⁴ Hzcorrect
What went wrong: Power-of-ten / division slip. ν = c/λ, not λ/c or c/2λ.

Solution

  1. ν₀ = c/λ₀ = 3 × 10⁸/5 × 10⁻⁷ = 6.0 × 10¹⁴ Hz.
Remember: Sanity check: visible light is about 4–8 × 10¹⁴ Hz, and 500 nm is visible.
Q31Slope of V₀–ν graphWrong · −1

In a photoelectric experiment the stopping potential is plotted against the frequency of the incident light. The slope of the resulting straight line is

  1. (A)1.06 × 10⁻³⁴ V s
  2. (B)2.07 × 10⁻¹⁵ V syour answer
  3. (C)4.14 × 10⁻¹⁴ V s
  4. (D)4.14 × 10⁻¹⁵ V scorrect
What went wrong: You halved h/e. Slope = h/e exactly.

Solution

  1. eV₀ = hν − φ → V₀ = (h/e)ν − φ/e.
  2. Slope = h/e = 6.63 × 10⁻³⁴/1.6 × 10⁻¹⁹ = 4.14 × 10⁻¹⁵ V s.
ν₀Aν₀B AB −φ/eνV₀ slope = h/e (same for all) V₀ = (h/e)ν − φ/e: x-intercept ν₀ = φ/h, y-intercept −φ/e
Remember: V₀–ν slope is h/e (V s); K–ν slope is h (J s).
Q32Intercept of V₀–ν graphWrong · −1

For a certain metal the stopping-potential-against-frequency line cuts the stopping-potential axis at −2.0 V. The threshold frequency of that metal is about

  1. (A)4.8 × 10¹⁴ Hzcorrect
  2. (B)3.0 × 10¹⁴ Hz
  3. (C)4.8 × 10¹⁵ Hzyour answer
  4. (D)1.2 × 10¹⁵ Hz
What went wrong: Right method, but the power of ten slipped (10¹⁵ instead of 10¹⁴).

Solution

  1. y-intercept = −φ/e = −2 V → φ = 2 eV = 3.2 × 10⁻¹⁹ J.
  2. ν₀ = φ/h = 3.2 × 10⁻¹⁹/6.63 × 10⁻³⁴ = 4.8 × 10¹⁴ Hz.
ν₀Aν₀B AB −φ/eνV₀ slope = h/e (same for all) V₀ = (h/e)ν − φ/e: x-intercept ν₀ = φ/h, y-intercept −φ/e
Remember: Check the region: a 2 eV threshold (≈ 620 nm) is visible light, ~10¹⁴ Hz.
Q35Planck's constant from dataWrong · −1

Light of wavelength 400 nm gives a stopping potential of 1.2 V and light of wavelength 300 nm gives a stopping potential of 2.2 V, on the same surface. The value of Planck's constant that follows from these two readings is

  1. (A)1.6 × 10⁻³⁴ J s
  2. (B)6.6 × 10⁻³⁴ J syour answer
  3. (C)3.2 × 10⁻³⁴ J s
  4. (D)6.4 × 10⁻³⁴ J scorrect
What went wrong: You picked the textbook value of h. The question wants the value these readings give.

Solution

  1. e(V₂ − V₁) = hc(1/λ₂ − 1/λ₁).
  2. 1.6 × 10⁻¹⁹ × 1.0 = h × 3 × 10⁸ × (1/300 − 1/400) × 10⁹ = h × 3 × 10⁸ × 8.33 × 10⁵.
  3. h = 1.6 × 10⁻¹⁹/2.5 × 10¹⁴ = 6.4 × 10⁻³⁴ J s.
Remember: “Follows from these readings” → calculate; don't recall.
Q36Assertion–ReasonWrong · −1

A: The photoelectric current increases when the intensity of the incident light is increased.
R: The energy carried by each photon is proportional to the frequency of the light.

  1. (A)Both A and R are true and R is the correct explanation of Ayour answer
  2. (B)Both A and R are true but R is not the correct explanation of Acorrect
  3. (C)A is true but R is false
  4. (D)A is false but R is true
What went wrong: Both are true, but R is about photon energy (frequency), while A is about photon number (intensity).

Solution

  1. A true: more photons/s → more electrons/s → more current.
  2. R true: E = hν.
  3. R doesn't explain A, because current depends on photon number, not photon energy → (B).
Remember: Intensity ↔ number of photons. Frequency ↔ energy per photon. An R about one can't explain an A about the other.
Q37Threshold wavelengthSkipped · 0

The work function of a metal is 2.5 eV. The longest wavelength that can eject photoelectrons from it is about

  1. (A)248 nm
  2. (B)310 nm
  3. (C)621 nm
  4. (D)497 nmcorrect
What went wrong: Left blank. This is a one-line question.

