Part of PH-01 — Dual Nature of Radiation & Matter

PYQ Pattern Analysis

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NEET Question Pattern Analysis — PH-01

PatternFrequencyYears AppearedTrap / Subtlety
"V0V_{0} vs intensity" conceptualVery High (≥5 times)2017, 2019, 2020, 2022, 2024V0V_{0} does NOT change with intensity. Answer is always "unchanged."
Graph interpretation: slope of KE vs νHigh (3-4 times)2018, 2021, 2023Slope = h (not h/e). The V0V_{0} vs ν slope is h/e — easily confused.
Calculate KE_max and V0V_{0} from λ and φVery High2018, 2019, 2020, 2023Must convert λ to energy in eV using E = hc/λ or E = 1240/λ(nm) eV
de Broglie λ for electron through VHigh2021, 2022, 2024Use λ = 1.227/√V nm. This ONLY works for electrons (m = m_e, q = e).
Electron vs proton λ comparison at same VModerate2019, 2021λ ∝ 1/√m at same V. λ_e/λ_p = √(m_p/m_e) ≈ 43.
Threshold wavelength from work functionModerate2018, 2022λ_{0} = hc/φ. Use hc = 1240 eV·nm for quick calculation: λ_{0} = 1240/φ(eV) nm
Davisson-Germer conceptualLow-Moderate2016, 2020Confirms wave nature of ELECTRONS (matter), not photons. Ni crystal, 54V, 50°.
Two-wavelength problem (find h and φ)Moderate2017, 2023eV01V_{01} = hν_{1} − φ and eV02V_{02} = hν_{2} − φ → subtract to get h; then get φ

Top 5 NEET Traps in This Chapter

  1. Intensity → V0V_{0} trap: "Doubled intensity → V0V_{0} doubles." Answer: NO. V0V_{0} is frequency-dependent only.
  2. Slope confusion: KE vs ν slope = h; V0V_{0} vs ν slope = h/e. Students mix these up under pressure.
  3. Proton shortcut trap: Using λ = 1.227/√V nm for protons/alpha particles (valid ONLY for electrons).
  4. Frequency-doubling KE: If ν doubles, KE_max does NOT double. KE_max = h(2ν) − φ = 2hν − φ ≠ 2KE_max.
  5. Work function units: φ must be in same units as hν when computing KE_max. If φ in eV and E in J — conversion error!

NEET Shortcut: hc = 1240 eV·nm

E(eV)=1240λ(nm)andλ0(nm)=1240ϕ(eV)E(\text{eV}) = \frac{1240}{\lambda(\text{nm})} \quad \text{and} \quad \lambda_0(\text{nm}) = \frac{1240}{\phi(\text{eV})}

This avoids the long calculation with h = 6.63×10346.63 \times 10^{-34} J s and c = 3×1083 \times 10^{8} m/s.

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