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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$5 \rightarrow 4$

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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 radiation corresponds to higher energy transitions (Lyman series),while infrared radiation corresponds to lower energy transitions (Paschen,Brackett,or Pfund series).Given that the transition $n=4 ightarrow n=3$ results in ultraviolet radiation for this specific atom,it implies that the energy gap $\Delta E_{4 ightarrow 3}$ is quite large.To obtain infrared radiation,we need a transition with a significantly smaller energy gap than $\Delta E_{4 ightarrow 3}$.Comparing the options:$(A)$ $2 ightarrow 1$: This is a transition between lower energy levels,resulting in a larger energy gap than $4 ightarrow 3$.$(B)$ $3 ightarrow 2$: This is also a transition between lower energy levels,resulting in a larger energy gap than $4 ightarrow 3$.$(C)$ $4 ightarrow 2$: This involves a larger jump in quantum numbers,resulting in a larger energy gap.$(D)$ $5 ightarrow 4$: This transition occurs between higher energy levels where the energy difference between consecutive levels is much smaller. Thus,$5 ightarrow 4$ will have the smallest energy gap and will result in infrared radiation.

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

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