Question

What is meant by the concept of matter-wave of de Broglie ? Establish the relation for de Broglie wavelength in terms of kinetic energy.

Hint:

This formula \( \lambda = h/\sqrt{2mK} \) is extremely useful. It can be further adapted for a charged particle (charge q) accelerated through a potential difference V, where its kinetic energy is \( K = qV \). The formula then becomes \( \lambda = h/\sqrt{2mqV} \).

Answer 1

Step 1: The Concept of Matter-Wave of de Broglie:
In 1924, Louis de Broglie proposed a revolutionary hypothesis about the dual nature of matter. He suggested that, just as radiation (like light) exhibits both wave-like and particle-like properties, matter should also exhibit this duality.
The concept of matter-waves (or de Broglie waves) states that every moving particle, regardless of its size (from an electron to a macroscopic object), has a wave associated with it. The properties of the particle (like mass and velocity) are related to the properties of its associated wave (like wavelength). This idea unified the physics of matter and energy, suggesting that the wave-particle duality is a universal principle of nature.
Step 2: Establishing the de Broglie Wavelength Relation:
De Broglie combined two fundamental equations from physics: \[ E = hf, \quad E = mc^2 \] For a photon, which has momentum \( p = E/c \), combining these gives \( p = hf/c \). Since \( c = f\lambda \), this becomes \( p = h/\lambda \).
De Broglie postulated that this relationship between momentum and wavelength should hold true for material particles as well. For a particle of mass \( m \) moving with velocity \( v \), the momentum is \( p = mv \).
Thus, the de Broglie wavelength (\(\lambda\)) associated with a moving particle is given by: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] Step 3: Relation in Terms of Kinetic Energy:
The kinetic energy of a non-relativistic particle is given by: \[ K = \tfrac{1}{2}mv^2 \] The momentum is given by: \[ p = mv \] We can write the kinetic energy as: \[ K = \frac{p^2}{2m} \] Rearranging: \[ p^2 = 2mK \quad \Rightarrow \quad p = \sqrt{2mK} \] Substitute this into the de Broglie relation: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \] Step 4: Final Answer:
The de Broglie wavelength of a particle is inversely proportional to the square root of its kinetic energy: \[ \lambda = \frac{h}{\sqrt{2mK}} \]