Physics · Atoms · NEET
Bohr had no proof — he only assumed it because it matched the hydrogen spectrum. De Broglie (1924) gave the missing reason 11 years later. He treated the electron as a wave with wavelength lambda = h/(mv). A wave on a closed loop survives only if it joins up smoothly with itself, which means the circumference must hold a whole number of wavelengths: n lambda = 2 pi r. Substitute lambda = h/(mv): n h/(mv) = 2 pi r, so mvr = nh/(2 pi). The strange 2 pi and the whole number n now have a clear physical meaning.
Then the wave overlaps itself out of step after each loop. The crests and troughs from successive turns cancel by destructive interference, so no steady wave can exist there. That orbit is not allowed. Only radii where the wave meets itself in phase (n = 1, 2, 3, ...) give a stable standing wave and a permitted orbit. This is exactly why the atom has discrete allowed orbits and not a continuous range.
Both descriptions apply. In de Broglie's picture the moving electron has a matter wavelength lambda = h/(mv). The whole-number-of-wavelengths rule is a wave idea, but the electron still carries mass, charge and momentum like a particle. This wave-particle duality is the key point: it is the wave nature that forces the orbits to be quantised, but the electron is not literally spread out like a ring of charge.
Because 2 pi r is the full circumference of the circular orbit (perimeter of a circle of radius r). The wave has to travel all the way around and return to its start point in step. So the total path length 2 pi r must equal a whole number of wavelengths n lambda. That is where the 2 pi in mvr = nh/(2 pi) comes from — it is the geometry of a full circle, nothing more.
The standing-wave condition n lambda = 2 pi r is general for any circular orbit, so it explains quantisation for hydrogen-like ions (He+, Li2+) too. But it still assumes a neat circular orbit, which is a simplification. For many-electron atoms and for the true 3D shape of orbitals, even this picture is not enough — full quantum mechanics (Schrodinger's equation) is needed. De Broglie's model is the correct bridge idea, not the final theory.
The de Broglie wavelength of an electron in the n = 2 state of hydrogen atom is close to: (Given Bohr radius = 0.052 nm)
In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is: [Given Bohr radius a0 = 52.9 pm]
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The electron acts as a standing wave, so a stable orbit must hold a whole number of its wavelengths: n lambda = 2 pi r, which directly gives Bohr's rule mvr = nh/(2 pi).
Put the de Broglie wavelength lambda = h/(mv) into n lambda = 2 pi r. You get n h/(mv) = 2 pi r. Rearranging gives mvr = nh/(2 pi), Bohr's quantisation of angular momentum.
n is the number of complete electron wavelengths that fit around the orbit. n = 1 is the ground orbit (one full wave), n = 2 fits two waves, and so on. It must be a whole number for a steady standing wave.
An orbit that does not fit a whole number of wavelengths makes the wave interfere destructively with itself and cancel out. Only whole-number fits survive, so only discrete orbits are allowed.
No. It correctly explains why orbits are quantised, but it still uses simple circular orbits. The full and accurate model comes from quantum mechanics (Schrodinger's wave equation).
