Question

The de Broglie wavelength of a particle is:

• A. \(\frac{h}{p}\)
• B. \(\frac{p}{h}\)
• C. \(\frac{h^2}{p}\)
• D. \sqrt{\frac{h}{p}}

Hint:

The de Broglie wavelength is inversely proportional to momentum: \( \lambda = \frac{h}{p} \). Because Planck's constant \( h \) is extremely small, wave-like properties are only observable for subatomic particles with tiny masses, like electrons.

Answer 1

The Correct Option is A

Concept: In 1924, French physicist Louis de Broglie introduced the hypothesis of wave-particle duality. He proposed that all moving matter exhibits wave-like characteristics. The wavelength associated with a material particle depends directly on its momentum.

Step 1: State the historical context and formula.
De Broglie adapted the expressions used for photons and applied them to matter particles. For a photon, the energy relations are given by Einstein and Planck as: \[ E = mc^2 \quad \text{and} \quad E = h\nu = \frac{hc}{\lambda} \] Equating these two energy expressions yields: \[ mc^2 = \frac{hc}{\lambda} \quad \Rightarrow \quad mc = \frac{h}{\lambda} \] Since the momentum of a photon traveling at light speed is \( p = mc \), this simplifies to: \[ p = \frac{h}{\lambda} \quad \Rightarrow \quad \lambda = \frac{h}{p} \]

Step 2: Extend the relation to massive particles.
De Broglie generalized this equation to any physical particle with a mass \( m \) moving at a velocity \( v \). The momentum of the particle is given by: \[ p = m \cdot v \] Substituting this classical momentum into the wave relation gives the de Broglie wavelength formula: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] Where \( h \) represents Planck's constant (\( 6.626 \times 10^{-34}\text{ J}\cdot\text{s} \)). This matches Option (A).