Chemistry · Structure Of Atom · NEET
n1 is the LOWER energy level (closer to the nucleus) and n2 is the HIGHER energy level. So n2 is always the bigger number. For example, in a 2 -> 3 jump, n1 = 2 and n2 = 3. Always put the smaller value in n1. If you swap them, you get a negative answer, which is wrong for a wavelength. Rule to remember: (small)^2 first, minus (big)^2 second.
R = 1.097 x 10^7 per metre (m^-1). This is the value NEET expects when the formula gives 1/lambda directly. Some books write R_H = 109677 cm^-1, which is the same number in different units. When the question uses energy (E = -R_H/n^2 in joules), R_H = 2.18 x 10^-18 J. Do not mix these two up: the 1.097 x 10^7 m^-1 value goes with the wavelength formula.
It directly gives the wave number, which is 1/lambda (how many waves fit in one metre). To get the actual wavelength lambda, take the reciprocal: lambda = 1 / [R(1/n1^2 - 1/n2^2)]. So the formula output is 1/lambda first; you flip it at the end to get lambda. This is why the left side is written as 1/lambda, not lambda.
When light is emitted, the electron falls from a higher level (n2) to a lower level (n1). Energy is released, so the wave number must be positive. Since 1/n1^2 is bigger than 1/n2^2 when n1 is smaller, the bracket (1/n1^2 - 1/n2^2) comes out positive. If you accidentally wrote the big level first, you would get a negative number, which has no physical meaning for a wavelength.
Each series is just a fixed value of n1. Lyman: n1 = 1 (UV). Balmer: n1 = 2 (visible). Paschen: n1 = 3 (infrared). Brackett: n1 = 4. Pfund: n1 = 5. You keep n1 fixed and change n2 to any higher level to get the different lines in that series. So the Rydberg formula is one master equation, and the series just tell you what to plug in for n1.
Yes, but only for one-electron (hydrogen-like) species such as He+, Li2+, Be3+. For these you add Z^2: 1/lambda = R Z^2 (1/n1^2 - 1/n2^2), where Z is the atomic number. For plain hydrogen, Z = 1, so Z^2 = 1 and it disappears. Never use this formula for atoms with two or more electrons, because their spectra are not simple.
No. When a NEET question asks for the RATIO of two wavelengths, R and Z cancel out. You only compare the brackets. For example, for 2 -> 3 the bracket is 1/4 - 1/9 = 5/36, and for 4 -> 6 it is 1/16 - 1/36 = 5/144. Since lambda is proportional to 1/bracket, the wavelength ratio is (5/144)/(5/36) = 1/4. This exact question came in NEET 2025.
The ratio of the wavelengths of the light absorbed by a hydrogen atom when it undergoes n=2 -> n=3 and n=4 -> n=6 transitions, respectively, is:
Which of the following series of transitions in the spectrum of the hydrogen atom falls in the visible region?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is one equation that tells you the wavelength of light given out or taken in when the electron in hydrogen jumps between levels: 1/lambda = R(1/n1^2 - 1/n2^2), with n1 the lower level and n2 the higher level.
R = 1.097 x 10^7 per metre (also written 109677 cm^-1). Use this with the wavelength form. The energy form uses R_H = 2.18 x 10^-18 J; keep the two separate.
It works for hydrogen and any one-electron species like He+ and Li2+ if you add Z^2. It does not work for atoms that have two or more electrons.
Because you put the smaller level (n1) first. Then 1/n1^2 is larger than 1/n2^2, so the difference is positive, giving a real, positive wavelength.
The formula gives 1/lambda first. Calculate that value, then take its reciprocal (flip it) to get lambda in metres.
