Looks like a typo. It should also be ##m_e##, the rest mass of the electron.Cocoleia said:
Photons have no rest mass. Setting it to zero in the equation in your problem yields p = K/c, where K is the energy of the photon. That's a valid result linking energy and momentum for a photon. To get at the wavelength appeal to the Planck relationship, ##E = h \nu##.Cocoleia said:
The DeBroglie wavelength based on kinetic energy is a concept in quantum mechanics that describes the wavelength of a particle in motion. It is based on the kinetic energy of the particle and is related to its momentum.
The DeBroglie wavelength is calculated using the formula: λ = h/mv, where λ is the wavelength, h is Planck's constant, m is the mass of the particle, and v is the velocity of the particle.
The DeBroglie wavelength is significant because it demonstrates the wave-particle duality of matter. It shows that particles, such as electrons, can exhibit both wave-like and particle-like behavior.
The DeBroglie wavelength is related to the uncertainty principle, which states that the more accurately we know a particle's position, the less accurately we can know its momentum, and vice versa. The DeBroglie wavelength is a measure of the uncertainty in a particle's momentum.
The DeBroglie wavelength is typically observed in subatomic particles, such as electrons, due to their small mass and high velocities. It is not observable in everyday objects because their mass and velocity are too large to exhibit significant wave-like behavior.