Which statement correctly describes this collision process?
The projectile electron must lose exactly 4.9 eV4.9\text{ eV}4.9 eV of its kinetic energy during the collision.
The projectile electron is captured by the mercury atom, which then emits a photon of energy 0.7 eV0.7\text{ eV}0.7 eV.
The ground-state mercury electron absorbs a photon of energy 4.9 eV4.9\text{ eV}4.9 eV directly from the projectile electron to undergo the transition.
The collision results in the ionization of the mercury atom, leaving the projectile electron with 0.7 eV0.7\text{ eV}0.7 eV of kinetic energy.
