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Structure of Atom - Dual Nature of Matter and Radiation

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Dual Nature of Radiation: Electromagnetic radiation exhibits both wave-like properties (interference, diffraction) and particle-like properties (photoelectric effect, black body radiation).

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Planck's Quantum Theory: Energy is emitted or absorbed in discrete packets called 'quanta'. For light, these are called 'photons'. The energy is given by E=hνE = h\nu.

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Photoelectric Effect: When light of a suitable frequency (ν>ν0\nu > \nu_0) strikes a metal surface, electrons are ejected. Einstein explained this using the particle nature of light.

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De Broglie's Hypothesis: Just as radiation has dual nature, matter also possesses dual nature. Any moving material particle has an associated wavelength λ\lambda.

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Heisenberg's Uncertainty Principle: It is impossible to determine simultaneously and accurately both the position (Δx\Delta x) and momentum (Δp\Delta p) of a microscopic moving particle.

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Significance of Dual Nature: While applicable to all objects, de Broglie's wavelength and Heisenberg's uncertainty are significant only for subatomic particles like electrons due to their small mass mm.

📐Formulae

E=hν=hcλE = h\nu = \frac{hc}{\lambda}

hν=hν0+12mev2h\nu = h\nu_0 + \frac{1}{2}m_e v^2

λ=hmv=hp\lambda = \frac{h}{mv} = \frac{h}{p}

λ=h2mEk\lambda = \frac{h}{\sqrt{2mE_k}}

Δx⋅Δp≥h4π\Delta x \cdot \Delta p \ge \frac{h}{4\pi}

Δx⋅(mΔv)≥h4π\Delta x \cdot (m \Delta v) \ge \frac{h}{4\pi}

💡Examples

Problem 1:

Calculate the de Broglie wavelength of an electron moving with a velocity of 2.05×107 m s−12.05 \times 10^7 \text{ m s}^{-1}. (Given: me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg}, h=6.626×10−34 Jsh = 6.626 \times 10^{-34} \text{ Js})

Solution:

Using de Broglie equation: λ=hmv\lambda = \frac{h}{mv}. Substituting the values: λ=6.626×10−349.1×10−31×2.05×107\lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 2.05 \times 10^7}. λ≈3.55×10−11 m\lambda \approx 3.55 \times 10^{-11} \text{ m}.

Explanation:

The wavelength is calculated by dividing Planck's constant by the momentum (mass times velocity) of the electron.

Problem 2:

A microscopic particle of mass 10−6 g10^{-6} \text{ g} has an uncertainty in position of 10−4 cm10^{-4} \text{ cm}. Calculate the uncertainty in its velocity.

Solution:

Convert units to SI: m=10−9 kgm = 10^{-9} \text{ kg}, Δx=10−6 m\Delta x = 10^{-6} \text{ m}. Using Δx⋅mΔv=h4π\Delta x \cdot m \Delta v = \frac{h}{4\pi}, we get Δv=6.626×10−344×3.1416×10−9×10−6\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 10^{-9} \times 10^{-6}}. Δv≈5.27×10−20 m s−1\Delta v \approx 5.27 \times 10^{-20} \text{ m s}^{-1}.

Explanation:

Heisenberg's uncertainty principle relates the uncertainty in position and velocity. Since the mass is relatively large for a 'microscopic' particle, the uncertainty in velocity is extremely small.

Dual Nature of Matter and Radiation Revision - Class 11 Chemistry ICSE