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(D) The energy of a photon emitted during a transition from state $n_i$ to $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2}

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$n=5$ to $n=3$

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(D) The energy of a photon emitted during a transition from state $n_i$ to $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.Ultraviolet radiation corresponds to the Lyman series $(n_f = 1)$.Visible radiation corresponds to the Balmer series $(n_f = 2)$.Infrared radiation corresponds to the Paschen series $(n_f = 3)$,Brackett series $(n_f = 4)$,or Pfund series $(n_f = 5)$.Looking at the options:Option $A$: $n=3$ to $n=1$ is Lyman series (Ultraviolet).Option $B$: $n=4$ to $n=2$ is Balmer series (Visible).Option $C$: $n=6$ to $n=2$ is Balmer series (Visible).Option $D$: $n=5$ to $n=3$ is Paschen series (Infrared).Therefore,the correct transition for infrared radiation is $n=5$ to $n=3$.

In a hydrogen atom,a transition from the state $n=6$ to $n=1$ results in ultraviolet radiation. Infrared radiation will be obtained in the transition:

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