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?
The energy of a photon of radiation with frequency 5.0 × 10¹⁴ Hz is (h = 6.626 × 10⁻³⁴ J·s):
Electromagnetic radiation does not require a medium for propagation. Which of the following correctly describes the orientation of the electric and magnetic field vectors?
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)
Which region of the electromagnetic spectrum has wavelengths longer than visible light but shorter than microwaves?
In the photoelectric effect, if the intensity of incident light is doubled (keeping frequency constant and above threshold), which of the following doubles?
The relationship between frequency (ν), wavelength (λ), and speed of light (c) for electromagnetic radiation is:
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:
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.) ---
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. 2Energy of nth orbit. Negative (bound). Ground state H: -13.6 eV.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E_n | orbit energy | eV |
| Z | nuclear charge | - |
| n | principal | - |
Radius of nth Bohr orbit for hydrogen-like atom of nuclear charge Z.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | principal quantum number | - |
| Z | nuclear charge | - |
| r_n | orbit radius | Å |
Wavelength associated with moving particle of momentum mv.
| Symbol | Quantity | SI Unit |
|---|---|---|
| h | Planck 6.626e-34 | J*s |
| m | mass | kg |
| v | velocity | m/s |
Position and momentum cannot both be known with arbitrary precision.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Δx | position uncertainty | m |
| Δp | momentum uncertainty | kg*m/s |
Spectral wavelengths of hydrogen-like atoms. Lyman (n1=1, UV), Balmer (n1=2, visible), Paschen (n1=3, IR).
| Symbol | Quantity | SI Unit |
|---|---|---|
| lambda | wavelength | m |
| R_H | Rydberg 1.097e7 | 1/m |
| Z | nuclear charge | - |
| n1, n2 | integers, n2>n1 | - |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student writes Cr as [Ar]3d⁴4s² (expected) instead of actual [Ar]3d⁵4s¹. Same for Cu: actual [Ar]3d¹⁰4s¹ (one e⁻ promoted from 4s to 3d).
Question asks for ground-state electronic configuration of Cr (Z=24) or Cu (Z=29).
Half-filled (d⁵) and fully filled (d¹⁰) configurations have extra stability from exchange energy and symmetry. Cr and Cu adopt these configurations by promoting one 4s electron.
Category: Similar Terms
Student forgets Z² scaling when applying Bohr formulas to He⁺ (Z=2) or Li²⁺ (Z=3).
Question involves hydrogen-like ion (He+, Li2+, etc.).
E_n = -13.6 × Z²/n² eV. r_n = (0.529/Z) × n² Å. He+: 4× more bound than H. Li²⁺: 9× more bound. Always include Z².
Root cause: concept gap
Half-filled (d⁵) and full-filled (d¹⁰) configurations have extra exchange-energy stability. Cr and Cu adopt these by promoting one 4s electron.
Root cause: formula misuse
Always include Z². E_n = -13.6 × Z²/n². For He+: 4× more energetic than H. For Li²⁺: 9×.
9 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
From the following pairs of ions which one is not an iso-electronic pair? Fe2+, Mn2+
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
misses cr cu anomaly
Writes 3d⁴4s² instead of 3d⁵4s¹
forgets z squared
Drops Z² for hydrogen-like
uses h instead of h over 4pi
Drops 4π factor
uses wrong l range
Uses l ≤ n instead of l < n
Test yourself on this topic with real past-paper questions:
Practice this topic →Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.