Which one of the following statements for the emission spectrum of hyd

Which one of the following statements for the emission spectrum of hydrogen is true?

[amp_mcq option1=”The Lyman series lies in the visible region and the Paschen series lies in the infrared region.” option2=”The Lyman series lies in the ultraviolet region and the Paschen series lies in the infrared region.” option3=”Both the Lyman and the Paschen series lie in the visible region.” option4=”The Lyman series lies in the ultraviolet region and the Paschen series lies in the visible region.” correct=”option2″]

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UPSC Geoscientist – 2023
The emission spectrum of hydrogen arises from the de-excitation of electrons from higher energy levels (n_initial) to lower energy levels (n_final). Different series are named based on the final energy level (n_final).
– Lyman series: n_final = 1. Transitions from n_initial = 2, 3, 4,… to n=1. These transitions involve the largest energy drops and thus result in the emission of high-energy photons, which fall in the ultraviolet (UV) region of the electromagnetic spectrum.
– Balmer series: n_final = 2. Transitions from n_initial = 3, 4, 5,… to n=2. These transitions correspond to visible light.
– Paschen series: n_final = 3. Transitions from n_initial = 4, 5, 6,… to n=3. These transitions involve smaller energy drops than Balmer series and fall in the infrared (IR) region of the electromagnetic spectrum.
– Brackett series: n_final = 4. Transitions from n_initial = 5, 6, 7,… to n=4. These are in the far-infrared region.
– Pfund series: n_final = 5. Transitions from n_initial = 6, 7, 8,… to n=5. These are also in the far-infrared region.
The statement that is true is that the Lyman series lies in the ultraviolet region and the Paschen series lies in the infrared region.
Different series in the hydrogen emission spectrum are defined by the principal quantum number of the final energy level of the electron transition. The energy and thus the region of the spectrum depend on the energy difference between the initial and final levels.
The energy of the emitted photon is given by the difference in energy between the initial and final states, E = R_H * (1/n_final² – 1/n_initial²), where R_H is the Rydberg constant. Larger energy differences (smaller n_final) correspond to shorter wavelengths (higher energy photons).