A stationary He ion emitted a photon corresponding to the first line of the Lyman series. That photo — Atomic Structure Chemistry Question

💡 Solution & Explanation

Energy of first Lyman line of He$^+$ ($2 \to 1$) is $E = 13.6 \times 2^2 \times \left(1 - \frac{1}{4}\right) = 40.8$ eV. H atom ground state IE is 13.6 eV. Kinetic energy of photoelectron $= 40.8 - 13.6 = 27.2$ eV. Velocity $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 27.2 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 3.09 \times 10^6$ m/s. Therefore, correct answer is A.