Quick Exam Notes
  • Formula: \( \lambda = \frac{h}{p} = \frac{h}{mv} \)
  • Applies to: All moving particles (electrons, protons, neutrons)
  • Key Point: Higher momentum → shorter wavelength
  • Special Case (in terms of voltage & energy): \( \lambda = \frac{h}{\sqrt{2meV}} \) & \( \lambda = \frac{h}{\sqrt{2mE}} \)

The de Broglie hypothesis, proposed by Louis de Broglie in 1924, states that all matter exhibits both wave-like and particle-like properties. This concept is a fundamental principle in quantum mechanics and led to the development of wave-particle duality.

The de Broglie wavelength (𝜆) of a particle is given by the equation:

\[ \lambda =\frac{h}{p} \]

  • 𝜆 = wavelength of the particle (in meters)
  • h = Planck’s constant (6.626×10-34 Js)
  • 𝑝 = momentum of the particle (𝑝=𝑚𝑣, where 𝑚 is mass and 𝑣 is velocity)

Derivation of Debrogle wavelength


From Einstein’s Mass-Energy Relation

\[ E = mc^2 \hspace{5cm} (1)\]

From Planck’s Quantum Theory

\[ E = \frac{hc}{\lambda} \hspace{5cm} (2) \]

Equating equations 1 and 2,

\[ mc^2 = \frac{hc}{\lambda} \]

Solving for \( \lambda \):

\[ \lambda = \frac{h}{mc} \]

If the particle is moving with velocity \( v \), replacing \( c \) with \( v \):

\[ \lambda = \frac{h}{mv} \]

In terms of momentum \( p \):

\[ \lambda = \frac{h}{p} \]

Derivation of Debrogle wavelength in terms of energy


Kinetic Energy of the Particle

\[ KE = \frac{1}{2} mv^2 \hspace{5cm} (3)\]

Multiplying by \( m \) on both sides:

\[ 2mKE = m^2 v^2 \hspace{5cm} (4)\]

De Broglie Wavelength

\[ \lambda = \frac{h}{mv} \]

Squaring both sides:

\[ \lambda^2 = \frac{h^2}{m^2 v^2} \]

Substituting Eqn 4 in the above Eqn:

\[ \lambda^2 = \frac{h^2}{2mKE} \]

Solving for \( \lambda \):

\[ \lambda = \frac{h}{\sqrt{2mKE}} \]

Derivation of Debrogle wavelength in terms of voltage


If a charged particle 'e' is accelerated through a potential difference 'V', the kinetic energy is given by

\[ KE = eV \hspace{5cm}(5)\]

kinetic energy of the particle:

\[ KE = \frac{1}{2} mv^2 \hspace{5cm} (6)\]

Equating equations 5 and 6,

\[ eV = \frac{1}{2} mv^2 \]

multiplying m on both sides:

\[ 2m eV = m^2v^2 \hspace{5cm} (7) \]

The debroglie wavelength is given by:

\[ \lambda = \frac{h}{mv} \]

Squaring both sides

\[ \lambda^2 = \frac{h^2}{m^2v^2} \]

Substituting Eqn 7 in above Eqn :

\[ \lambda^2 = \frac{h^2}{2meV} \]

\[ \lambda = \frac{h}{\sqrt{2meV}} \hspace{5cm} (7) \]

The final equation we get after substituting the all constant values (h, m and e) in Eqn 7,

\[ \lambda = \frac{12.26}{\sqrt{V}} Å \]

Summary

The De-Broglie Hypothesis introduced the revolutionary idea of wave-particle duality, stating that every moving particle has an associated wavelength called the De-Broglie wavelength. This wavelength is inversely proportional to momentum, meaning lighter and faster particles have shorter wavelengths. The hypothesis successfully explained why electrons in atoms show diffraction patterns like waves while behaving as particles in experiments. It laid the foundation for quantum mechanics and was later experimentally verified by the Davisson–Germer experiment.

Multiple Choice Questions (MCQs)

  1. According to De-Broglie, the wavelength of a particle is inversely proportional to:
    • a) Energy
    • b) Mass
    • c) Momentum
    • d) Charge
    Answer

    c) Momentum

  2. The De-Broglie wavelength of an electron accelerated through a potential V is:
    • a) h/p
    • b) h/√(2meV)
    • c) h/mc
    • d) hv/E
    Answer

    b) h/√(2meV)

  3. Which experiment confirmed the wave nature of electrons?
    • a) Hertz experiment
    • b) Davisson–Germer experiment
    • c) Michelson–Morley experiment
    • d) Rutherford experiment
    Answer

    b) Davisson–Germer experiment

  4. For heavier particles, the De-Broglie wavelength is:
    • a) Longer
    • b) Independent of mass
    • c) Very short
    • d) Infinite
    Answer

    c) Very short

  5. The De-Broglie hypothesis is significant for:
    • a) Only light particles
    • b) Only photons
    • c) Only heavy particles
    • d) Stationary objects
    Answer

    a) Only light particles