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Structure of Atom - Dual Nature of Matter and Radiation

Grade 11CBSEChemistry

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

🔑Concepts

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The Dual Nature of Matter was proposed by Louis de Broglie, suggesting that just like radiation, matter also exhibits both particle and wave-like properties. This means moving particles like electrons have an associated wavelength λ\lambda.

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The Photoelectric Effect provides evidence for the particle nature of light. It states that when light of frequency ν\nu higher than a threshold frequency ν0\nu_0 strikes a metal surface, electrons are ejected. The energy of the photon is E=hνE = h\nu.

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Heisenberg's Uncertainty Principle states that it is impossible to determine simultaneously the exact position and exact momentum (or velocity) of a microscopic moving particle like an electron. The mathematical limit is given by Δx⋅Δp≥h4π\Delta x \cdot \Delta p \ge \frac{h}{4\pi}.

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Black Body Radiation: An ideal black body is a perfect absorber and emitter of radiation. The distribution of intensity of radiation emitted by a black body depends only on its temperature, which could not be explained by classical wave theory but was explained by Planck's Quantum Theory.

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The wave nature of the electron was experimentally confirmed by the Davisson-Germer experiment, which showed that electrons undergo diffraction, a property characteristic of waves.

📐Formulae

E=hν=hcλE = h\nu = \frac{hc}{\lambda}

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

Δx⋅Δp≥h4π or Δx⋅mΔv≥h4π\Delta x \cdot \Delta p \ge \frac{h}{4\pi} \text{ or } \Delta x \cdot m\Delta v \ge \frac{h}{4\pi}

hν=hν0+K.E.max  ⟹  hν=hν0+12mev2h\nu = h\nu_0 + K.E._{max} \implies h\nu = h\nu_0 + \frac{1}{2}m_e v^2

λ=h2mK.E.\lambda = \frac{h}{\sqrt{2mK.E.}}

💡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 mass of electron me=9.11×10−31 kgm_e = 9.11 \times 10^{-31} \text{ kg} and h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s}.

Solution:

Using de Broglie's equation: λ=hmv\lambda = \frac{h}{mv}. λ=6.626×10−34 kg m2 s−1(9.11×10−31 kg)×(2.05×107 m s−1)=6.626×10−341.867×10−23≈3.55×10−11 m\lambda = \frac{6.626 \times 10^{-34} \text{ kg m}^2 \text{ s}^{-1}}{(9.11 \times 10^{-31} \text{ kg}) \times (2.05 \times 10^7 \text{ m s}^{-1})} = \frac{6.626 \times 10^{-34}}{1.867 \times 10^{-23}} \approx 3.55 \times 10^{-11} \text{ m}.

Explanation:

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

Problem 2:

A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 A˚0.1 \text{ \AA}. What is the uncertainty involved in the measurement of its velocity?

Solution:

Given Δx=0.1 A˚=10−11 m\Delta x = 0.1 \text{ \AA} = 10^{-11} \text{ m}. From Heisenberg's Uncertainty Principle: Δv=h4πmΔx\Delta v = \frac{h}{4\pi m \Delta x}. Δv=6.626×10−34 J s4×3.14×9.11×10−31 kg×10−11 m≈5.79×106 m s−1\Delta v = \frac{6.626 \times 10^{-34} \text{ J s}}{4 \times 3.14 \times 9.11 \times 10^{-31} \text{ kg} \times 10^{-11} \text{ m}} \approx 5.79 \times 10^6 \text{ m s}^{-1}.

Explanation:

Because the position is known with high precision (0.1 A˚0.1 \text{ \AA}), the uncertainty in velocity becomes quite significant for a subatomic particle like an electron.

Dual Nature of Matter and Radiation Class 11 Notes & Examples