When electron jumps from the fourth orbit to the second orbit in He ion, the radiation emitted out w — Atomic Structure Chemistry Question

💡 Solution & Explanation

Energy emitted $\Delta E = 13.6 \times 2^2 \times \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 54.4 \times \frac{3}{16} = 10.2$ eV. Wavelength $\lambda = \frac{1240}{10.2} \approx 121.6$ nm. This falls in the ultraviolet region. Therefore, correct answer is A.