Physics · Atoms · NEET
The last line means the SHORTEST wavelength. In NEET, 'last line' or 'series limit' is the transition from n = infinity down to n1. This is the biggest energy drop, so the photon has the highest energy and the shortest wavelength. The 'first line' is the longest wavelength (smallest jump, n1+1 to n1). Students often reverse these, so remember: last line = series limit = shortest wavelength.
Use the Rydberg formula with n2 = infinity, so the second term vanishes: 1/lambda = R (1/n1^2 - 1/infinity) = R/n1^2. So the shortest wavelength is lambda_min = n1^2 / R. For Lyman n1=1: lambda = 1/R. For Balmer n1=2: lambda = 4/R. For Paschen n1=3: lambda = 9/R. Just square the lower level and divide by R.
The longest wavelength comes from the smallest allowed jump: n2 = n1 + 1. So 1/lambda_max = R (1/n1^2 - 1/(n1+1)^2). Example, Lyman first line (2 to 1): 1/lambda = R(1/1 - 1/4) = 3R/4, so lambda = 4/(3R). Balmer first line, the H-alpha line (3 to 2): 1/lambda = R(1/4 - 1/9) = 5R/36, so lambda = 36/(5R).
n1 is the LOWER (final) level and n2 is the HIGHER (initial) level, with n2 greater than n1. The formula is 1/lambda = R(1/n1^2 - 1/n2^2). n1 is fixed for a series: Lyman n1=1, Balmer n1=2, Paschen n1=3, Brackett n1=4, Pfund n1=5. Only n2 changes to give the different lines of that series.
Wavelength and photon energy are inversely related: E = hc/lambda. A longer wavelength means less energy. The first line (n1+1 to n1) is the smallest energy gap in the series, so it releases the least energy, hence the longest wavelength. As you go up the series toward the series limit, gaps shrink at the top but the total drop from infinity is largest, giving the shortest wavelength.
A bigger n1 gives a LONGER series-limit wavelength because lambda_min = n1^2 / R grows with n1^2. That is why Lyman (n1=1) sits in ultraviolet with the shortest wavelengths, Balmer (n1=2) in visible, and Paschen, Brackett, Pfund (n1=3,4,5) go into the infrared with longer wavelengths.
Given the value of Rydberg constant is 10^7 per metre, the wave number of the last line of the Balmer series in hydrogen spectrum will be:
The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:
In hydrogen spectrum, the shortest wavelength in the Balmer series is lambda. The shortest wavelength in the Brackett series is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The series limit is the shortest-wavelength line of a series, produced when the electron falls from n = infinity to the fixed lower level n1. Its wave number is 1/lambda = R/n1^2. Beyond this limit the lines merge into a continuum.
For Lyman, n1 = 1 and the series limit is n2 = infinity. So 1/lambda = R(1/1^2) = R, giving lambda = 1/R which is about 91.2 nm, in the ultraviolet region.
The longest wavelength is the first line, the H-alpha line, from n2 = 3 to n1 = 2. 1/lambda = R(1/4 - 1/9) = 5R/36, so lambda = 36/(5R) which is about 656 nm (red light).
Series-limit wavelength is proportional to n1^2. So the ratio of shortest wavelengths of two series equals the ratio of the squares of their lower levels: lambda_a/lambda_b = n1a^2 / n1b^2. For Balmer to Lyman it is 2^2/1^2 = 4.
Maximum energy. Since E = hc/lambda, the shortest wavelength (series limit, infinity to n1) is the largest energy drop and thus the most energetic photon of that series. The longest wavelength is the least energetic photon.
