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PhysicsDual Nature of Radiation
For a metal of work function 6·6 eV, which of the following wavelengths of incident radiation does not give rise to the photoelectric effect? (Take Planck's constant h = 6·6 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s)
Options
1
100 nm
2
150 nm
3
200 nm
4
50 nm
Correct Answer
Option 3 : 200 nm
Step-by-Step Solution
1

Find threshold wavelength λ₀:

Work function φ = 6.6 eV = 6.6 × 1.6×10⁻¹⁹ = 10.56×10⁻¹⁹ J

λ₀ = hc / φ = (6.6×10⁻³⁴ × 3×10⁸) / (10.56×10⁻¹⁹)

λ₀ = 19.8×10⁻²⁶ / 10.56×10⁻¹⁹ = 1.875 × 10⁻⁷ m ≈ 188 nm

Photoelectric effect occurs only when λ < λ₀ = 188 nm

2

Check photon energy at each wavelength:

λ (nm)E = hc/λ (eV)E vs φEffect?
50 nm24.75 eV> 6.6 eVYES ✓
100 nm12.375 eV> 6.6 eVYES ✓
150 nm8.25 eV> 6.6 eVYES ✓
200 nm6.19 eV< 6.6 eVNO ✗
3

Verify 200 nm calculation:

E = hc/λ = (6.6×10⁻³⁴ × 3×10⁸) / (200×10⁻⁹)

E = 19.8×10⁻²⁶ / 200×10⁻⁹ = 9.9×10⁻¹⁹ J

E in eV = 9.9×10⁻¹⁹ / 1.6×10⁻¹⁹ = 6.19 eV

Since 6.19 eV < 6.6 eV (work function), photon energy is insufficient — no photoelectric effect at 200 nm

Theory: Photoelectric Effect
1. Discovery and Background

The photoelectric effect is the emission of electrons from a metal surface when light of sufficient frequency (or short enough wavelength) falls on it. Hertz first observed it in 1887. Classical wave theory completely failed to explain it — it predicted any frequency of light should eventually eject electrons given enough intensity, which is not what happens. Einstein explained it in 1905 by proposing that light consists of discrete energy packets called photons, earning him the Nobel Prize in 1921.

2. Einstein's Photoelectric Equation

Each photon transfers all its energy to a single electron. Part of this energy is used to overcome the work function (the binding energy of the electron to the metal surface), and the rest becomes the kinetic energy of the emitted electron:

KE_max = hν − φ = hc/λ − φ

Threshold: hν₀ = φ → λ₀ = hc/φ

Condition for emission: λ < λ₀ (or hν > φ)

If photon energy hν is less than work function φ, no electron is emitted — regardless of how intense or prolonged the illumination. This was the key experimental result that classical wave theory could not explain, but quantum theory handled perfectly.

3. Key Experimental Observations

📌 No emission below threshold frequency ν₀ — regardless of intensity

📌 Emission is instantaneous — no time lag even at very low intensity

📌 Maximum KE of electrons depends on frequency, NOT on intensity

📌 Number of emitted electrons (photocurrent) ∝ intensity of light

📌 Stopping potential V₀ = KE_max / e — independent of intensity

📌 Higher frequency → higher maximum KE → higher stopping potential

4. Threshold Wavelength and Frequency

ν₀ = φ/h (threshold frequency)

λ₀ = hc/φ (threshold wavelength)

λ < λ₀ → effect occurs  |  λ > λ₀ → no effect

The threshold wavelength is the maximum wavelength (minimum frequency) of light that can cause photoelectric emission. For this metal with φ = 6.6 eV, λ₀ = 188 nm. Only UV radiation below 188 nm can cause emission — 200 nm is just above this threshold, so it cannot cause the effect even though it is very close.

5. Stopping Potential

The stopping potential V₀ is the minimum reverse voltage applied to the collector that stops even the most energetic emitted electrons completely. At stopping potential, all kinetic energy of the fastest electrons is converted to electrical potential energy:

eV₀ = KE_max = hν − φ

V₀ = (hν − φ)/e = h(ν − ν₀)/e

The slope of V₀ vs frequency graph = h/e (Planck's constant divided by electron charge). Millikan used this to experimentally measure h and confirmed Einstein's equation. Increasing light intensity does NOT change V₀ — it only increases the number of emitted electrons (photocurrent).

6. Work Function of Common Metals

📌 Caesium (Cs): 2.0 eV — threshold ~620 nm (visible red light)

📌 Potassium (K): 2.3 eV — threshold ~540 nm (visible green)

📌 Sodium (Na): 2.75 eV — threshold ~450 nm (visible blue/violet)

📌 Zinc (Zn): 3.6 eV — requires UV (λ₀ ~345 nm)

📌 Copper (Cu): 4.5 eV — requires deep UV (λ₀ ~275 nm)

📌 Platinum (Pt): 5.65 eV — requires far UV (λ₀ ~220 nm)

📌 This metal: 6.6 eV — threshold λ₀ = 188 nm (very short UV)

7. Photon — Particle Nature of Light

A photon is the quantum of electromagnetic energy. It has no rest mass, travels at speed c, and carries energy E = hν and momentum p = h/λ = E/c. Photons are not deflected by electric or magnetic fields (no charge). In photon-electron collisions, both energy and momentum are conserved. This is demonstrated in the Compton effect — X-rays scattered by electrons shift to longer wavelengths, proving photons carry momentum.

