Electromagnetic radiation travels as oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation. It does not need a medium — it travels through vacuum at 3 × 10⁸ m/s. This is the starting point for understanding atomic structure in NCERT Class 11 Chemistry Chapter 2.
The wave model works — until it doesn't. James Clerk Maxwell's wave theory explains interference, diffraction, and polarisation. A wave is described by wavelength (λ), frequency (ν), and the relation c = νλ. The electromagnetic spectrum ranges from radio waves (long λ, low ν) to gamma rays (short λ, high ν). For NEET, you must be able to rank regions by wavelength and frequency without hesitation.
Where the wave model fails: Black-body radiation and the photoelectric effect. Classical wave theory predicts that increasing light intensity should eject electrons with higher kinetic energy. Experimentally, that does not happen. Increasing intensity increases the number of ejected electrons, not their kinetic energy. Kinetic energy depends on frequency.
Planck's quantum theory resolved the black-body crisis: energy is emitted or absorbed in discrete packets (quanta). E = hν, where h = 6.626 × 10⁻³⁴ J·s. Einstein extended this to explain the photoelectric effect: each photon carries energy hν. If hν ≥ hν₀ (the work function, threshold energy), an electron is ejected. The excess energy becomes kinetic energy: KE = hν − hν₀.
Common confusion: Students conflate intensity with frequency. Intensity is energy per unit area per unit time — proportional to the number of photons, not to the energy per photon. A high-intensity beam below threshold frequency ejects zero electrons. A low-intensity beam above threshold frequency ejects electrons immediately.
Watch-out for NEET: Questions may ask you to identify which property of the photoelectric effect cannot be explained by wave theory, or to calculate KE given frequency and threshold frequency. Keep c = νλ, E = hν, and KE = hν − hν₀ ready.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following properties of the photoelectric effect CANNOT be explained by the classical wave theory of light?
Answer: A. The existence of a threshold frequency is inexplicable by classical wave theory, which predicts that any frequency should eject electrons if given enough intensity. This is a key failure of the wave model (NCERT Class 11 Chemistry Chapter 2, page 41).
Why B is wrong: B — Wave theory does predict that oscillating electric fields can dislodge surface electrons, so emission itself is not the problematic observation.
Why C is wrong: C — Wave theory correctly predicts that higher intensity (more energy delivered) should liberate more electrons.
Why D is wrong: D — Wave theory allows for material-dependent response to radiation, so dependence on the metal's nature is not the issue.
The energy of a photon of radiation with frequency 5.0 × 10¹⁴ Hz is (h = 6.626 × 10⁻³⁴ J·s):
Answer: D. E = hν = 6.626 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.313 × 10⁻¹⁹ J. Direct application of Planck's equation (NCERT Class 11 Chemistry Chapter 2).
Why A is wrong: A — This value is roughly h × 2 × 10¹⁴, suggesting the student divided ν by an extra factor or misread the frequency.
Why B is wrong: B results from a power-of-ten error (10⁻³⁴ × 10¹⁴ = 10⁻²⁰ is wrong; correct exponent sum is −34 + 14 = −20, but the coefficient product 6.626 × 5.0 = 33.13, shifting to 3.313 × 10⁻¹⁹).
Why C is wrong: C — This is the value of h × 10⁵, suggesting the student used ν = 10¹⁵ instead of 5.0 × 10¹⁴.
Electromagnetic radiation does not require a medium for propagation. Which of the following correctly describes the orientation of the electric and magnetic field vectors?
Answer: A. In electromagnetic radiation, the electric and magnetic field vectors are mutually perpendicular and both are perpendicular to the direction of propagation — this defines a transverse wave (NCERT Class 11 Chemistry Chapter 2, page 38).
Why B is wrong: B — If both fields were parallel to propagation, the wave would be longitudinal, which EM radiation is not.
Why C is wrong: C — This mixed arrangement does not occur in EM waves; both fields are transverse.
Why D is wrong: D — If the fields were parallel to each other, they would not generate the self-sustaining oscillation that characterises EM propagation. They must be perpendicular to each other.
A photon of wavelength 300 nm strikes a metal surface with a work function of 3.5 eV. What is the maximum kinetic energy of the emitted photoelectron? (h = 6.626 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J)
Answer: D. E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(300 × 10⁻⁹) = 6.626 × 10⁻¹⁹ J = 4.14 eV. KE = E − φ = 4.14 − 3.5 = 0.64 eV. Multi-step: first compute photon energy, then subtract work function (NCERT Class 11 Chemistry Chapter 2).
Why A is wrong: A — The photon energy (4.14 eV) exceeds the work function (3.5 eV), so emission does occur. This error arises from incorrectly computing photon energy below threshold.
Why B is wrong: B — This is the photon energy itself (hν), not the kinetic energy. The student forgot to subtract the work function.
Why C is wrong: C — Likely an arithmetic slip in the energy conversion or subtraction step.
Which region of the electromagnetic spectrum has wavelengths longer than visible light but shorter than microwaves?
Answer: C. The electromagnetic spectrum in order of increasing wavelength: gamma rays < X-rays < UV < visible < infrared < microwaves < radio waves. Infrared lies between visible and microwave regions (NCERT Class 11 Chemistry Chapter 2).
Why A is wrong: A — Ultraviolet has wavelengths shorter than visible light, not longer.
Why B is wrong: B — X-rays have even shorter wavelengths than UV, placing them far from the microwave end.
Why D is wrong: D — Gamma rays have the shortest wavelengths in the EM spectrum, at the opposite end from microwaves.
