The de Broglie wavelength $\lambda$ of a particle with mass $m$ and kinetic energy $K$ is given by which expression?

Solution

1. The de Broglie wavelength is $\lambda =\frac{h}{p}$, where $p$ is momentum.
2. Kinetic energy $K$ is related to momentum by $K=\frac{p^2}{2m}$.
3. Rearranging for momentum gives $p=\sqrt{2mK}$.
4. Substituting $p$ into the wavelength formula gives $\lambda =\frac{h}{\sqrt{2mK}}$.
5. Hence the answer is (A).