A photon, an electron and a uranium nucleus all have same wavelength. The one with the most energy \(\ldots \ldots \ldots \ldots\) (A) is the photon (B) is the electron (C) is the uranium nucleus (D) depends upon the wavelength and properties of the particle.

The de Broglie wavelength formula, derived by French physicist Louis de Broglie, relates the wavelength (λ) of a particle to its momentum (p). The formula is given by: \[ λ = \frac{h}{p} \] where h is the Planck's constant (\( 6.626 \times 10^{-34} Js \)) and p is the momentum of the particle, which is the product of its mass (m) and velocity (v): \( p = mv \).

Since the particles all have the same wavelength, we can write the energies of the photon (E_photon), electron (E_electron), and uranium nucleus (E_uranium) as follows: \[ E_{photon} = \frac{h \cdot c}{λ} \] \[ E_{electron} = \frac{1}{2}m_{electron}v_{electron}^2 \] \[ E_{uranium} = \frac{1}{2}m_{uranium}v_{uranium}^2 \] where c is the speed of light (\( 3 \times 10^8 m/s\)), and \(m_{electron}\) and \(m_{uranium}\) are the masses of the electron and uranium nucleus, respectively.

To determine which particle has the most energy, we compare the energy expressions obtained in Step 2. We note that in the case of the photon, the energy depends only on the wavelength (λ). As the wavelength is the same for all three particles, the energy of the photon remains constant. Now, let's consider the electron and the uranium nucleus. For each of these particles, the energy depends on the square of their velocities and their masses. As the mass of the uranium nucleus is much greater than that of the electron, and both have the same wavelength (meaning their velocities are proportional to their inverse masses), it is clear that the energy of the uranium nucleus will be much greater than that of the electron. Thus, the particle with the most energy is the uranium nucleus.

Based on our comparison in Step 3, the correct answer is: (C) is the uranium nucleus