Final answer:
The ratio of the de Broglie wavelength of an electron to the wavelength of a photon is 1:1 when both particles carry the same energy, because for both particles, the momentum is calculated using the same energy-momentum relationship.
Step-by-step explanation:
The question pertains to calculating the ratio of the de Broglie wavelength of an electron to the wavelength of a photon, where both have the same energy E.
According to de Broglie's hypothesis, both the massive electron and the massless photon can be described in terms of waves, with energy and momentum being related by the same set of equations. The energy of a particle is given by the equation E² = (pc)² + (mc²)², where p is the momentum and m is the rest mass.
For a photon, the mass m is zero, resulting in the momentum p = E/c, where c is the speed of light and the de Broglie wavelength λ of a particle with momentum p is given by λ = h/p, where h is Planck's constant.
For a photon, since E = pc, the photon's wavelength λ is simply h/E. For the electron, the same energy E implies the same momentum magnitude p, and thus its de Broglie wavelength is also h/p.
Therefore, given that the momentum p for both the photon and electron corresponding to the energy E is the same, the ratio of the de Broglie wavelength associated with the electron to the wavelength of the photon is 1:1.