Solution

  1. λ₀ = hc/φ = 1243 eV nm/2.5 eV = 497 nm.
Remember: Memorise hc ≈ 1240 eV nm for fast conversions.
Q40Reading V₀–ν graphsWrong · −1

The figure shows stopping potential plotted against frequency for two metals A and B (parallel lines; A cuts the ν-axis at ν₀A, to the left of ν₀B). Which statement is correct?

  1. (A)A has the larger work function, because its line lies above B's
  2. (B)B has the larger work function, because its line meets the frequency axis further to the rightcorrect
  3. (C)A has the larger work function, because its threshold frequency is the smaller one
  4. (D)the two work functions are equal, because the lines are parallelyour answer
What went wrong: Parallel lines only mean the slope (h/e) is the same. The work function is read from the intercept.

Solution

  1. x-intercept = ν₀ = φ/h.
  2. ν₀B > ν₀A → φ_B > φ_A → B has the larger work function.
ν₀Aν₀B AB −φ/eνV₀ slope = h/e (same for all) V₀ = (h/e)ν − φ/e: x-intercept ν₀ = φ/h, y-intercept −φ/e
Remember: Slope → h/e (universal). Intercept → work function (depends on the metal).
Q43Reading I–V graphsWrong · −1

The figure shows photocurrent against retarding potential for three metals X, Y, Z illuminated by light of one frequency (Z cuts off furthest, then Y, then X). The metal with the smallest work function is

  1. (A)Y
  2. (B)X
  3. (C)Zcorrect
  4. (D)they all have the same work functionyour answer
What went wrong: The curves have the same saturation current, but that only tells you the intensity is the same. The work function shows up in the cut-off.

Solution

  1. Same ν for all three. The cut-off gives V₀ (stopping potential).
  2. Z has the largest V₀ → largest K_max = hν − φ → smallest φ.
ZYX Z cuts off furthest → largest V₀ → largest K_max → smallest φ eV₀ = hν − φ, with ν the same for all three metals.
Remember: Further cut-off → bigger V₀ → smaller φ (for a fixed ν).
Q45Doubling intensity on a graphWrong · −1

In the figure (three frequencies ν₃ > ν₂ > ν₁ with equal saturation current), the intensity of the beam of frequency ν₂ alone is doubled, its frequency unchanged. Its curve would then show

  1. (A)twice the saturation current, meeting the axis further to the right
  2. (B)twice the saturation current, meeting the axis further to the leftyour answer
  3. (C)the same saturation current, meeting the axis further to the left
  4. (D)twice the saturation current, meeting the axis at the same pointcorrect
What went wrong: Doubling intensity doesn't change the cut-off. The stopping potential depends only on frequency.

Solution

  1. Intensity × 2 → saturation current × 2.
  2. ν unchanged → V₀ unchanged → curve meets the axis at the same point.
2I (doubled)I −V₀collector potential → Same ν → same −V₀; intensity changes only the saturation level.
Remember: Same as Q4. This mistake came up twice, so it's worth revising.
Q48Photon flux from a point sourceWrong · −1

A 100 W lamp radiates light of wavelength 500 nm equally in all directions. The number of photons crossing one square metre each second at a distance of 2 m from the lamp is about

  1. (A)1.3 × 10¹⁹
  2. (B)2.5 × 10²⁰your answer
  3. (C)5.0 × 10²⁰
  4. (D)5.0 × 10¹⁸correct
What went wrong: 2.5 × 10²⁰ is the total photons per second from the lamp. You didn't spread them over the sphere of radius 2 m.

Solution

  1. N = Pλ/hc = 100 × 5 × 10⁻⁷/1.989 × 10⁻²⁵ = 2.5 × 10²⁰ s⁻¹.
  2. Area = 4πr² = 4π(2)² = 50.3 m².
  3. Per m² = 2.5 × 10²⁰/50.3 = 5.0 × 10¹⁸ m⁻² s⁻¹.
r = 2 m 100 W area = 4πr²= 50.3 m² photons per m² per s = (P/E_photon)/(4πr²)
Remember: ‘Per square metre at distance r’ → divide by 4πr².
Q54Photon speed graphWrong · −1

The graph of the speed of a photon in vacuum against its frequency is

  1. (A)a straight line parallel to the speed axis
  2. (B)a straight line parallel to the frequency axiscorrect
  3. (C)a straight line through the origin at 45° to the frequency axis
  4. (D)a rectangular hyperbolayour answer
What went wrong: A rectangular hyperbola would be λ vs ν (λν = c). The speed itself never changes.

