An electron with kinetic energy 20.0 eV collides with a ground-state hydrogen atom and ionises it. Neglect the recoil kinetic energy of the proton. After the collision, the electron removed from the atom has kinetic energy 2.0 eV. If the scattered incident electron has kinetic energy (N ÷ 10) eV, find N.
Physics · Atomic Structure and X-rays · atomic structure and bohr atomic model
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Question

An electron with kinetic energy 20.0 eV collides with a ground-state hydrogen atom and ionises it. Neglect the recoil kinetic energy of the proton. After the collision, the electron removed from the atom has kinetic energy 2.0 eV. If the scattered incident electron has kinetic energy (N ÷ 10) eV, find N.

Physics · Atomic Structure and X-rays · atomic structure and bohr atomic model

G1 Physics · Accepted Answer

Ionisation of ground-state hydrogen requires 13.6 eV. Initial projectile kinetic energy: Kᵢ = 20.0 eV. After supplying the ionisation energy, the total kinetic energy available to the two outgoing electrons is K(available) = 20.0−13.6 = 6.4 eV. The ejected atomic electron carries 2.0 eV. Therefore the scattered incident electron carries K(scattered) = 6.4−2.0 = 4.4 eV. Thus, 4.4 eV = 44 ÷ 10 eV. Hence N = 44.