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Dual Nature of Radiation and Matter

Physics · Class 12 · Part 2NCERT pp. 274–2862 sets × 25 questionsFree

Key facts and formulas for Dual Nature of Radiation and Matter, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.

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Formula sheet for this chapter

IdeaRule or fact
Photon energyE = hν = hc/λ; hc = 1240 eV nm = 12400 eV Å
Photon momentump = h/λ = E/c; equivalent mass m = hν/c²
Einstein equationK_max = hν − φ₀ = e V₀; φ₀ = hν₀
PhotocurrentSaturation current ∝ intensity (number of photons); V₀ does not depend on intensity
V₀ and K_max graphsK_max vs ν: slope h, intercept ν₀; V₀ vs ν: slope h/e; V₀ vs 1/λ: slope hc/e
de Broglie wavelengthλ = h/p = h/√(2mK); λ ∝ 1/√K
Electron through V voltλ = 12.27/√V Å; for a particle, λ = h/√(2mqV)
X-ray cut-offλ_min = hc/(eV): depends only on the tube voltage
Radiation force and pressureAbsorbed: F = P/c, pressure = I/c; reflected back: twice as much
Key experimentsMillikan: charge e; Davisson–Germer: electron diffraction; Compton: photon particle nature

Three sample questions

Q1. Read the statements. I. Crookes discovered cathode rays in 1870. II. Thomson confirmed that they are negative particles and measured their speed. III. Millikan measured the specific charge e/m of the electron. Which are correct?
  1. I and II only
  2. II and III only
  3. I and III only
  4. I, II and III
Hint and solution

Hint: Recall who measured the charge e alone and who measured the ratio e/m. They are different experiments.

  1. I: Crookes 1870, cathode rays — Crookes discovered cathode rays in 1870, so statement I is correct.
  2. II: Thomson, fields at right angles — Thomson used crossed electric and magnetic fields, showed the rays are negative particles and measured their speed. Statement II is correct.
  3. III: Millikan measured charge e — Millikan's oil drop experiment measured the electron charge e, not the ratio e/m. Statement III is false.
  4. Correct: I and II only — Only I and II are correct. Option A.
Answer: A. Millikan measured the electron charge (oil drop, 1913), not e/m. Statements I and II are correct.
Q2. An oil drop carries a charge of 9.612 × 10⁻¹⁹ C. The number of excess electrons on it is
  1. 5
  2. 6
  3. 7
  4. 8
Hint and solution

Hint: Charge is quantised: total charge equals a whole number times e. Divide by e = 1.602 × 10⁻¹⁹ C.

  1. q = n e — Charge on a drop is always a whole number n times the electron charge e.
  2. n = 9.612 × 10⁻¹⁹ / 1.602 × 10⁻¹⁹ — Divide the given charge by e. The powers of ten cancel.
  3. n = 6 — 9.612 divided by 1.602 gives exactly 6.
  4. Answer: 6 electrons — So the drop has six excess electrons. Option B.
Answer: B. n = 9.612 × 10⁻¹⁹ / 1.602 × 10⁻¹⁹ = 6.
Q3. Electrons are pulled out of a sharp metal tip by an extremely strong electric field of about 10⁸ V m⁻¹. This process is called
  1. thermionic emission
  2. photoelectric emission
  3. field emission
  4. Compton scattering
Hint and solution

Hint: The energy is supplied by a very strong electric field, not by heat or light.

  1. Electron emission needs energy ≥ φ₀ — Electrons leave a metal only if they get energy more than the work function.
  2. Heat → thermionic; light → photoelectric; field → field emission — The three emissions are named by where the energy comes from.
  3. Here the source is a field of 10⁸ V m⁻¹ — The problem says a strong electric field pulls electrons out, so the process is field emission.
  4. Answer: field emission — So the answer is field emission. Option C.
Answer: C. A field of about 10⁸ V/m tearing electrons from the metal is field emission (a spark plug is an example).

How to use this chapter

Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.

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