Question

Write the de Broglie wavelengths in terms of kinetic energy.

Hint:

This formula is extremely important in quantum mechanics, especially for problems involving electrons accelerated through a potential difference V. In that case, the kinetic energy gained is \(K = qV\) (where q is the charge), so the wavelength becomes \(\lambda = \frac{h}{\sqrt{2mqV}}\).

Answer 1

Step 1: Understanding the Concept:
The de Broglie hypothesis proposes that all matter, such as electrons and protons, exhibits wave-like properties. A particle with momentum \(p\) has an associated wavelength \(\lambda\), known as the de Broglie wavelength. Kinetic energy (K) is the energy a particle possesses due to its motion. The task is to derive an expression for the wavelength \(\lambda\) in terms of the particle's kinetic energy \(K\).

Step 2: Key Formula or Approach:
The de Broglie wavelength is fundamentally defined as: \[ \lambda = \frac{h}{p} \] where \(h\) is Planck's constant and \(p\) is the momentum of the particle (\(p=mv\)).
The classical kinetic energy (K) of a particle of mass \(m\) is given by \(K = \frac{1}{2}mv^2\). This can be related to momentum by the formula: \[ K = \frac{p^2}{2m} \]

Step 3: Detailed Explanation:
Our goal is to substitute momentum \(p\) in the de Broglie equation with an expression involving kinetic energy \(K\). We start with the kinetic energy formula and solve for \(p\): \[ K = \frac{p^2}{2m} \] Multiplying both sides by \(2m\): \[ p^2 = 2mK \] Taking the square root of both sides gives the momentum: \[ p = \sqrt{2mK} \] Now, we substitute this expression for \(p\) into the de Broglie wavelength formula: \[ \lambda = \frac{h}{\sqrt{2mK}} \] This equation gives the de Broglie wavelength directly from the particle's kinetic energy and mass.

Step 4: Final Answer:
The de Broglie wavelength (\(\lambda\)) in terms of kinetic energy (\(K\)) for a particle of mass \(m\) is given by the expression: \[ \lambda = \frac{h}{\sqrt{2mK}} \]