The transition from the state n = 3 to n = 1 | vedclass

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(D) The energy of a photon emitted during a transition is given by $\Delta E = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.U

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$4 	o 3$

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(D) The energy of a photon emitted during a transition is given by $\Delta E = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.Ultraviolet $(UV)$ radiation corresponds to higher energy transitions (Lyman series),while Infrared $(IR)$ radiation corresponds to lower energy transitions (Paschen,Brackett,or Pfund series).For a transition to result in infrared radiation,the energy difference must be significantly smaller than that of the $n=3 	o n=1$ transition.Let's evaluate the transitions:$1. 2 	o 1$: This is part of the Lyman series,which is in the $UV$ region.$2. 3 	o 1$: This is also part of the Lyman series,which is in the $UV$ region.$3. 4 	o 2$: This is part of the Balmer series,which is in the visible region.$4. 4 	o 3$: This is part of the Paschen series,which corresponds to the infrared $(IR)$ region.Since the energy gap for $4 	o 3$ is the smallest among the given options,it corresponds to the longest wavelength,which falls in the infrared region.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

The transition from the state $n = 3$ to $n = 1$ in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

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