The given diagram indicates the energy levels of a certain atom. When the system moves from level to — Atomic Structure Chemistry Question
Question
The given diagram indicates the energy levels of a certain atom. When the system moves from $2E$ level to $E$ level, a photon of wavelength $\lambda$ is emitted. The wavelength of the photon emitted during its transition from $4E/3$ level to $E$ level is. [Diagram: Three energy levels at $2E$, $4E/3$, and $E$. Arrow from $2E$ to $E$ emits $\lambda$. Arrow from $4E/3$ to $E$ is unknown.]
Answer: D
💡 Solution & Explanation
Energy difference for the first transition is $\Delta E_1 = 2E - E = E$. Thus, $\lambda = \frac{hc}{E}$. Energy difference for the second transition is $\Delta E_2 = \frac{4E}{3} - E = \frac{E}{3}$. Wavelength $\lambda_2 = \frac{hc}{E/3} = 3 \left(\frac{hc}{E}\right) = 3\lambda$. Therefore, correct answer is D.
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