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Step-by-Step Explanation
Step 1: Understand the Energy Levels in a Hydrogen-like Atom
Hydrogen-like atoms (e.g., hydrogen, singly ionized helium, etc.) have discrete electronic energy levels labeled by the principal quantum number $n = 1, 2, 3, \ldots$. The energy associated with level $n$ is inversely proportional to $n^2$, so higher $n$ corresponds to higher energy.
Step 2: Recognize How Frequency Relates to Energy Difference
When an electron transitions from a higher energy level $n_2$ to a lower energy level $n_1$, it emits a photon of frequency $\nu$ given by the energy difference:
$ \Delta E = E_{n_2} - E_{n_1} \quad \text{and} \quad \Delta E = h \nu \quad (\text{where } h \text{ is Planck's constant}).$
The larger the energy difference, the higher the frequency of the emitted radiation. Conversely, smaller energy differences yield lower-frequency radiation.
Step 3: Relate Specific Transitions to UV or IR Regions
Ultraviolet (UV) Region: A transition with a larger energy gap (for example, $n=3 \to n=1$) will emit higher energy photons, typically in the ultraviolet region.
Infrared (IR) Region: A transition with a smaller energy gap (for example, $n=4 \to n=3$) will emit lower energy photons, typically in the infrared region.
Step 4: Identify the Correct Transition for Infrared Radiation
The question states that a transition from $n = 3$ to $n = 1$ produces ultraviolet radiation (a large energy difference). Thus, to get infrared radiation, we look for a transition with a smaller energy gap. Among the given options, $4 \to 3$ has a smaller energy gap compared to $3 \to 1$. Therefore, the transition from $n = 4$ to $n = 3$ corresponds to infrared radiation.
Conclusion
The correct answer is the transition from $n = 4$ to $n = 3$, which emits infrared radiation.
