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In the case when hen a photon hits a moving atom and stops it, if we consider that while equating energy, we ignore transmission of energy into other forms, and thus, momentum conservation is better thing to do as it is particle nature of light. But why do we ignore the same transmission of energy into other forms in the case of photoelectric effect where photon collides the electron?

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closed as unclear what you're asking by Gilbert, sammy gerbil, Jon Custer, Emilio Pisanty, ZeroTheHero Jul 17 '18 at 6:41

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We don't ignore the transmission of energy.

The easy way to understand collisions is usually in the center of momentum frame. And for motivation here, assume it is all in one line. In that case, before the collision the photon is coming in from, say, the left, and the atom from the right. Their momenta will be equal but opposite. After, the photon goes back out the way it came, and so does the atom. They simply reflect off each other.

Knowing that, you just transform back to the lab frame. In the lab frame the atom is coming in from the right and finishes stopped. It means, in this frame, the photon had to start with less energy, and finish up with all the (kinetic) energy and momentum.

So you need to work out what the original speed of the atom is, and what the energy of the photon has to be to correspond to stopping the atom. Hint: don't forget to do relativistic velocity addition and kinetic energy for the atom.

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When a photon interacts with an atom three things can happen:

  • elastic scattering, the photon keeps its energy and changes direction

  • inelastic scattering, the photon gives part of its energy to the atom and changes direction

  • absorption-reemission, the photon gives all its energy to the atom, and the valence electron moves to a higher energy level as per QM

In the case of absorption, there can be two ways:

  • the energy level of the photon is the same as the difference between energy levels of the atom, and the photon gets absorbed.

  • the energy level of the photon is too high, and it kicks the electron off the atom, that is the photoelectric effect

You are asking in this case why they ignore the transmission of momentum in the case of the photoelectric effect. They do not. If the photon energy is absorbed, some of the energy liberates the electron from the atom, and the rest contributes to the kinetic energy of the electron as a free particle.

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Your question reveals a misunderstanding. You ask "Why do we take the momentum of a photon rather than energy...". For a photon, its total energy is equal to its momentum. Given the relativistic Energy equation: $$ E^{2} = (pc)^{2} + (m_0 c²)^{2} $$ For a photon, $m_0 = 0$ so, $E = p c$. Often in physics, the speed of light is set identically to $1$ so $E = p$. [Note: $p$ is momentum].

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