# A continuous band of radiation having all wavelengths from about 1000 Angstrom to 10000 Angstrom is passed through a gas of monoatomic hydrogen. In the emission spectrum one can observe the entire: a)Lyman series b)Balmer series c)Paschen series d)Pfund series I didn’t understand the question. If the radiation containing wavelengths from 1000 Angstrom to 10000 Angstrom is passed through a gas of monoatomic hydrogen, isn’t that the absorption spectrum of hydrogen atom? If that is the case, then no spectrum will be produced because the series limit of Lyman, Balmer and Paschen series comes in the range from 1000 Angstrom to 10000 Angstrom and all these lines will be missing in the spectrum. Please correct me if I have wrongly understood the question.
You need to determine which of the series lie entirely inside the spectrum of the incident radiation.
Strictly speaking, one can actually never observe the entire series as the lines will become weaker and weaker the closer they get to the series limit. Usually one can observe only a couple of lines. However, only the Balmer series would be completely contained within 1000-10000 A (the Lyman series would require 911 A as the lower limit). This holds both for emission and absorption. Thomas
Baulmer indeed!, First convert the spectrum to equivalent energy: 1000 Angstrom = 100 nm, and E=hc/lambda, where hc = 1240eV/nm. This gives a photon energy of 1240/100 = 12.4 eV at 1000 Angstrom. At 10000 Angstrom, the energy is exactly 1/10 as much, of course, 1.24eV. The photon energies in each series must correspond exactly to the difference between two Hydrogen energy levels. Since the nth energy level in H is -13.6eV/(n^2), you get the photon energies as 13.6eV(1/n_final^2 - 1/n_initial^2). The line series you mentioned are defined by the value of n_final. For Lyman, Balmer, Paschen, Pfund, n_final = 1, 2, 3, 4, in that order. For each series, the lowest possible energy is when the electron jumps from just one level higher (n_initial = 1 + n_final). In that case, the energy is 13.6eV(1/(n^2) - 1/(n+1)^2). Putting in each n values gives Lyman = 10.2, Balmer = 1.889, Paschen = .6611, Pfund = .306. For each series, the highest possible energy is when the electron jumps from an extremely high level (n goes to infinity) down to n_final. In that case, the energy is 13.6eV(1/n^2). Putting in each n values gives Lyman = 13.6, Balmer = 3.4, Paschen = 1.511, Pfund = .85. Therefore the Lyman series extends from 10.2 to 13.6eV, Balmer from 1.889 to 3.4eV, Paschen from .6611 to 1.511,eV and Pfund from .306 to .85eV. Now it is obvious which of these lie entirely within the range from 1.24 to 12.4eV. The whole Pfund series is too low energy, so NONE of its lines lie in the range. Paschen has some lines with too low energy, so it is only partially in the range. Balmer fits completely within the range. Lyman has its lowest lines in the range, but has some lines with too high energy. Therefore the answer is b) Balmer. Hope this helps!
Baulmer indeed!, First convert the spectrum to equivalent energy: 1000 Angstrom = 100 nm, and E=hc/lambda, where hc = 1240eV/nm. This gives a photon energy of 1240/100 = 12.4 eV at 1000 Angstrom. At 10000 Angstrom, the energy is exactly 1/10 as much, of course, 1.24eV. The photon energies in each series must correspond exactly to the difference between two Hydrogen energy levels. Since the nth energy level in H is -13.6eV/(n^2), you get the photon energies as 13.6eV(1/n_final^2 - 1/n_initial^2). The line series you mentioned are defined by the value of n_final. For Lyman, Balmer, Paschen, Pfund, n_final = 1, 2, 3, 4, in that order. For each series, the lowest possible energy is when the electron jumps from just one level higher (n_initial = 1 + n_final). In that case, the energy is 13.6eV(1/(n^2) - 1/(n+1)^2). Putting in each n values gives Lyman = 10.2, Balmer = 1.889, Paschen = .6611, Pfund = .306. For each series, the highest possible energy is when the electron jumps from an extremely high level (n goes to infinity) down to n_final. In that case, the energy is 13.6eV(1/n^2). Putting in each n values gives Lyman = 13.6, Balmer = 3.4, Paschen = 1.511, Pfund = .85. Therefore the Lyman series extends from 10.2 to 13.6eV, Balmer from 1.889 to 3.4eV, Paschen from .6611 to 1.511,eV and Pfund from .306 to .85eV. Now it is obvious which of these lie entirely within the range from 1.24 to 12.4eV. The whole Pfund series is too low energy, so NONE of its lines lie in the range. Paschen has some lines with too low energy, so it is only partially in the range. Balmer fits completely within the range. Lyman has its lowest lines in the range, but has some lines with too high energy. Therefore the answer is b) Balmer. Hope this helps!
