The transition from the state n = 4 to n = 3 | vedclass

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(D) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 Z^2 (\frac{1}{n_f^2} - \frac

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$5 	o 4$

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(D) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 Z^2 (\frac{1}{n_f^2} - \frac{1}{n_i^2}) 	ext{ eV}$.Ultraviolet $(UV)$ radiation corresponds to higher energy transitions (Lyman series),while infrared $(IR)$ radiation corresponds to lower energy transitions (Paschen,Brackett,or Pfund series).Since the transition $4 	o 3$ results in $UV$ radiation,and $IR$ radiation requires significantly lower energy transitions,we look for a transition where the energy gap is smaller than that of $4 	o 3$.The energy difference decreases as $n$ increases. Among the given options,the transition $5 	o 4$ involves the smallest energy change,which corresponds to the infrared region of the electromagnetic spectrum.

The transition from the state $n = 4$ to $n = 3$ in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition:

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