Chemistry · Structure Of Atom · NEET
Only the Balmer series is in the visible region. In the Balmer series, the electron falls from a higher level (n = 3, 4, 5, ...) down to n = 2. Lyman series falls to n = 1 and is in the ultraviolet (UV). Paschen, Brackett and Pfund fall to n = 3, 4, 5 and are in the infrared (IR). NEET loves this exact fact.
Both are line series in the hydrogen spectrum, but they differ by where the electron lands. Lyman series: electron ends at n = 1, region is UV, so we cannot see it. Balmer series: electron ends at n = 2, region is visible, so we can see it. Simple rule: land on 1 = UV (Lyman), land on 2 = visible (Balmer).
Each jump gives out light of a certain energy, and energy decides the region (UV, visible, or IR). Jumps ending at n = 1 (Lyman) give very high energy light = UV. Jumps ending at n = 2 (Balmer) give medium energy light, which happens to land in the visible range (about 400 to 700 nm). Jumps ending at n = 3 or higher (Paschen, Brackett) give low energy light = IR. So only Balmer's energies match what human eyes can see.
In the formula 1/lambda = R(1/n1^2 - 1/n2^2), n1 is the LOWER level (where the electron lands) and n2 is the HIGHER level (where it starts). Always keep n1 smaller than n2 so the answer stays positive. For the Balmer series, n1 = 2 always, and n2 = 3, 4, 5, and so on. For Lyman, n1 = 1.
Within any series, the smallest jump (like n=3 to n=2 in Balmer) gives the LONGEST wavelength, and the jump from n = infinity (the series limit) gives the SHORTEST wavelength. Across series, the Lyman series has the shortest wavelengths overall because its jumps release the most energy. Remember: more energy = shorter wavelength.
Use the order Lyman, Balmer, Paschen, Brackett, Pfund for n1 = 1, 2, 3, 4, 5. For regions: Lyman = UV, Balmer = Visible, Paschen and everything after = IR. Sentence to memorise: 'Little Boys Prefer Bright Paint' for the names, and just remember only the second one (Balmer) is visible.
Which of the following series of transitions in the spectrum of the hydrogen atom falls in the visible region?
The ratio of the wavelengths of the light absorbed by a hydrogen atom when it undergoes n=2 to n=3 and n=4 to n=6 transitions, respectively, is:
Which one is the wrong statement?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The formula is 1/lambda = R(1/n1^2 - 1/n2^2), where R is the Rydberg constant (109677 cm^-1 for hydrogen), n1 is the lower level, and n2 is the higher level. It gives the wavenumber (1/lambda) of each spectral line.
The Lyman series lies in the ultraviolet (UV) region. Its electrons all fall to n = 1, which releases the most energy, giving short-wavelength UV light that our eyes cannot see.
The Rydberg constant for hydrogen is 109677 cm^-1. In energy terms, the related constant R_H is about 2.18 x 10^-18 J. NCERT uses 109677 cm^-1 in the wavenumber formula.
Because the electron can only exist in fixed energy levels, its jumps release only certain fixed energies. Each fixed energy is one sharp line. This line (discrete) spectrum is direct proof that energy levels in the atom are quantised.
The Paschen (ends at n = 3), Brackett (ends at n = 4) and Pfund (ends at n = 5) series all lie in the infrared (IR) region. They come from low-energy jumps to higher landing levels.
