The de-Broglie wavelength of electron of He ion is 3.329 \AA. If the photon emitted upon de-excitati — Atomic Structure Chemistry Question

💡 Solution & Explanation

First determine the orbit of He$^+$: $2\pi r = n\lambda \implies 2\pi\left(0.529 \frac{n^2}{2}\right) = n(3.329) \implies n \approx 2$. De-excitation from $n=2 \to 1$ emits $\Delta E = 54.4 \left(1 - \frac{1}{4}\right) = 40.8$ eV. This hits H atom (I.E. = 13.6 eV). Photoelectron K.E. $= 40.8 - 13.6 = 27.2$ eV. Wavelength $\lambda = \frac{12.27}{\sqrt{27.2}} \text{ \AA} \approx \frac{12.27}{5.21} \approx 2.35$ \AA. Therefore, correct answer is A.