Question

The de-Broglie wavelength ($\lambda$) of a particle is related to its kinetic energy ($E$) as

• A. $\lambda \propto E$
• B. $\lambda \propto E^{-1}$
• C. $\lambda \propto E^{\frac{1}{2}}$
• D. $\lambda \propto E^{-\frac{1}{2}}$

Hint:

Physics Tip: This relationship implies that if the kinetic energy of a particle increases, its associated de-Broglie wavelength decreases.

Answer 1

The Correct Option is D

Concept: Physics (Dual Nature of Radiation and Matter) - de-Broglie Wavelength.

Step 1: State the de-Broglie wavelength formula. The de-Broglie wavelength ($\lambda$) of a particle is given by the ratio of Planck's constant ($h$) to its momentum ($p$): $$\lambda = \frac{h}{p} \text{}$$

Step 2: Relate momentum to kinetic energy. The relationship between momentum ($p$) and kinetic energy ($E$) for a particle of mass $m$ is: $$p = \sqrt{2mE} \text{}$$

Step 3: Substitute and find the proportionality. Substituting the expression for momentum into the wavelength formula: $$\lambda = \frac{h}{\sqrt{2mE}} \text{}$$ Since $h$ and $m$ are constants, we can see that: $$\lambda \propto \frac{1}{\sqrt{E}} \text{ or } \lambda \propto E^{-\frac{1}{2}} \text{}$$ $$ \therefore \text{The correct relationship is } \lambda \propto E^{-\frac{1}{2}}. \text{} $$