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Dual Nature of Radiation and Matter - De Broglie Hypothesis

Grade 12CBSEPhysics

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

🔑Concepts

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The de Broglie hypothesis states that every moving material particle has a wave associated with it, known as a matter wave or de Broglie wave.

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The dual nature of matter implies that particles like electrons, protons, and even atoms can exhibit interference and diffraction under specific conditions.

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The de Broglie wavelength λ\lambda is inversely proportional to the momentum pp of the particle, given by λ=hp\lambda = \frac{h}{p}.

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For a particle of mass mm moving with velocity vv, the wavelength is λ=hmv\lambda = \frac{h}{mv}.

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If a particle of charge qq is accelerated from rest through a potential difference VV, its kinetic energy KK is qVqV, and its wavelength is λ=h2mqV\lambda = \frac{h}{\sqrt{2mqV}}.

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For an electron, substituting the values of mem_e, ee, and hh, the wavelength simplifies to λ=1.227V nm\lambda = \frac{1.227}{\sqrt{V}} \text{ nm}, where VV is in volts.

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The wave nature of particles is only significant for microscopic objects (like electrons) because hh (6.63×10−34 J s6.63 \times 10^{-34} \text{ J s}) is extremely small, making λ\lambda negligible for macroscopic bodies.

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The Davisson-Germer experiment provided the first experimental proof of the wave nature of electrons by demonstrating diffraction using a nickel crystal.

📐Formulae

λ=hp\lambda = \frac{h}{p}

λ=hmv\lambda = \frac{h}{mv}

K=p22m  ⟹  p=2mKK = \frac{p^2}{2m} \implies p = \sqrt{2mK}

λ=h2mK\lambda = \frac{h}{\sqrt{2mK}}

λ=h2mqV\lambda = \frac{h}{\sqrt{2mqV}}

λelectron=1.227V nm=12.27V A˚\lambda_{electron} = \frac{1.227}{\sqrt{V}} \text{ nm} = \frac{12.27}{\sqrt{V}} \text{ \AA}

💡Examples

Problem 1:

Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V100 \text{ V}.

Solution:

Using the specific formula for an electron: λ=1.227V nm\lambda = \frac{1.227}{\sqrt{V}} \text{ nm}. Given V=100 VV = 100 \text{ V}, λ=1.227100=1.22710=0.1227 nm\lambda = \frac{1.227}{\sqrt{100}} = \frac{1.227}{10} = 0.1227 \text{ nm}.

Explanation:

This wavelength is in the same order as the interatomic spacing in crystals, allowing electrons to undergo diffraction.

Problem 2:

A proton and an alpha particle have the same kinetic energy. What is the ratio of their de Broglie wavelengths?

Solution:

The wavelength is given by λ=h2mK\lambda = \frac{h}{\sqrt{2mK}}. Since KK is the same for both, λ∝1m\lambda \propto \frac{1}{\sqrt{m}}. Let mpm_p be the mass of the proton. The mass of the alpha particle is mα=4mpm_{\alpha} = 4m_p. Therefore, λpλα=mαmp=4mpmp=4=2\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_p}} = \sqrt{\frac{4m_p}{m_p}} = \sqrt{4} = 2. The ratio is 2:12:1.

Explanation:

Because the alpha particle is heavier than the proton, it has a shorter de Broglie wavelength for the same kinetic energy.

Problem 3:

Determine the momentum of a photon with a wavelength of 500 nm500 \text{ nm}.

Solution:

From λ=hp\lambda = \frac{h}{p}, we have p=hλp = \frac{h}{\lambda}. Substituting h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s} and λ=500×10−9 m\lambda = 500 \times 10^{-9} \text{ m}, p=6.63×10−345×10−7=1.326×10−27 kg m/sp = \frac{6.63 \times 10^{-34}}{5 \times 10^{-7}} = 1.326 \times 10^{-27} \text{ kg m/s}.

Explanation:

This formula relates the particle-like property (momentum) to the wave-like property (wavelength) for both light and matter.