Solution

  1. In vacuum every photon travels at c, whatever ν is.
  2. Speed = constant → horizontal line, parallel to the frequency axis.
c νspeed c = 3 × 10⁸ m s⁻¹ for every frequency
Remember: Read the axes: speed vs ν → constant; λ vs ν → hyperbola.
Q59de Broglie (macroscopic)Wrong · −1

The de Broglie wavelength of a ball of mass 0.05 kg moving at 10 m s⁻¹ is

  1. (A)1.33 × 10⁻³³ mcorrect
  2. (B)1.33 × 10⁻³² m
  3. (C)6.63 × 10⁻³⁴ m
  4. (D)1.33 × 10⁻³⁴ myour answer
What went wrong: Power-of-ten slip: mv = 0.5, and 6.63 × 10⁻³⁴/0.5 = 1.33 × 10⁻³³.

Solution

  1. p = mv = 0.05 × 10 = 0.5 kg m s⁻¹.
  2. λ = h/p = 6.63 × 10⁻³⁴/0.5 = 1.33 × 10⁻³³ m.
Remember: Dividing by a number < 1 makes the answer bigger, so the exponent rises.
Q62Same λ → compare KWrong · −1

A proton and an α-particle have the same de Broglie wavelength. The ratio of their kinetic energies, K_p : K_α, is

  1. (A)1 : 4your answer
  2. (B)4 : 1correct
  3. (C)2 : 1
  4. (D)1 : 2
What went wrong: Inverted ratio. Same λ means same p, so the lighter particle has more KE.

Solution

  1. Same λ → same p.
  2. K = p²/2m → K ∝ 1/m → K_p/K_α = m_α/m_p = 4 → 4 : 1.
Remember: Same p → K ∝ 1/m. Same K → λ ∝ 1/√m. Same V → λ ∝ 1/√(mq).
Q63Same K → compare λWrong · −1

A proton and an α-particle have the same kinetic energy. The ratio of their de Broglie wavelengths, λ_p : λ_α, is

  1. (A)2 : 1correct
  2. (B)1 : 2your answer
  3. (C)4 : 1
  4. (D)1 : 4
What went wrong: Inverted again. The lighter particle has the longer wavelength.

Solution

  1. λ = h/√(2mK) → λ ∝ 1/√m at equal K.
  2. λ_p/λ_α = √(m_α/m_p) = √4 = 2 → 2 : 1.
Remember: After a ratio, check: lighter → longer λ.
Q64Same speedWrong · −1

An electron and a proton move with the same speed. Then

  1. (A)their de Broglie wavelengths are equal
  2. (B)the proton has the longer de Broglie wavelengthyour answer
  3. (C)the electron has the longer de Broglie wavelengthcorrect
  4. (D)the proton has the longer wavelength only if its charge is larger
What went wrong: λ = h/mv: a bigger mass gives a shorter wavelength.

Solution

  1. Same v → λ ∝ 1/m.
  2. m_e ≪ m_p → λ_e ≫ λ_p → electron has the longer wavelength (≈ 1836×).
Remember: Charge never appears in h/mv.
Q66Nature of matter wavesWrong · −1

Electrons from a gun pass through a double slit and build up a fringe pattern on the screen. This shows that matter waves

  1. (A)are longitudinal pressure waves in the air of the tubeyour answer
  2. (B)are electromagnetic waves of very short wavelength
  3. (C)travel through the apparatus at the speed of light
  4. (D)are neither electromagnetic nor mechanical wavescorrect
What went wrong: The tube is evacuated, so there is no air for pressure waves. Matter waves are a separate kind of wave.

Solution

  1. No medium is needed, so they're not mechanical.
  2. They're associated with charged or neutral particles of any mass, not EM radiation, and they don't travel at c.
  3. → neither electromagnetic nor mechanical.
Remember: Matter waves are probability waves: no medium, no EM fields, speed ≠ c.
Q69KE from λSkipped · 0

An electron has a de Broglie wavelength of 1.0 Å. Its kinetic energy is about

  1. (A)302 eV
  2. (B)75 eV
  3. (C)151 eVcorrect
  4. (D)15 eV
What went wrong: Left blank. You had the momentum right in Q68, so only one more step was needed.

