Physics · Atoms · NEET
They are closely linked but not the same. Bohr's frequency condition is the physics rule: photon energy = energy difference, hf = E_i - E_f. The Rydberg formula (1/lambda = R(1/n1^2 - 1/n2^2)) is what you get after you put Bohr's energy levels E_n = -13.6/n^2 eV into that condition. So the frequency condition is the parent idea; the Rydberg formula is its numerical result for hydrogen.
Both use the same equation hf = E_higher - E_lower. In emission the electron starts in a higher level and falls to a lower one, releasing a photon of that exact energy. In absorption the electron starts low and jumps up by swallowing a photon of that exact energy. The frequency of the light is identical for the same pair of levels - only the direction of the electron jump (and the photon) is reversed.
This is Bohr's break from classical physics. Classically a circling electron should radiate at its own revolution frequency and spiral in. Bohr said no radiation happens while the electron stays in an orbit (stationary state). Light only comes out during a jump, and its frequency is fixed by the ENERGY GAP divided by h, not by how fast the electron goes around. So orbital frequency and photon frequency are different quantities.
Only on the gap (the difference). A jump from n=3 to n=2 and a jump from n=300 to n=299 both give a photon, but only the energy difference E_i - E_f decides f. Two different level pairs with the same energy gap would give the exact same frequency of light.
It is Bohr's third postulate, the frequency condition. It comes from energy conservation plus Planck's photon idea (E = hf). The atom loses energy E2 - E1, and that energy leaves as one photon whose energy hf must equal that loss. Rearranged, f = (E2 - E1)/h, and wavelength lambda = c/f = hc/(E2 - E1).
If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength lambda. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
When an electron jumps from a higher energy level E_i to a lower level E_f, the atom emits a photon whose frequency f satisfies h times f = E_i - E_f. For absorption the electron jumps up and takes in a photon of the same energy. Here h is Planck's constant, 6.63 x 10^-34 J s.
It is the third postulate. The first is stationary orbits (no radiation while orbiting), the second is quantisation of angular momentum (mvr = nh/2pi), and the third is this frequency condition for emission and absorption during a jump.
Since f = (E_i - E_f)/h and c = f times lambda, the wavelength is lambda = hc/(E_i - E_f). If energy is in electron-volts, a handy shortcut is lambda (in nm) = 1240 / (E_i - E_f in eV).
Yes. The rule hf = E_i - E_f is general. You just use the level energies for that ion, E_n = -13.6 Z^2/n^2 eV, where Z is the nuclear charge (Z = 2 for He+). Larger Z means bigger energy gaps and higher-frequency photons.
In absorption, atoms remove exactly those photons whose energy matches a gap, leaving dark lines in a continuous background. In emission, excited atoms release photons at exactly those energies, giving bright coloured lines. Both sets of lines occur at the same frequencies because both obey hf = E_i - E_f.
