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(D) The energy of a photon emitted during a transition is given by $\Delta E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of

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$D$

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(D) The energy of a photon emitted during a transition is given by $\Delta E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.From this relation,we have $\lambda = \frac{hc}{\Delta E}$,which implies $\lambda \propto \frac{1}{\Delta E}$.For the shortest wavelength,the energy difference $\Delta E$ must be the maximum.Looking at the provided energy level diagram:Transition $A$ is from $n=4$ to $n=3$.Transition $B$ is from $n=4$ to $n=2$.Transition $C$ is from $n=3$ to $n=2$.Transition $D$ is from $n=3$ to $n=1$.The energy gap is largest for the transition from $n=3$ to $n=1$,which corresponds to transition $D$.Therefore,the correct option is $D$.

The energy levels of a hydrogen atom are shown below. The transition corresponding to the emission of the shortest wavelength is

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