The final answer is: $\boxed{2.2}$
Show that the wavelength of a particle of mass $m$ and kinetic energy $K$ is $\lambda = \frac{h}{\sqrt{2mK}}$. The de Broglie wavelength of a particle is $\lambda = \frac{h}{p}$, where $p$ is the momentum of the particle. 2: Express the momentum in terms of kinetic energy For a nonrelativistic particle, $K = \frac{p^2}{2m}$. Solving for $p$, we have $p = \sqrt{2mK}$. 3: Substitute the momentum into the de Broglie wavelength $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$. The final answer is: $\boxed{2
Please provide the problem number, chapter and specific question from the book "Introductory Nuclear Physics" by Kenneth S. Krane that you would like me to look into. I'll do my best to assist you. $K = \frac{p^2}{2m}$. Solving for $p$