An electron of kinetic energy 45.0 eV collides with a stationary He⁺ ion whose electron is initially in the ground state. In a successful inelastic collision, the He⁺ electron is excited to the highest bound state energetically accessible at this incident energy. Neglect recoil. If the scattered incident electron is left with 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 of kinetic energy 45.0 eV collides with a stationary He⁺ ion whose electron is initially in the ground state. In a successful inelastic collision, the He⁺ electron is excited to the highest bound state energetically accessible at this incident energy. Neglect recoil. If the scattered incident electron is left with kinetic energy (N ÷ 10) eV, find N.

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

G1 Physics · Accepted Answer

For He⁺, Eₙ = −54.4 ÷ n² eV. Excitation to n = 2 requires ΔE₂ = 54.4[1−1 ÷ 4] = 40.8 eV. Excitation to n = 3 requires ΔE₃ = 54.4[1−1 ÷ 9] ≈ 48.36 eV. Since 45.0 eV is less than 48.36 eV, the third orbit is inaccessible. The highest accessible bound state is n = 2. Remaining projectile energy: K_f = 45.0−40.8 = 4.2 eV. Thus, 4.2 eV = 42 ÷ 10 eV. Hence N = 42.