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Dual Nature of Radiation and Matter - Matter Waves (De Broglie Wavelength)

Grade 12ICSEPhysics

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 nature associated with it. These waves are called matter waves or de Broglie waves.

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The wavelength λ\lambda of a matter wave is inversely proportional to the momentum pp of the particle: λ=hp\lambda = \frac{h}{p}, where hh is Planck's constant (6.63×10−34 J s6.63 \times 10^{-34} \text{ J s}).

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Matter waves are not electromagnetic in nature. Unlike electromagnetic waves, they are not produced by accelerated charges and can travel with velocities greater than the speed of light (phase velocity), though the particle itself (group velocity) cannot.

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The de Broglie wavelength associated with a particle of mass mm and kinetic energy KK is given by λ=h2mK\lambda = \frac{h}{\sqrt{2mK}}.

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For a charged particle of charge qq accelerated from rest through a potential difference VV, the work done qVqV is converted into kinetic energy (K=qVK = qV), resulting in λ=h2mqV\lambda = \frac{h}{\sqrt{2mqV}}.

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For an electron (me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg} and q=1.6×10−19 Cq = 1.6 \times 10^{-19} \text{ C}), the wavelength is simplified to λ≈12.27V A˚\lambda \approx \frac{12.27}{\sqrt{V}} \text{ \AA} or 1.227V nm\frac{1.227}{\sqrt{V}} \text{ nm}.

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Davisson-Germer Experiment: This provided the first experimental evidence of the wave nature of electrons by demonstrating electron diffraction, which is a wave phenomenon.

📐Formulae

λ=hp\lambda = \frac{h}{p}

λ=hmv\lambda = \frac{h}{mv}

λ=h2mK\lambda = \frac{h}{\sqrt{2mK}}

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

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

λ=h3mkT\lambda = \frac{h}{\sqrt{3mkT}} (for a gas molecule at temperature TT)

💡Examples

Problem 1:

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

Solution:

Given V=100 VV = 100 \text{ V}. Using the simplified formula for electrons: λ=12.27V A˚\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA}. λ=12.27100=12.2710=1.227 A˚\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \text{ \AA}.

Explanation:

Since the particle is an electron, we can use the shortcut formula derived from constants hh, mem_e, and ee. The result is in Angstroms (10−10 m10^{-10} \text{ m}).

Problem 2:

A proton and an alpha particle (He2+He^{2+}) have the same kinetic energy. Which one has a shorter de Broglie wavelength?

Solution:

The wavelength is given by λ=h2mK\lambda = \frac{h}{\sqrt{2mK}}. Since hh and KK are constant, λ∝1m\lambda \propto \frac{1}{\sqrt{m}}. Mass of alpha particle mα≈4mpm_{\alpha} \approx 4m_p. Therefore, λpλα=mαmp=4mpmp=2\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_p}} = \sqrt{\frac{4m_p}{m_p}} = 2. This implies λα=λp2\lambda_{\alpha} = \frac{\lambda_p}{2}.

Explanation:

The alpha particle, being heavier (44 times the mass of a proton), will have a shorter de Broglie wavelength when both have the same kinetic energy.

Problem 3:

Find the momentum of a particle if its de Broglie wavelength is 2 A˚2 \text{ \AA}.

Solution:

Given λ=2×10−10 m\lambda = 2 \times 10^{-10} \text{ m} and h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s}. From λ=hp\lambda = \frac{h}{p}, we get p=hλ=6.63×10−342×10−10=3.315×10−24 kg m s−1p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{2 \times 10^{-10}} = 3.315 \times 10^{-24} \text{ kg m s}^{-1}.

Explanation:

Direct application of the de Broglie relation to find momentum from wavelength.

Matter Waves (De Broglie Wavelength) Revision - Class 12 Physics ICSE