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Let's first understand what photon energy is. Photons are the fundamental particles of light and other electromagnetic radiation. They are unique because they have energy but no mass and travel at the speed of light. The energy of a photon is directly related to its frequency, denoted by \( u \), and can be calculated using the formula:

• \( E_{ph} = hu \)

Here, \( h \) represents Planck’s constant, approximately \( 6.626 \times 10^{-34} \text{ Js} \). The relationship can also be expressed with the photon's wavelength \( \lambda \), using the speed of light \( c \):

• \( E_{ph} = \frac{hc}{\lambda} \)

This formula tells us that higher frequency light has more energetic photons. In the context of our problem, the photon's energy is twice the kinetic energy of an electron, which indicates a much higher energy per particle for photons compared to electrons, despite both having wave-like properties.

Kinetic energy is the energy an object possesses due to its motion. For objects with mass, such as electrons, kinetic energy is calculated using their velocity and mass. The standard formula is:

• \( K = \frac{1}{2} mv^2 \)

Where \( m \) is the mass and \( v \) is the velocity of the object. In our discussion, we focus on electrons, which have mass but move much slower than light. Their kinetic energy is less straightforward when compared to photons because electrons also exhibit wave-like properties. Despite having a much slower speed, when electrons exhibit this behaviour, their kinetic energy is a small fraction of what their photon counterparts hold in energy.The exercise further shows that the relationship between photon energy and electron kinetic energy is such that the photon's energy, given by \( c \cdot m_0 v \), is twice the kinetic energy of the electron, \( K = \frac{1}{2} m_0 v^2 \). This illustrates that photon energy surpasses the kinetic energy of the electron even though both create the same diffraction pattern.

Momentum is a crucial concept when discussing the behavior of particles, especially in quantum mechanics. Classical momentum is simply the product of an object's mass and velocity, \( p = mv \). However, in the context of quantum particles like photons and electrons, we often use their wave-like properties to define momentum.For photons, which lack mass, the momentum \( p_{ph} \) derives from their wave characteristics:

• \( p_{ph} = \frac{h}{\lambda} \)

This shows how momentum is directly linked to wavelength and, thus, energy. For non-relativistic electrons, momentum is traditionally given as:

• \( p_e = m_0 v \)

Given that both photon and electron display the same angular diffraction pattern, their momenta equal each other at certain incidences, directly tying into their wavelength relationship in de Broglie's equation \( \lambda = \frac{h}{p} \). Understanding this, we realize that the different forms of momentum allow us to compare the energies and understand why photon energy is greater than the kinetic energy of electrons, indicating a fascinating crossover of classical and quantum mechanics.