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(A) The energy of an emitted photon is given by $\Delta E = E_{initial} - E_{final}$.For emission,the electron must move from a higher energy level to

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$III$

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(A) The energy of an emitted photon is given by $\Delta E = E_{initial} - E_{final}$.For emission,the electron must move from a higher energy level to a lower energy level.Transition $I$ is an absorption transition (upward arrow).Transitions $II$,$III$,and $IV$ are emission transitions (downward arrows).Comparing the energy gaps:Transition $II$: $n=4$ to $n=3$Transition $III$: $n=2$ to $n=1$Transition $IV$: $n=4$ to $n=2$The energy difference between levels is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$ (for Hydrogen-like atoms).The energy gap $\Delta E$ increases as we move to lower $n$ values.The gap between $n=2$ and $n=1$ (Transition $III$) is the largest among the emission transitions shown.Therefore,transition $III$ corresponds to the emission of a photon with the most energy.

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

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