In the photoelectric effect, if the intensity of incident light is doubled (keeping frequency constant and above threshold), which of the following doubles?
Answer: B. Intensity is proportional to the number of photons per unit time. Each photon ejects at most one electron. Doubling intensity doubles photon count, hence doubles the number of emitted electrons. KE depends on frequency, not intensity (NCERT Class 11 Chemistry Chapter 2).
Why A is wrong: A — KE_max = hν − hν₀, which depends on frequency alone. Changing intensity does not change the energy per photon. This is exactly the confusion the classical wave model creates.
Why C is wrong: C — Threshold frequency is a property of the metal surface, not of the incident light. It cannot change by altering intensity.
Why D is wrong: D — Work function is an intrinsic property of the metal; it does not depend on incident light properties.
The relationship between frequency (ν), wavelength (λ), and speed of light (c) for electromagnetic radiation is:
Answer: B. The fundamental wave equation for EM radiation is c = νλ, where c is the speed of light, ν is frequency, and λ is wavelength (NCERT Class 11 Chemistry Chapter 2, page 38).
Why A is wrong: A — c = ν/λ is dimensionally incorrect. Frequency (s⁻¹) divided by wavelength (m) gives m⁻¹s⁻¹, not m/s.
Why C is wrong: C — c = λ/ν has units of m/(s⁻¹) = m·s, which is not speed.
Why D is wrong: D — ν = cλ gives (m/s)(m) = m²/s, which is not frequency (s⁻¹). The correct rearrangement is ν = c/λ.
The threshold frequency for a metal is 5.0 × 10¹⁴ Hz. Light of frequency 6.0 × 10¹⁴ Hz strikes the surface. If the frequency is changed to 7.0 × 10¹⁴ Hz (keeping intensity constant), the maximum kinetic energy of emitted electrons will:
Answer: C. KE₁ = h(6.0 × 10¹⁴ − 5.0 × 10¹⁴) = h × 1.0 × 10¹⁴. KE₂ = h(7.0 × 10¹⁴ − 5.0 × 10¹⁴) = h × 2.0 × 10¹⁴. The increase = KE₂ − KE₁ = h × 1.0 × 10¹⁴, which is exactly the energy of one quantum of frequency 1.0 × 10¹⁴ Hz. The relationship KE = hν − hν₀ is linear in ν (NCERT Class 11 Chemistry Chapter 2).
Why A is wrong: A — KE depends linearly on frequency. Changing frequency changes KE. This error comes from confusing intensity (which does not affect KE) with frequency (which does).
Why B is wrong: B — KE₁ = h × 1.0 × 10¹⁴ and KE₂ = h × 2.0 × 10¹⁴. While KE₂ is indeed 2 × KE₁, the question asks about the change in KE, not whether KE doubled. The increase equals h × 1.0 × 10¹⁴, not 2 × KE₁. However, if the student chose B thinking KE doubles, they would be conflating 'doubles' with 'increases by h × Δν', which is correct only coincidentally at these values — the general principle is linear increase, not doubling.
Why D is wrong: D — Increasing frequency above threshold always increases KE. It cannot decrease.
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Given
Light of wavelength 250 nm falls on a metal surface. The threshold wavelength for the metal is 400 nm.
h = 6.626 × 10⁻³⁴ J·s, c = 3.0 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J
Required
Maximum kinetic energy of the emitted photoelectron (in eV).
Concept
Einstein's photoelectric equation: KE_max = hν − hν₀. Since c = νλ, this becomes KE_max = hc/λ − hc/λ₀ = hc(1/λ − 1/λ₀).
Formula
KE_max = hc(1/λ − 1/λ₀)
Substitution
KE_max = (6.626 × 10⁻³⁴)(3.0 × 10⁸) × (1/(250 × 10⁻⁹) − 1/(400 × 10⁻⁹))
Calculation
hc = 6.626 × 10⁻³⁴ × 3.0 × 10⁸ = 1.988 × 10⁻²⁵ J·m
1/λ = 1/(250 × 10⁻⁹) = 4.0 × 10⁶ m⁻¹
1/λ₀ = 1/(400 × 10⁻⁹) = 2.5 × 10⁶ m⁻¹
Difference = 1.5 × 10⁶ m⁻¹
KE_max = 1.988 × 10⁻²⁵ × 1.5 × 10⁶ = 2.982 × 10⁻¹⁹ J
Converting: 2.982 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 1.86 eV
Exact constants note: The conversion factor 1 eV = 1.6 × 10⁻¹⁹ J is a defined constant and does not limit significant figures. The answer precision is governed by the given data (2–3 significant figures).
Final answer
KE_max ≈ 1.86 eV
Common trap
Forgetting to convert both wavelengths to the same unit (metres) before substituting. Another trap: computing hc/λ and hc/λ₀ separately in eV and then subtracting — this works but students sometimes forget to convert one of them, mixing joules and eV.
Similar NEET-style question
"Light of wavelength 200 nm is incident on a metal with work function 4.2 eV. Calculate the maximum kinetic energy of the photoelectrons emitted." (Same method: E = hc/λ, convert to eV, subtract work function.)
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Electron (Thomson, 1897, e/m by cathode-ray); proton (Goldstein, 1886, anode rays); neutron (Chadwick, 1932). Charges: e⁻ = -1.6×10⁻¹⁹ C, p⁺ = +1.6×10⁻¹⁹ C, n = neutral.
-- NCERT Class 11 Chemistry, Ch. 2, p. 30More in Structure of Atom: 4 exam traps and mistakes · 5 formulas · 3 question patterns from its other lessons.
1 question from NEET 2021. Answers verified against NTA official keys.
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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