📌 Photon energy: E = hν = hc/λ

📌 Photon momentum: p = h/λ = E/c

📌 Rest mass of photon = 0

📌 1 eV = 1.6×10⁻¹⁹ J

📌 hc = 1240 eV·nm (useful shortcut: E(eV) = 1240/λ(nm))

8. de Broglie Wavelength — Matter Waves

Louis de Broglie proposed that just as waves (light) can show particle behaviour, particles (electrons, protons) should also exhibit wave behaviour. Every moving particle has an associated de Broglie wavelength:

λ = h/p = h/mv

For electron through voltage V: λ = h/√(2meV)

λ(nm) = 1.227/√V (V in volts, for electrons)

Davisson and Germer confirmed this experimentally in 1927 by showing electron diffraction — electrons scattered off a nickel crystal produced interference patterns, exactly like X-ray diffraction. This wave-particle duality is a cornerstone of quantum mechanics.

9. Compton Effect

When X-rays scatter off electrons, the scattered X-rays have a longer wavelength than the incident ones. This Compton shift can only be explained by treating photons as particles with momentum h/λ. The Compton wavelength shift:

Δλ = (h/mₑc)(1 − cosθ)

h/mₑc = 2.43 pm (Compton wavelength of electron)

Maximum shift at θ = 180°: Δλ_max = 4.86 pm

⚠️ Shorter wavelength = higher frequency = higher photon energy

⚠️ λ < λ₀ → photoelectric effect | λ > λ₀ → no effect

⚠️ Quick formula: E(eV) = 1240/λ(nm) saves calculation time in NEET

⚠️ KE_max depends on frequency only — NOT on intensity

⚠️ Photocurrent depends on intensity — NOT on frequency

Frequently Asked Questions
1. Why does 200 nm not cause the photoelectric effect?
Photon energy at 200 nm = hc/λ = 1240/200 = 6.2 eV. Work function = 6.6 eV. Since 6.2 eV < 6.6 eV, photon energy is insufficient to liberate an electron from the metal surface. No matter how many such photons hit the surface, no electron is emitted.
2. What is the quick eV formula for photon energy?
E(eV) = 1240/λ(nm). This comes from E = hc/λ with hc = 6.626×10⁻³⁴ × 3×10⁸ = 1.988×10⁻²⁵ J·m = 1240 eV·nm. For 200 nm: E = 1240/200 = 6.2 eV. For 150 nm: E = 1240/150 = 8.27 eV. Very fast in NEET!
3. What is the threshold wavelength for this metal?
λ₀ = hc/φ = 1240/6.6 ≈ 188 nm. Any wavelength less than 188 nm causes the photoelectric effect. 200 nm > 188 nm so no effect. 150 nm, 100 nm, 50 nm are all less than 188 nm so they all cause the effect.
4. What is the KE of electron emitted by 100 nm radiation?
E at 100 nm = 1240/100 = 12.4 eV. KE_max = E − φ = 12.4 − 6.6 = 5.8 eV = 5.8 × 1.6×10⁻¹⁹ = 9.28×10⁻¹⁹ J. Stopping potential = 5.8 V.
5. Why does intensity not affect the threshold?
Each photon interacts with exactly one electron — they cannot pool their energies. If one photon has energy less than the work function, no electron is ejected regardless of how many photons hit per second (intensity). Intensity only controls how many electrons are emitted per second (photocurrent).
6. What is stopping potential for 150 nm on this metal?
E at 150 nm = 1240/150 = 8.27 eV. KE_max = 8.27 − 6.6 = 1.67 eV. Stopping potential V₀ = KE_max/e = 1.67 V. This means a 1.67V reverse potential completely stops all emitted electrons.
7. What is de Broglie wavelength of electron emitted at 100 nm?
KE = 5.8 eV = 9.28×10⁻¹⁹ J. p = √(2mKE) = √(2 × 9.11×10⁻³¹ × 9.28×10⁻¹⁹) = √(1.692×10⁻⁴⁸) = 1.3×10⁻²⁴ kg·m/s. λ_dB = h/p = 6.6×10⁻³⁴/1.3×10⁻²⁴ ≈ 5.1×10⁻¹⁰ m = 0.51 nm.
8. What was Millikan's contribution to photoelectric effect?
Millikan (1916) carefully measured stopping potential for different light frequencies. Plotting V₀ vs ν gave a straight line with slope = h/e. This gave an accurate value of Planck's constant h and confirmed Einstein's equation experimentally. Despite initially trying to disprove Einstein, Millikan's precise measurements validated quantum theory completely.
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