Question
Show that the de Broglie wavelength of a particle of mass $m$ and kinetic energy KE is given by$$\lambda=\frac{h c}{\sqrt{K E\left(K E+2 m c^{2}\right)}}$$
Step 1
Step 1: The energy of a particle is given by the equation $$ E=\sqrt{p^{2} c^{2}+m^{2} c^{4}} $$ where $E$ is the energy, $p$ is the momentum, $c$ is the speed of light, and $m$ is the mass of the particle. Show more…
Show all steps
Your feedback will help us improve your experience
Ankur S and 79 other Physics 103 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Show that the de Broglie wavelength of a particle of mass $m$ and kinetic energy KE is given by $$ \lambda=\frac{h c}{\sqrt{K E\left(K E+2 m c^{2}\right)}} $$
Show that the wavelength of a particle of mass $m$ with kinetic energy $K$ is given by the relativistic formula $\lambda=h c / \sqrt{K^{2}+2 m c^{2} K}$
Determine the de Broglie wavelength of a particle of mass m and kinetic energy K. Do this for both (a) a relativistic and (b) a nonrelativistic particle.
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD