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(D) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{initial} - E_{final}$.Since the energy levels are $E_n =

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$(III)$

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(D) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{initial} - E_{final}$.Since the energy levels are $E_n = -\frac{13.6 Z^2}{n^2} 	ext{ eV}$,the energy difference for a transition from $n_i$ to $n_f$ is $\Delta E = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.Emission occurs when an electron moves from a higher energy level to a lower energy level (downward arrows).Looking at the diagram:Transition $(II)$ is from $n=4$ to $n=3$.Transition $(III)$ is from $n=2$ to $n=1$.Transition $(IV)$ is from $n=3$ to $n=2$.Transition $(I)$ is an absorption (upward arrow),so it does not represent emission.Comparing the energy differences:For $(II)$: $\Delta E \propto (\frac{1}{3^2} - \frac{1}{4^2}) = (\frac{1}{9} - \frac{1}{16}) = \frac{7}{144} \approx 0.0486$.For $(III)$: $\Delta E \propto (\frac{1}{1^2} - \frac{1}{2^2}) = (1 - 0.25) = 0.75$.For $(IV)$: $\Delta E \propto (\frac{1}{2^2} - \frac{1}{3^2}) = (\frac{1}{4} - \frac{1}{9}) = \frac{5}{36} \approx 0.1389$.Comparing the values,the transition $(III)$ has the largest energy difference,corresponding to the emission of the most energetic photon.

The diagram shows different transitions across the energy levels for an electron in a certain atom. Among these,which transition represents the emission of a photon with the most energy?

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