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(D) The energy of an emitted photon is given by $\Delta E = E_2 - E_1 = \frac{hc}{\lambda}$.From this relation,we see that the wavelength $\lambda = \

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$II$

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(D) The energy of an emitted photon is given by $\Delta E = E_2 - E_1 = \frac{hc}{\lambda}$.From this relation,we see that the wavelength $\lambda = \frac{hc}{\Delta E}$.Therefore,for the wavelength $\lambda$ to be maximum,the energy difference $\Delta E$ must be minimum.Among the emission transitions shown $(II, III, IV)$,the transition $II$ occurs between $n=4$ and $n=3$,which has the smallest energy gap $(\Delta E = 0.66 \ eV)$.Thus,transition $II$ corresponds to the emission of a photon with the maximum wavelength.

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

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