Final answer:
For particles with the same wavelength, the electron in a uranium nucleus will have the maximum kinetic energy because it has a much lower mass compared to a proton or neutron, requiring a higher velocity to maintain the same de Broglie wavelength, thus resulting in higher kinetic energy.
Step-by-step explanation:
The question is asking which particle—a proton, neutron, or electron—in a uranium nucleus with the same wavelength would have the maximum kinetic energy.
According to de Broglie's hypothesis, all particles have wave-like properties, and the associated wavelength λ is inversely proportional to the momentum p of the particle (λ = h/p, where h is Planck's constant).
Now considering the formula for kinetic energy (KE) for non-relativistic particles, KE = (1/2)mv^2, and that momentum p = mv, we can observe that as the mass (m) of the particle increases, for a fixed wavelength, its velocity (v) must decrease, and thus so does its kinetic energy.
However, since an electron is much less massive than a proton or a neutron, for the same de Broglie wavelength, its velocity must be much higher to account for its smaller mass.
Consequently, the electron will have a higher kinetic energy compared to a proton or a neutron with the same wavelength.