Physics · Atoms · NEET
Think of the electron as a wave wrapped around the circular orbit. For the wave to survive, its crest and trough must line up perfectly after one full loop, so the pattern repeats. This only works if the circle length fits an exact whole number of wavelengths. If it does not fit exactly, the wave overlaps out of step with itself, cancels out (destructive interference), and cannot exist. So only orbits with 2*pi*r = n*lambda are allowed. NCERT (Section 12.6) shows this for n = 4, where 2*pi*r = 4*lambda.
Start with the standing wave condition: 2*pi*r = n*lambda. The de Broglie wavelength is lambda = h/p = h/(m*v). Substitute it in: 2*pi*r = n*h/(m*v). Now multiply both sides by (m*v)/(2*pi): m*v*r = n*h/(2*pi). That is exactly Bohr's second postulate (quantised angular momentum). So de Broglie's wave idea is the reason behind Bohr's rule, not a separate assumption.
The wave does not join smoothly after one loop. On the next round it arrives out of step with itself. Peaks land on troughs, and after many loops the wave adds up to zero (destructive interference). Such a wave cannot maintain itself, so that orbit is not allowed. Only 'resonant' standing waves (whole-number fits) persist, which is why orbits are quantised.
The standing wave picture is a model that explains WHY only certain orbits exist. The electron has wave-particle duality (de Broglie), and Davisson-Germer proved electrons behave as waves. The circular standing wave is the correct intuition for quantisation, but remember Bohr's full planet-like orbit picture is only an approximation; the exact modern description uses quantum mechanics and orbitals.
Rearrange the standing wave condition 2*pi*r = n*lambda to get lambda = 2*pi*r_n / n. Since r_n = a0*n^2 (a0 = Bohr radius), lambda = 2*pi*a0*n^2 / n = 2*pi*a0*n. So the wavelength grows in direct proportion to n. For n = 1 it is 2*pi*a0, for n = 2 it is 4*pi*a0, and so on.
In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is: [Given Bohr radius a0 = 52.9 pm]
The de Broglie wavelength of an electron in the n = 2 state of hydrogen atom is close to: (Given Bohr radius = 0.052 nm)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The circumference of the orbit must equal a whole number of de Broglie wavelengths: 2*pi*r = n*lambda, where n = 1, 2, 3, ... . This is the condition for a stable (resonant) standing wave and it explains why only certain orbits are allowed.
Louis de Broglie, in 1923, ten years after Bohr's model. He proposed that electrons have a wave nature (lambda = h/mv), and that stable orbits are circular standing waves. This was later supported by the Davisson-Germer experiment (1927).
No. It gives the same result mvr = nh/2*pi as Bohr's postulate, so all the radius (r_n = a0*n^2) and energy (E_n = -13.6/n^2 eV) formulas stay the same. De Broglie's picture just explains WHY the quantisation rule holds.
Exactly one full wavelength fits the circumference in the ground state: 2*pi*r_1 = 1*lambda. For n = 2 it is two wavelengths, for n = 3 it is three, and so on.