Solution

  1. p = h/λ = 6.63 × 10⁻²⁴ kg m s⁻¹ (from Q68).
  2. K = p²/2m = (6.63 × 10⁻²⁴)²/(2 × 9.1 × 10⁻³¹) = 2.42 × 10⁻¹⁷ J = 151 eV.
  3. Check: λ = 12.27/√V Å → V = (12.27)² ≈ 151 V ✔.
Remember: Shortcut for electrons: K(eV) = (12.27/λ(Å))².
Q75λ from accelerating potentialWrong · −1

A proton accelerated from rest through 200 V has a de Broglie wavelength λ₀. An α-particle accelerated from rest through 400 V has a de Broglie wavelength

  1. (A)λ₀/4correct
  2. (B)λ₀/2your answer
  3. (C)λ₀/8
  4. (D)λ₀/16
What went wrong: You likely forgot one factor: the α-particle has 4× mass, 2× charge and 2× voltage.

Solution

  1. λ = h/√(2mqV).
  2. α: m × 4, q × 2, V × 2 → product × 16 → λ ÷ √16 = ÷ 4.
  3. λ_α = λ₀/4.
Remember: List the three ratios (m, q, V), multiply them, then take the square root.
Q77Davisson–GermerSkipped · 0

In the Davisson–Germer apparatus, the electron beam strikes the nickel crystal along the normal to its face and the detector is swung round to different values of φ. The accelerating potential V is used to

  1. (A)fix the number of electrons leaving the filament each second
  2. (B)fix the de Broglie wavelength of the electrons in the beamcorrect
  3. (C)deflect the scattered electrons towards the detector
  4. (D)heat the nickel crystal to the required temperature
What went wrong: Left blank. The whole Davisson–Germer section (Q77–80) was skipped, giving up 16 marks.

Solution

  1. V gives each electron energy eV → momentum √(2meV).
  2. λ = h/√(2meV) → V sets the de Broglie wavelength. (The filament current sets the number of electrons.)
filament + anode (V) nickel crystal detector φ V sets the electron energy eV → momentum → λ = h/√(2meV) λ = 12.27/√V Å → at 54 V, λ = 1.67 Å
Remember: Filament current → number of electrons; accelerating V → λ.
Q78Davisson–Germer peakSkipped · 0

The polar graph shows the scattered electron intensity at an accelerating potential of 54 V. The pronounced peak near φ = 50° arises from

  1. (A)constructive interference between electron waves scattered from different layers of atoms in the crystalcorrect
  2. (B)destructive interference between electron waves scattered from different layers of atoms in the crystal
  3. (C)constructive interference between electron waves scattered from the same atom of the crystal
  4. (D)resonance between the incident electrons and the vibrating nickel atoms
What went wrong: Left blank.

Solution

  1. The atomic planes act like a diffraction grating (as with X-rays).
  2. At 50°, waves from successive layers arrive in phase → constructive interference → peak.
φ = 0° peak at φ ≈ 50° Waves scattered fromdifferent atomic layersarrive in phase Constructive interference → wave nature of electrons (54 V)
Remember: A peak means constructive interference, and it takes different layers to make interference.
Q79λ at 54 VSkipped · 0

In the Davisson–Germer experiment the electrons were accelerated through 54 V. Their de Broglie wavelength was about

  1. (A)0.167 Å
  2. (B)1.67 Åcorrect
  3. (C)16.7 Å
  4. (D)1.67 nm
What went wrong: Left blank.

Solution

  1. λ = 12.27/√V Å = 12.27/√54 = 12.27/7.35 = 1.67 Å.
  2. This matched the value from X-ray diffraction (1.65 Å), confirming de Broglie's hypothesis.
filament + anode (V) nickel crystal detector φ V sets the electron energy eV → momentum → λ = h/√(2meV) λ = 12.27/√V Å → at 54 V, λ = 1.67 Å
Remember: 12.27/√V Å for electrons: learn it.
Q80What D–G provedSkipped · 0

The experiments of Davisson and Germer and of G. P. Thomson established

  1. (A)the particle nature of electrons
  2. (B)the wave nature of electronscorrect
  3. (C)that the electron carries a negative charge
  4. (D)the value of the specific charge e/m of the electron
What went wrong: Left blank. This is pure recall and worth 4 free marks.

Solution

  1. Electron diffraction is a wave property → the wave nature of electrons.
Remember: Particle nature of electrons → J.J. Thomson (e/m). Wave nature → Davisson–Germer and G.P. Thomson.