Dual Nature of Radiation and Matter for NEET UG
Light sometimes behaves like a stream of particles, and electrons sometimes behave like waves. The photoelectric effect and de Broglie waves show this dual nature.
PhysicsClass 1261 MCQs2 lessons11 flashcards3 memory tricks
Practise Dual Nature of Radiation and Matter free
What you'll learn
- The photoelectric effect
Use Einstein's photoelectric equation to find stopping potential, work function and threshold frequency. - Matter waves (de Broglie)
Calculate the de Broglie wavelength of particles, especially electrons accelerated through a voltage.
Key formulas
| Photon energy | E = hν = hc/λ |
|---|---|
| Einstein's equation | K_max = hν − φ₀ |
| Stopping potential | eV₀ = K_max |
| Threshold frequency | ν₀ = φ₀/h |
| Handy conversion | E (eV) ≈ 1240/λ (nm) |
| de Broglie wavelength | λ = h/p = h/mv |
| From kinetic energy | λ = h/√(2mK) |
| Electron accelerated through V volts | λ ≈ 1.227/√V nm |
Memory tricks
Light below the threshold frequency ejects no electrons, however bright it is. Above it, a higher frequency means more energetic electrons, and a higher intensity means more electrons.
K_max = hν − φ. Stopping potential V₀ = K_max/e.
λ = h/p = h/√(2mK). Heavy, fast objects have unmeasurably small wavelengths; electrons have wavelengths similar to atomic spacings.
Try 3 NEET-style questions
Q1. A 3 W source emits light of wavelength 663 nm. The number of photons it emits per second is about (h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s)
- 10¹⁸
- 10²⁰
- 9 × 10⁻¹⁹
- 10¹⁹
Show answer
Answer: (4) 10¹⁹
E per photon = hc/λ = 6.63 × 10⁻³⁴ × 3 × 10⁸ / 663 × 10⁻⁹ = 3 × 10⁻¹⁹ J. N = 3/(3 × 10⁻¹⁹) = 10¹⁹ per second.
Q2. Light of wavelength 310 nm falls on a metal of work function 2 eV. The stopping potential is (hc = 1240 eV nm)
- 4 V
- 2 V
- 1 V
- 6 V
Show answer
Answer: (2) 2 V
E = 1240/310 = 4 eV; K_max = 4 − 2 = 2 eV, so V₀ = 2 V.
Q3. The threshold frequency of a metal is the frequency below which
- electrons are emitted with maximum energy
- no photoelectrons are emitted
- the stopping potential is maximum
- the photon energy is zero
Show answer
Answer: (2) no photoelectrons are emitted
Below the threshold frequency ν₀, the photon energy hν is less than the work function, so no electrons are emitted.
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