A proton and an electron are both accelerated to the same final kinetic energy. If $\lambda _ { p }$ is the de Broglie wavelength of the proton and $\lambda _ { \mathrm { e } }$ is the de Broglie wavelength of the electron, then A) $\lambda _ { p } = \lambda _ { e }$. B) $\lambda _ { p } \lambda _ { e }$.

Question

A proton and an electron are both accelerated to the same final kinetic energy. If $\lambda _ { p }$ is the de Broglie wavelength of the proton and $\lambda _ { \mathrm { e } }$ is the de Broglie wavelength of the electron, then
A) $\lambda _ { p } = \lambda _ { e }$.
B) $\lambda _ { p } < \lambda _ { e }$.
C) $\lambda _ { p } > \lambda _ { e }$.

Quizplus · Accepted Answer

The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum of the particle. For the same kinetic energy, the momentum $p = \sqrt{2mK}$, where $m$ is the mass of the particle and $K$ is the kinetic energy. Since the mass of a proton is much larger than the mass of an electron, the momentum of the proton will be smaller for the same kinetic energy, leading to a larger wavelength for the electron. Therefore, $\lambda _ { p } < \lambda _ { e }$.

A Proton and an Electron Are Both Accelerated to the Same