Wavelength of photon which have energy equal to average of energy of photons with Å and Å will be — Atomic Structure Chemistry Question

💡 Solution & Explanation

Average energy $E_{avg} = \frac{E_1 + E_2}{2} = \frac{1}{2}\left(\frac{hc}{4000} + \frac{hc}{6000}\right) = \frac{hc}{2} \left(\frac{3 + 2}{12000}\right) = \frac{hc}{2} \frac{5}{12000} = \frac{5hc}{24000} = \frac{hc}{4800}$. Thus $\lambda = 4800$ Å. Therefore, correct answer is B.