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Edexcel A Level (IAL) Physics-2.22 The de Broglie Equation- Study Notes- New Syllabus

Edexcel A Level (IAL) Physics -2.22 The de Broglie Equation- Study Notes- New syllabus

Edexcel A Level (IAL) Physics -2.22 The de Broglie Equation- Study Notes -Edexcel A level Physics – per latest Syllabus.

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Edexcel A level Physics-Study Notes- All Topics

Using the de Broglie Equation \( \lambda = \dfrac{h}{p} \)

The de Broglie hypothesis states that all matter exhibits wave-like behaviour. Particles such as electrons, neutrons, atoms, and even large molecules have a wavelength associated with their motion.

 de Broglie Equation

Louis de Broglie proposed that the wavelength of a particle is given by:

\( \lambda = \dfrac{h}{p} \)

  • \( \lambda \) = de Broglie wavelength (m)
  • \( h \) = Planck’s constant \( 6.63\times10^{-34}\ \mathrm{Js} \)
  • \( p \) = momentum of particle \( = mv \)
  • \( m \) = mass (kg)
  • \( v \) = velocity (m s\(^{-1}\))

Key idea: Particles with higher momentum have smaller wavelengths.

 When Do de Broglie Wavelengths Matter?

  • Wave effects become noticeable when \( \lambda \) is comparable to the size of obstacles or gaps.
  • Electrons (very small mass) have measurable wavelengths → show diffraction.
  • Large everyday objects have extremely tiny wavelengths → no observable wave behaviour.

Thus, quantum wave behaviour is important only at atomic or subatomic scales.

 Alternative Forms of the Equation

If velocity is known:

\( \lambda = \dfrac{h}{mv} \)

If electrons are accelerated through a potential difference \( V \):

\( \lambda = \dfrac{h}{\sqrt{2meV}} \)

(Not always required, but useful in advanced questions.)

Important Observations

  • Increasing velocity → increases momentum → decreases wavelength.
  • Higher accelerating voltage for electrons → smaller wavelength → less diffraction.
  • Electron diffraction experiments confirm wave–particle duality.

Example (Easy)

Find the de Broglie wavelength of an electron moving at \( 2.0\times10^{6}\ \mathrm{m\,s^{-1}} \). (The mass of an electron is \( 9.11\times10^{-31}\ \mathrm{kg} \))

▶️ Answer / Explanation

Step 1: Find momentum.

\( p = mv = (9.11\times10^{-31})(2.0\times10^{6}) = 1.82\times10^{-24}\ \mathrm{kg\,m\,s^{-1}} \)

Step 2: Apply de Broglie equation.

\( \lambda = \dfrac{6.63\times10^{-34}}{1.82\times10^{-24}} = 3.64\times10^{-10}\ \mathrm{m} \)

Wavelength = \( 0.364\ \mathrm{nm} \)

Example (Medium)

A neutron (mass \( 1.67\times10^{-27}\ \mathrm{kg} \)) has momentum \( 5.0\times10^{-24}\ \mathrm{kg\,m\,s^{-1}} \). Find its wavelength.

▶️ Answer / Explanation

\( \lambda = \dfrac{6.63\times10^{-34}}{5.0\times10^{-24}} = 1.33\times10^{-10}\ \mathrm{m} \)

Wavelength = \( 0.133\ \mathrm{nm} \)

Example (Hard)

An electron is accelerated through a potential difference of \( 150\ \mathrm{V} \). Calculate its de Broglie wavelength.

▶️ Answer / Explanation

Step 1: Find kinetic energy.

\( E = eV = (1.6\times10^{-19})(150) = 2.4\times10^{-17}\ \mathrm{J} \)

Step 2: Momentum from KE.

\( p = \sqrt{2mE} = \sqrt{2(9.11\times10^{-31})(2.4\times10^{-17})} \)

\( p = 6.62\times10^{-24}\ \mathrm{kg\,m\,s^{-1}} \)

Step 3: Apply de Broglie equation.

\( \lambda = \dfrac{6.63\times10^{-34}}{6.62\times10^{-24}} = 1.00\times10^{-10}\ \mathrm{m} \)

Wavelength = \( 0.10\ \mathrm{nm} \)

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