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(C) The energy of an emitted photon is given by $\Delta E = E_2 - E_1 = \frac{hc}{\lambda}$.Since $\lambda = \frac{hc}{\Delta E}$,the wavelength $\lam

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$C$ and $D$ respectively

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(C) The energy of an emitted photon is given by $\Delta E = E_2 - E_1 = \frac{hc}{\lambda}$.Since $\lambda = \frac{hc}{\Delta E}$,the wavelength $\lambda$ is inversely proportional to the energy difference $\Delta E$.For the shortest wavelength,the energy difference $\Delta E$ must be maximum. Looking at the diagram,transition $C$ (from $n=4$ to $n=2$) covers a larger energy gap compared to $D$ (from $n=4$ to $n=3$) or $B$ (from $n=3$ to $n=2$).For the longest wavelength,the energy difference $\Delta E$ must be minimum. Transition $D$ (from $n=4$ to $n=3$) represents the smallest energy gap among the given transitions.Therefore,the shortest wavelength corresponds to transition $C$ and the longest wavelength corresponds to transition $D$.

Consider the following electronic energy level diagram of $H-$ atom. Photons associated with shortest and longest wavelengths would be emitted from the atom by the transitions labelled:

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